Many familiar physical quantities can be specified completely by giving a single number and the appropriate unit. For example, “a class period lasts 50 min” or “the gas tank in my car holds 65 L” or “the distance between two posts is 100 m.” A physical quantity that can be specified completely in this manner is called a scalar quantity. Scalar is a synonym of “number.” Time, mass, distance, length, volume, temperature, and energy are examples of scalar quantities.

Scalar quantities that have the same physical units can be added or subtracted according to the usual rules of algebra for numbers. For example, a class ending 10 min earlier than 50 min lasts (50 min – 10 min) = 40 min. Similarly, a 60-cal serving of corn followed by a 200-cal serving of donuts gives (60 cal + 200 cal) = 260 cal of energy. When we multiply a scalar quantity by a number, we obtain the same scalar quantity but with a larger (or smaller) value. For example, if yesterday’s breakfast had 200 cal of energy and today’s breakfast has four times as much energy as it had yesterday, then today’s breakfast has 4(200 cal) = 800 cal of energy. Two scalar quantities can also be multiplied or divided by each other to form a derived scalar quantity. For example, if a train covers a distance of 100 km in 1.0 h, its speed is 100.0 km/1.0 h = 27.8 m/s, where the speed is a derived scalar quantity obtained by dividing distance by time.

Many physical quantities, however, cannot be described completely by just a single number of physical units. For example, when the U.S. Coast Guard dispatches a ship or a helicopter for a rescue mission, the rescue team must know not only the distance to the distress signal, but also the direction from which the signal is coming so they can get to its origin as quickly as possible. Physical quantities specified completely by giving a number of units (magnitude) and a direction are called vector quantities. Examples of vector quantities include displacement, velocity, position, force, and torque. In the language of mathematics, physical vector quantities are represented by mathematical objects called vectors. We can add or subtract two vectors, and we can multiply a vector by a scalar or by another vector, but we cannot divide by a vector. The operation of division by a vector is not defined.

Source: University Physics Volume 1, OpenStax CNX https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/2-1-scalars-and-vectors

1st Law:

Newton’s first law states that: “A body at rest will remain at rest unless acted on by an unbalanced force. A body in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.”

This law, also sometimes called the “law of inertia”, means that bodies maintain their current velocity unless a force is applied to change that velocity. If an object is at rest with zero velocity, it will remain at rest until some force begins to change that velocity, and if an object is moving at a set speed and in a set direction, it will remain at that same velocity until some force begins to change that velocity.

Net Forces: It is important to note that the net force is what will cause a change in velocity. The net force is the sum of all forces acting on the body. For example, we can imagine gently pushing on the rock in the figure below and observing that the rock does not move. This is because we will have a friction force equal in magnitude and opposite in direction opposing our gentle pushing force. The sum of these two forces will be equal to zero; therefore, the net force is zero, and the change in velocity is zero.

Rotational Motion: Newton’s first law also applies to moments and rotational velocities. A body will maintain its current rotational velocity until a net moment is exerted to change that rotational velocity. This can be seen in things like toy tops, flywheels, stationary bikes, and other objects that will continue spinning once started until brakes or friction stop them.

In the absence of friction in space, this space capsule will maintain its current velocity until some outside force causes that velocity to change. Public Domain image by NASA.

This rock is at rest with zero velocity and will remain at rest until a net force causes the rock to move. The net force on the rock is the sum of any force pushing the rock and the friction force of the ground on the rock opposing that force. Image by Liz Gray CC-BY-SA 2.0.

In the absence of friction, this spinning top would continue to spin forever, but the small frictional moment exerted at the point of contact between the top and the ground will slow the tops spinning over time. Image by Carrotmadman6 CC-BY-2.0

Source: Engineering Mechanics, Jacob Moore et al., Mechanics Map – Newton’s First Law

2nd Law:

Newton’s second law states that: “When a net force acts on any body with mass, it produces an acceleration of that body. The net force will be equal to the mass of the body times the acceleration of the body.”

You will notice that the force and the acceleration in the equation above have an arrow above them. This means that they are vector quantities, having both a magnitude and a direction. Mass, on the other hand, is a scalar quantity having only a magnitude. Based on the above equation, you can infer that the magnitude of the net force acting on the body will be equal to the mass of the body times the magnitude of the acceleration, and that the direction of the net force on the body will be equal to the direction of the acceleration of the body.

Rotational Motion: Newton’s second law also applies to moments and rotational velocities. The revised version of the second law equation states that the net moment acting on the object will be equal to the mass moment of inertia of the body about the axis of rotation (I) times the angular acceleration of the body.

You should again notice that the moment and the angular acceleration of the body have arrows above them, indicating that they are vector quantities with both a magnitude and direction. The mass moment of inertia, on the other hand, is a scalar quantity having only a magnitude. The magnitude of the net moment will be equal to the mass moment of inertia times the magnitude of the angular acceleration, and the direction of the net moment will be equal to the direction of the angular acceleration.

Source: Engineering Mechanics, Jacob Moore et al., Mechanics Map – Newton’s Second Law

3rd Law:

Newton’s Third Law states, “For any action, there is an equal and opposite reaction.” By “action,” Newton meant a force, so for every force one body exerts on another body, that second body exerts a force of equal magnitude but opposite direction back on the first body. Since all forces are exerted by bodies (either directly or indirectly), all forces come in pairs, one acting on each of the bodies interacting.

Though there may be two equal and opposite forces acting on a single body, it is important to remember that for each of the forces, a Third Law pair acts on a separate body. This can sometimes be confusing when there are multiple Third Law pairs at work. Below are some examples of situations where multiple Third Law pairs occur.

This volleyball resting on a surface has two pairs of Third Law forces. The first consists of the gravitational forces (one force on the ball and one force on the ground). The second consists of the normal forces at the point of contact (one force on the ball and one force on the ground). Image adapted from Public Domain image, no author listed.

If we ignore the weight of the two objects, this clamp will also have two pairs of Third Law forces. The first will be a set of normal forces at the top point of contact (one force on the wood and one force on the clamp) and the second will be another set of normal forces at the bottom point of contact (one force on the wood and one force on the clamp) Image adapted from Public Domain image, no author listed.

Source: Engineering Mechanics, Jacob Moore et al., Mechanics Map – Newton’s Third Law

Giving numerical values for physical quantities and equations for physical principles allows us to understand nature much more deeply than qualitative descriptions alone. To comprehend these vast ranges, we must also have accepted units in which to express them. We shall find that even in the potentially mundane discussion of meters, kilograms, and seconds, a profound simplicity of nature appears: all physical quantities can be expressed as combinations of only seven base physical quantities.

We define a physical quantity either by specifying how it is measured or by stating how it is calculated from other measurements. For example, we might define distance and time by specifying methods for measuring them, such as using a meter stick and a stopwatch. Then, we could define average speed by stating that it is calculated as the total distance traveled divided by time of travel.

Measurements of physical quantities are expressed in terms of units, which are standardized values. For example, the length of a race, which is a physical quantity, can be expressed in units of meters (for sprinters) or kilometers (for distance runners). Without standardized units, it would be extremely difficult for scientists to express and compare measured values in a meaningful way.

Two major systems of units are used in the world: SI units (for the French Système International d’Unités), also known as the metric system, and English units (also known as the customary or imperial system). English units were historically used in nations once ruled by the British Empire and are still widely used in the United States. English units may also be referred to as the foot–pound–second (fps) system, as opposed to the centimeter–gram–second (cgs) system. 

SI Units: Base and Derived Units

In any system of units, the units for some physical quantities must be defined through a measurement process. These are called the base quantities for that system and their units are the system’s base units. All other physical quantities can then be expressed as algebraic combinations of the base quantities. Each of these physical quantities is then known as a derived quantity and each unit is called a derived unit. The choice of base quantities is somewhat arbitrary, as long as they are independent of each other and all other quantities can be derived from them. Typically, the goal is to choose physical quantities that can be measured accurately to a high precision as the base quantities. The reason for this is simple. Since the derived units can be expressed as algebraic combinations of the base units, they can only be as accurate and precise as the base units from which they are derived.

Based on such considerations, the International Standards Organization recommends using seven base quantities, which form the International System of Quantities (ISQ). These are the base quantities used to define the SI base units. The following table lists these seven ISQ base quantities and the corresponding SI base units.

You are probably already familiar with some derived quantities that can be formed from the base quantities. For example, the geometric concept of area is always calculated as the product of two lengths. Thus, area is a derived quantity that can be expressed in terms of SI base units using square meters (m x m = m²).

Similarly, volume is a derived quantity that can be expressed in cubic meters (m³). Speed is length per time; so in terms of SI base units, we could measure it in meters per second (m/s). Volume mass density (or just density) is mass per volume, which is expressed in terms of SI base units such as kilograms per cubic meter (kg/m³). Angles can also be thought of as derived quantities because they can be defined as the ratio of the arc length subtended by two radii of a circle to the radius of the circle. This is how the radian is defined. Depending on your background and interests, you may be able to come up with other derived quantities, such as the mass flow rate (kg/s) or volume flow rate (m³/s) of a fluid, electric charge (A·s), mass flux density kg/(m²·s), and so on. We will see many more examples throughout this text. For now, the point is that every physical quantity can be derived from the seven base quantities, and the units of every physical quantity can be derived from the seven SI base units.

Source: University Physics Volume 1, OpenStax CNX, https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/1-2-units-and-standards/

It is often necessary to convert from one unit to another. For example, if you are reading a European cookbook, some quantities may be expressed in units of liters and you need to convert them to cups. Or perhaps you are reading walking directions from one location to another and you are interested in how many miles you will be walking. In this case, you may need to convert units of feet or meters to miles.

Let’s consider a simple example of how to convert units. Suppose we want to convert 80 m to kilometers. The first thing to do is to list the units you have and the units to which you want to convert. In this case, we have units in meters and we want to convert to kilometers. Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio that expresses how many of one unit are equal to another unit. For example, there are 12 in. in 1 ft, 1609 m in 1 mi, 100 cm in 1 m, 60 s in 1 min, and so on. In this case, we know that there are 1000 m in 1 km. Now we can set up our unit conversion. We write the units we have and then multiply them by the conversion factor so the units cancel out, as shown:

Note that the unwanted meter unit cancels, leaving only the desired kilometer unit. You can use this method to convert between any type of unit. Now, the conversion of 80 m to kilometers is simply the use of a metric prefix, as we saw in the preceding section, so we can get the same answer just as easily by noting that

Since “kilo-” means 10³, however, using conversion factors is handy when converting between units that are not metric or when converting between derived units, as the following examples illustrate.

Source: University Physics Volume 1, OpenStax CNX, https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/1-3-unit-conversion/

Source: Introductory Physics, Ryan Martin et al., https://openlibrary.ecampusontario.ca/catalogue/item/?id=4c3c2c75-0029-4c9e-967f-41f178bebbbb page 106

Right triangle: a triangle containing a 90° angle.
Pythagorean theorem: a relation among the three sides of a right triangle which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides (legs).

Using the Pythagorean theorem can find the length of the missing side in a right triangle.

▪ c is the longest side of the triangle (hypotenuse).
▪ Other two sides (legs) of the triangle a and b can be exchanged.

Source: Key Concepts of Intermediate Level  Math, Meizhong Wang and the College of New Caledonia, https://openlibrary.ecampusontario.ca/catalogue/item/?id=d8bdc88b-5439-4652-b4bb-2948f0d5c625, page 136.

The Six Basic Trigonometric Functions

Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.

To define the trigonometric functions, first consider the unit circle centred at the origin and a point P = (x,y) on the unit circle. Let θ be an angle with an initial side that lies along the positive x-axis and with a terminal side that is the line segment OP. We can then define the values of the six trigonometric functions for θ  in terms of the coordinates x and y.

Let P=(x,y) be a point on the unit circle centered at the origin O. Let θ be an angle with an initial side along the positive x-axis and a terminal side given by the line segment OP. The trigonometric functions are then defined as

If x=0, secθ and tanθ are undefined. If y=0, then cotθ and cscθ are undefined.

We can see that for a point P=(x,y) on a circle of radius r with a corresponding angle θ,the coordinates x and y satisfy:

The values of the other trigonometric functions can be expressed in terms of x, y, and r:

The table below shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of sinθ and cosθ:

Trigonometric Identities

A trigonometric identity is an equation involving trigonometric functions that is true for all angles θ for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next.

1.2.1 Cartesian Coordinate Frame in 2D

Vectors are usually described in terms of their components in a coordinate system. Even in everyday life we naturally invoke the concept of orthogonal projections in a rectangular coordinate system. For example, if you ask someone for directions to a particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction 37° north of east.

In a rectangular (Cartesian) x-y coordinate system in a plane, a point in a plane is described by a pair of coordinates (x, y). In a similar fashion, a vector  in a plane is described by a pair of its vector coordinates. The x-coordinate of vector A is called its x-component, and the y-coordinate of vector A is called its y-component. The vector x-component is a vector denoted by A_x. The vector y-component is a vector denoted by A_Y. In the Cartesian system, the x and y vector components of a vector are the orthogonal projections of this vector onto the x– and y-axes, respectively. In this way, following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum of its vector components:

As illustrated in the figure below, vector is the diagonal of the rectangle where the x-component is the side parallel to the x-axis and the y-component  is the side parallel to the y-axis. Vector component is orthogonal to vector component .

The figure shows the vector A in a plane in the Cartesian coordinate system, represented as the vector sum of its x- and y-components. The x-component vector, A_x, is the orthogonal projection of Vector A onto the x-axis, and the y-component vector, A_y, is the orthogonal projection onto the y-axis. The numbers A_x and A_x that multiply the unit vectors are the scalar components of the vector.

It is customary to denote the positive direction on the x-axis by the unit vector i and the positive direction on the y-axis by the unit vector j. Unit vectors of the axes, i and j, define two orthogonal directions in the plane. As shown in the figure above, the x– and y–components of a vector can now be written in terms of the unit vectors of the axes:

The vectors A_x and A_y defined by the figure above are the vector components of vector . The numbers A_x and A_y that define the vector components above are the scalar components of vector . Combining the diagram above with the equations above, we obtain the component form of a vector:

If we know the coordinates b(x_b, y_b) of the origin point of a vector (where b stands for “beginning”) and the coordinates e(x_e, y_e) of the end point of a vector (where e stands for “end”), we can obtain the scalar components of a vector simply by subtracting the origin point coordinates from the end point coordinates:

1.2.2. Cartesian Coordinate Frame in 3D

To specify the location of a point in space, we need three coordinates (x, y, z), where coordinates x and y specify locations in a plane, and coordinate z gives a vertical position above or below the plane. Three-dimensional space has three orthogonal directions, so we need not two but three unit vectors to define a three-dimensional coordinate system. In the Cartesian coordinate system, the first two unit vectors are the unit vector of the x-axis i and the unit vector of the y-axis j. The third unit vector k is the direction of the z-axis, as can be seen below. The order in which the axes are labeled, which is the order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The order x–y–z, which is equivalent to the order i-j-k, defines the standard right-handed coordinate system (positive orientation).

In three-dimensional space, vector  has three vector components: the x-component , which is the part of vector along the x-axis; the y-component , which is the part of  along the y-axis; and the z-component , which is the part of the vector along the z-axis. A vector in three-dimensional space is the vector sum of its three vector components:

If we know the coordinates of its origin b(x_b, y_b, z_b) and of its end e(x_e, y_e, z_e) its scalar components are obtained by taking their differences, and the z-component is given by:

Magnitude A is obtained by the following equation:

This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in the figure below, the diagonal in the x-y plane has length  and its square adds to the square A_z² to give A² . Note that when the z-component is zero, the vector lies entirely in the x-y plane and its description is reduced to two dimensions.

Source: University Physics Volume 1, OpenStax CNX, https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/2-2-coordinate-systems-and-components-of-a-vector/

In 2d & 3d:

Source: University Physics Volume 1, OpenStax CNX, https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/2-2-coordinate-systems-and-components-of-a-vector/

Vectors can be added together and multiplied by scalars. Vector addition is associative and commutative, and vector multiplication by a sum of scalars is distributive. Also, scalar multiplication by a sum of vectors is distributive:

In this equation, α is any number (a scalar). For example, a vector antiparallel to vector can be expressed simply by multiplying by the scalar α=-1:

The generalization of the number zero to vector algebra is called the null vector, denoted by . All components of the null vector are zero , so the null vector has no length and no direction.

Two vectors  and  are equal vectors if and only if their difference is the null vector:

This vector equation means we must have simultaneously , , and . Hence, we can write if and only if the corresponding components of vectors  and are equal:

   if  

Two vectors are equal when their corresponding scalar components are equal.

Resolving vectors into their scalar components (i.e., finding their scalar components) and expressing them analytically in vector component form allows us to use vector algebra to find sums or differences of many vectors analytically (i.e., using graphical methods). For example, to find the resultant of two vectors  and , we simply add them component by component, as follows:

In this way, scalar components of the resultant vector: .

A vector can be multiplied by another vector, but it may not be divided by another vector. There are two kinds of products of vectors used broadly in physics and engineering. One kind of multiplication is a scalar multiplication of two vectors. Taking a scalar product of two vectors results in a number (a scalar), as its name indicates. Scalar products are used to define work and energy relations. For example, the work that a force (a vector) performs on an object while causing its displacement (a vector) is defined as a scalar product of the force vector with the displacement vector. A quite different kind of multiplication is a vector multiplication of vectors. Taking a vector product of two vectors returns as a result a vector, as its name suggests. Vector products are used to define other derived vector quantities. For example, in describing rotations, a vector quantity called torque is defined as a vector product of an applied force (a vector) and its distance from pivot to force (a vector). It is important to distinguish between these two kinds of vector multiplications because the scalar product is a scalar quantity and a vector product is a vector quantity.

Scalar multiplication of two vectors yields a scalar product.

Dot Product

The scalar product of two vectors is a number defined by the equation:

where ϕ is the angle between the vectors. The scalar product is also called the dot product because of the dot notation that indicates it.

When the vectors are given in their vector component forms:

$$\vec A=A_x\underline{\hat{i}}+A_y\underline{\hat{j}}+A_z\underline{\hat{k}}\text{  and  }\vec B=B_x\underline{\hat{i}}+B_y\underline{\hat{j}}+B_z\underline{\hat{k}}$$

we can compute their scalar product as follows:

$$\vec A\cdot\vec B=(A_x\underline{\hat{i}}+A_y\underline{\hat{j}}+A_z\underline{\hat{k}})\cdot(B_x\underline{\hat{i}}+B_y\underline{\hat{j}}+B_z\underline{\hat{k}})\\=A_xB_x\underline{\hat{i}}\cdot\underline{\hat{i}}+A_xB_y\underline{\hat{i}}\cdot\underline{\hat{j}}+A_xB_z\underline{\hat{i}}\cdot\underline{\hat{k}}\\+A_yB_x\underline{\hat{j}}\cdot\underline{\hat{i}}+A_yB_y\underline{\hat{j}}\cdot\underline{\hat{j}}+A_yB_z\underline{\hat{j}}\cdot\underline{\hat{k}}\\+A_zB_x\underline{\hat{k}}\cdot\underline{\hat{i}}+A_zB_y\underline{\hat{k}}\cdot\underline{\hat{j}}+A_zB_z\underline{\hat{k}}\cdot\underline{\hat{k}}$$

Since scalar products of two different unit vectors of axes give zero, and scalar products of unit vectors with themselves give one, there are only three nonzero terms in this expression. Thus, the scalar product simplifies to:

We can use the equation below to find the angle between two vectors. When we divide by , we obtain the equation for cos(ϕ), into which we substitute the equation from above:

$$\text{cos}\,\phi =\frac{\overset{\to }{A}·\overset{\to }{B}}{AB}=\frac{{A}_{x}{B}_{x}+{A}_{y}{B}_{y}+{A}_{z}{B}_{z}}{AB}$$

The angle ϕ, between vectors is obtained by taking the inverse cosine of the expression above. 

Source: University Physics Volume 1, OpenStax CNX, https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/2-4-products-of-vectors (many examples at this page).

The vector product of two vectors  is denoted by  and is often referred to as a cross product. The vector product is a vector that has its direction perpendicular to both vectors . In other words, vector is perpendicular to the plane that contains vectors .

Source: University Physics Volume 1, OpenStax CNX, https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/2-4-products-of-vectors/

Unit vectors allow for a straightforward calculation of the cross product of two vectors under even the most general circumstances, e.g. circumstances in which each of the vectors is pointing in an arbitrary direction in a three-dimensional space. To take advantage of the method, we need to know the cross product of the Cartesian coordinate axis unit vectors i, j, and k with each other. First off, we should note that any vector crossed into itself gives zero. This is evident from the  equation:

.

because if A and B are in the same direction, then θ = 0°, and since sin 0° = 0, we have . Regarding the unit vectors, this means that:

$$\underline{\hat{i}}\times\underline{\hat{i}}=0\\\underline{\hat{j}}\times\underline{\hat{j}}=0\\\underline{\hat{k}}\times\underline{\hat{k}}=0$$

Next we note that the magnitude of the cross product of two vectors that are perpendicular to each other is just the ordinary product of the magnitudes of the vectors. This is also evident from the equation:

$$|\vec {A}\times\vec{B}|=|\vec A||\vec B|\sin\theta$$

because if is perpendicular to  then θ = 90° and sin90° = 1 so:

$$|\vec A\times\vec B|=|\vec A||\vec B|$$

Now if A and B are unit vectors, then their magnitudes are both 1, so, the product of their magnitudes is also 1. Furthermore, the unit vectors i, j, and k are all perpendicular to each other so the magnitude of the cross product of any one of them with any other one of them is the product of the two magnitudes, that is, 1.

Now how about the direction? Let’s use the right-hand rule to get the direction of i×j:

[To find the direction, we use the right-hand rule which we will cover more in section 3.1. Here is an overview]. With the fingers of the right hand pointing directly away from the right elbow, and in the same direction as i, (the first vector in “”) to make it so that if one were to close the fingers, they would point in the same direction as , the palm must be facing in the +y direction. That being the case, the extended thumb must be pointing in the +z direction. Putting the magnitude (the magnitude of each unit vector is 1) and direction (+z) information together we see that:

One way of remembering this is to write twice in succession:

Then, crossing any one of the first three vectors into the vector immediately to its right yields the next vector to the right. But crossing any one of the last three vectors into the vector immediately to its left yields the negative of the next vector to the left (left-to-right “+“, but right-to-left “−“).

Now we’re ready to look at the general case. Any vector   can be expressed in terms of unit vectors:

$$\vec{\textbf{A}}=A_x\hat{\underline{i}}+A_y\hat{\underline{j}}+A_z\hat{\underline{k}}$$

Doing the same for a vector    then allows us to write the cross product as:

$$\vec{\textbf{A}}\times\vec{\textbf{B}}=(A_x\hat{\underline{i}}+A_y\hat{\underline{j}}+A_z\hat{\underline{k}})\times(B_x\hat{\underline{i}}+B_y\hat{\underline{j}}+B_z\hat{\underline{k}})$$

Using the distributive rule for multiplication we can write this as:

$$\begin{aligned}\vec{\textbf{A}}\times\vec{\textbf{B}}=&A_x\hat{\underline{i}}\times(B_x\hat{\underline{i}}+B_y\hat{\underline{j}}+B_z\hat{\underline{k}})+\\&A_y\hat{\underline{j}}\times(B_x\hat{\underline{i}}+B_y\hat{\underline{j}}+B_z\hat{\underline{k}})+\\&A_z\hat{\underline{k}}\times(B_x\hat{\underline{i}}+B_y\hat{\underline{j}}+B_z\hat{\underline{k}})\end{aligned}$$

$$\begin{aligned}\vec{\textbf{A}}\times\vec{\textbf{B}}=&A_x\hat{\underline{i}}\times B_x\hat{\underline{i}}+A_x\hat{\underline{i}}\times B_y\hat{\underline{j}}+A_x\hat{\underline{i}}\times  B_z\hat{\underline{k}}+\\&A_y\hat{\underline{j}}\times B_x\hat{\underline{i}}+A_y\hat{\underline{j}}\times B_y\hat{\underline{j}}+A_y\hat{\underline{j}}\times B_z\hat{\underline{k}}+\\&A_z\hat{\underline{k}}\times B_x\hat{\underline{i}}+A_z\hat{\underline{k}}\times B_y\hat{\underline{j}}+A_z\hat{\underline{k}}\times B_z\hat{\underline{k}}\end{aligned}$$

Using, in each term, the commutative rule and the associative rule for multiplication we can write this as:

$$\begin{aligned}\vec{\textbf{A}}\times\vec{\textbf{B}}=&A_xB_x(\hat{\underline{i}}\times\hat{\underline{i}})+A_xB_y(\hat{\underline{i}}\times\hat{\underline{j}})+A_xB_z(\hat{\underline{i}}\times\hat{\underline{k}})+\\&A_yB_x(\hat{\underline{j}}\times\hat{\underline{i}})+A_yB_y(\hat{\underline{j}}\times\hat{\underline{j}})+A_yB_z(\hat{\underline{j}}\times\hat{\underline{k}})+\\&A_zB_x(\hat{\underline{k}}\times\hat{\underline{i}})+A_zB_y(\hat{\underline{k}}\times\hat{\underline{j}})+A_zB_z(\hat{\underline{k}}\times\hat{\underline{k}})\end{aligned}$$

Now we evaluate the cross product that appears in each term:

$$\begin{aligned}\vec A\times\vec B=&A_xB_x(0)+A_xB_y(\underline{\hat{k}})+A_xB_z(-\underline{\hat{j}})+\\&A_yB_x(-\underline{\hat{k}})+A_yB_y(0)+A_yB_z(\underline{\hat{i}})+\\&A_zB_x(\underline{\hat{j}})+A_zB_y(-\underline{\hat{i}})+A_zB_z(0)\end{aligned}$$

Eliminating the zero terms and grouping the terms with i together, the terms with j together, and the terms with k together yields:

$$\begin{aligned}\vec A\times\vec B=&A_yB_z(\underline{\hat{i}})+A_zB_y(-\underline{\hat{i}})+\\&A_zB_x(\underline{\hat{j}})+A_xB_z(-\underline{\hat{j}})+\\&A_xB_y(\underline{\hat{k}})+A_yB_x(-\underline{\hat{k}})\end{aligned}$$

Factoring out the unit vectors yields:

$$\begin{aligned}\vec A\times\vec B=&(A_yB_z-A_zB_y)\underline{\hat{i}}+\\&(A_zB_x-A_xB_z)\underline{\hat{j}}+\\&(A_xB_y-A_yB_x)\underline{\hat{k}}\end{aligned}$$

which can be written on one line as:

$$\vec A\times\vec B=(A_yB_z-A_zB_y)\underline{\hat{i}}+(A_zB_x-A_xB_z)\underline{\hat{j}}+(A_xB_y-A_yB_x)\underline{\hat{k}}$$

This is our end result. We can arrive at this result much more quickly if we borrow a tool from that branch of mathematics known as linear algebra (the mathematics of matrices).

We form the 3×3 matrix:

$$ \begin{bmatrix}
\underline{\hat{i}} & \underline{\hat{j}} & \underline{\hat{k}} \\
A_x & A_y & A_z \\
B_x & B_y & B_z
\end{bmatrix} $$

by writing i, j, k as the first row, then the components of the first vector that appears in the cross product as the second row, and finally the components of the second vector that appears in the cross product as the last row. It turns out that the cross product is equal to the determinant of that matrix. We use absolute value signs on the entire matrix to signify “the determinant of the matrix.” So we have:

$$ \vec A\times\vec B=\begin{bmatrix}
\underline{\hat{i}} & \underline{\hat{j}} & \underline{\hat{k}} \\
A_x & A_y & A_z \\
B_x & B_y & B_z
\end{bmatrix} $$

To take the determinant of a 3×3 matrix you work your way across the top row. For each element in that row you take the product of the elements along the diagonal that extends down and to the right, minus the product of the elements down and to the left; and you add the three results (one result for each element in the top row) together. If there are no elements down and to the appropriate side, you move over to the other side of the matrix (see below) to complete the diagonal.

For the first element of the first row, the i, take the product down and to the right,

( this yields iA_yB_z) minus the product down and to the left

( the product down-and-to-the-left is iA_zB_y).

For the first element in the first row, we thus have: A_yB_z − A_zB_y which can be written as: (A_yB_z − A_zB_y). Repeating the process for the second and third elements in the first row (the j and the k) we get (A_zB_x − A_xB_z) and (A_xB_y − AyB_x) respectively. Adding the three results, to form the determinant of the matrix results in:

$$\vec A\times\vec B=(A_yB_z-A_zB_y)\underline{\hat{i}}+(A_zB_x-A_xB_z)\underline{\hat{j}}+(A_xB_y-A_yB_x)\underline{\hat{k}}$$

as we found before, “the hard way.” Below the diagram shows the direction of each part of the cross product:

Source: Calculus Based Physics, Jeffrey W. Schnick, https://openlibrary.ecampusontario.ca/catalogue/item/?id=ce74a181-ccde-491c-848d-05489ed182e7 page 136–141

A moment (sometimes called a torque) is defined as the “tendency of a force to rotate a body”. Where forces cause linear accelerations, moments cause angular accelerations. In this way, moments can be thought of as twisting forces.

Imagine two boxes on an icy surface. The force on box A would simply cause the box to begin accelerating, but the force on box B would cause the box to both accelerate and to begin to rotate. The force on box B is exerting a moment, whereas the force on box A is not.

The Vector Representation of a Moment:

Moments, like forces, can be represented as vectors and have a magnitude, a direction, and a “point of application”. For moments, however, a better name for the point of application is the axis of rotation. This will be the point or axis where we will determine all the moments.

Magnitude:

The magnitude of a moment is the degree to which the moment will cause angular acceleration in the body it is acting on. It is represented by a scalar (a single number). The magnitude of the moment can be thought of as the strength of the twisting force exerted on the body. When a moment is represented as a vector, the magnitude of the moment is usually explicitly labelled. Though the length of the moment vector also often corresponds to the relative magnitude of the moment.

The magnitude of the moment is measured in units of force times distance. The standard metric units for the magnitude of moments are newton meters, and the standard English units for a moment are foot pounds.

.

Direction:

In a two-dimensional problem, the direction can be thought of as a scalar quantity corresponding to the direction of rotation the moment would cause. A moment that would cause a counter-clockwise rotation is a positive moment, and a moment that would cause a clockwise rotation is a negative moment.

In a three-dimensional problem, however, a body can rotate about an axis in any direction. If this is the case, we need a vector to represent the direction of the moment. The direction of the moment vector will line up with the axis of rotation that moment would cause, but to determine which of the two directions we can use along that axis we have available, we use the right-hand rule. To use the right-hand rule, align your right hand as shown so that your thumb lines up with the axis of rotation for the moment and your curled fingers point in the direction of rotation for your moment. If you do this, your thumb will be pointing in the direction of the moment vector.

If we look back at two-dimensional problems, all rotations occur about an axis pointing directly into or out of the page (the z-axis). Using the right-hand rule, counter-clockwise rotations are represented by a vector in the positive z direction and clockwise rotations are represented by a vector in the negative z direction.

Axis of Rotation:

In engineering statics problems, we can choose any point/axis as the axis of rotation. The choice of this point will affect the magnitude and direction of the resulting moment, however, and the moment is only valid about that point.

The magnitude and direction of a moment depend upon the chosen axis of rotation. For example, the single force above would cause different moments about Point A and Point B, because it would cause different rotations depending on the point we fix in place.

Though we can take the moment about any point in a statics problem, if we are adding together the moments from multiple forces, all the moments must be taken about a common axis of rotation. Moments taken about different points cannot be added together to find a ‘net moment’

Additionally, if we move into the subject of dynamics, where bodies are moving, we will want to relate moments to angular accelerations. For this to work, we will need to take the moments either about a single point that does not move (such as the hinge on a door) or we will need to take the moments about the center of mass of the body. Summing moments about other axes of rotation will not result in valid calculations.

Calculating Moments:

We will have two main options to calculate the moment that a force exerts on a body: scalar methods and vector methods. Scalar methods are generally faster for two-dimensional problems where a body can only rotate clockwise or counterclockwise, while vector methods are generally faster for three-dimensional problems where the axis of rotation is more complex.

Given any point on an extended body, if there is a force acting on that body that does not travel through that point, then that force will cause a moment about that point. As discussed on the moments page, a moment is a force’s tendency to cause rotation.

Source: Engineering Mechanics, Jacob Moore et al., http://mechanicsmap.psu.edu/websites/1_mechanics_basics/1-5_moments/moments.html

In discussing how to calculate the moment of a force about a point via scalar quantities, we will begin with the example of a force on a simple lever as shown below. In this simple lever there is a force on the end of the lever, distance d away from the center of rotation for the lever (point A) where the force has a magnitude F.

When using scalar quantities, the magnitude of the moment will be equal to the perpendicular distance between the line of action of the force and the point we are taking the moment about.

$$M=F\ast d$$

To determine the sign of the moment, we determine what type of rotation the force would cause. In this case we can see that the force would cause the lever to rotate counterclockwise about point A. Counterclockwise rotations are caused by positive moments while clockwise rotations are caused by negative moments.

Another important factor to remember is that the value d is the perpendicular distance from the force to the point we are taking the moment about. We could measure the distance from point A to the head of the force vector, or the tail of the force vector, or really any point along the line of action of force F. The distance we need to use for the scalar moment calculation however is the shortest distance between the point and the line of action of the force. This will always be a line perpendicular to the line of action of the force, going to the point we are taking the moment about.

Source: Engineering Mechanics, Jacob Moore et al., http://mechanicsmap.psu.edu/websites/3_equilibrium_rigid_body/3-1_moment_scalar/moment_scalar.html

An alternative to calculating the moment via scalar quantities is using the vector or cross product methods. For simple two-dimensional problems, using scalar quantities is usually easier, but for more complex problems, using the cross product method is usually easier. The cross product method for calculating moments says that the moment vector of a force about a point will be equal to the cross product of a vector r from the point to anywhere on the line of action of the force and the force vector itself.

$$\vec M=\vec r\times\vec F$$

A big advantage of this method is that r does not have to be the shortest distance between the point and the line of action; it goes from the point to any part of the line of action. For any problem, there are many possible r vectors, though, because of the way the cross product works, they should all result in the same moment vector in the end.

The moment vector of the force F about point A will be equal to the cross product of the r vector and the force vector. The r vector is a vector from point A to any point along the line of action of the force.

It is important to note here that all quantities (r, F and M) are vectors. Before you can solve for the cross product, you will need to write out r and F in vector component form. Also, even for two-dimensional problems, you will need to write out all three components of the r and F vectors. For two-dimensional problems, the z components of the r and F vectors will simply be zero, but those values are necessary for the calculations.

The moment vector you get will line up with the axis of rotation for the moment, where you can use the right-hand rule to determine if the moment is going clockwise or counterclockwise about that axis.

The result of r cross F will give us the moment vector. For this two-dimensional problem, the moment vector is pointing in the positive z direction. We can use the right-hand rule to determine the direction of rotation from the moment (line our right thumb up with the moment vector, and our curled fingers will point in the direction of rotation from the moment).

Finally, it is also important to note that the cross product, unlike multiplication, is not commutative. This means that the order of the vectors matters, and r cross F will not be the same as F cross r. It is important to always use r cross F when calculating moments.

Source: Engineering Mechanics, Jacob Moore et al., http://mechanicsmap.psu.edu/websites/3_equilibrium_rigid_body/moment_vector/momentvector.html

1. Problem

After a long day of studying, a student sitting at their computer moves the cursor from the bottom left of the screen to the top right in order to close a web browser. The computer mouse was displaced 6 cm along the x-axis and 3.5 cm along the y-axis. Draw the resultant vector and calculate the distance traveled.

 

Source: Black Computer Mouse on Table | Old dusty black computer mou… | Flickr

2. Draw

3. Knowns and Unknowns

Known:

• x = 6 cm
• y = 3.5 cm

Unknown:

• r
• θ

4. Approach

Use SOH CAH TOA, first find θ, then r

5. Analysis

\begin{aligned}
&\tan \theta=\frac{y}{x} \\
&\tan \theta=\frac{3.5 \mathrm{~cm}}{6 \mathrm{~cm}} \\
&\theta=\tan ^{-1}\left(\frac{3.5}{6}\right) \\
&\theta=30.256^{\circ} \\
&\sin \theta=\frac{y}{r} \\
&r=\frac{y}{\sin \theta} \\
&r=\frac{3.5 \mathrm{~cm}}{\sin \left(30.256^{\circ}\right)} \\
&r=6.946 \mathrm{~cm} \\
&r=6.9 \mathrm{~cm}
\end{aligned}

6. Review

It makes sense that the angle is less that 45, because y is smaller than x. Also, if you use Pythagorean theorem to find r, you get the same answer.

1. Problem

Mark is fishing in the ocean with his favourite fishing rod. The distance between the tip of the rod and the reel is 8 ft and the length of the reel handle is 0.25 ft. The angle between the fishing rod and fishing line is 45 degrees. If Mark catches a fish when 25 ft of the fishing line is released while the fish is diving down with a force of 180 N, how much force does Mark need to apply (push down) to the reel handle to bring in the fish? Draw the position vector of the fish relative to the reel.

Assumptions:

• Mark can reel in the fish when he generates more torque with the handle than the amount of torque that the fish is applying to the reel while pulling on the line.
• The fishing line comes out of the reel in a straight line at a 90-degree angle.

Source: File:Deepsea.JPG – Wikimedia Commons

2. Draw

Sketch:

Free-body diagram:

3. Knowns and Unknowns

Known:

• r_AB = 0.25 ft
• r_BC = 8 ft
• r_CD = 25 ft
• F_D = 180 N
• θ = 45°

Unknown:

• F_A
• vector r_AD

4. Approach

Convert inches to meters, then use the equation below.

5. Analysis

Step 1: Convert inches to meters

Step 2: Solve for T_D

Step 3: Solve for F_A

Vector r_AD:

6. Review

Though yielding a very large number, the answer appears correct from the information given. 13,000 N of force is the amount of force Mark would need to apply to the reel handle to generate the same amount of force that the fish creates. 13,000 N in reality is too much for one to generate, but also in a real scenario, one would not have to generate the same amount of force to reel in the fish, to reel gearing, the amount of torque generated by the fishing rod itself, etc. In other words 13 000 N of force is too high in a real scenario, but with the assumptions given in the problem, the number seems reasonable. The answer also has the correct unit, N.

1. Problem

2. Draw

N/A

3. Knowns and Unknowns

Known:

• a
• b

Unknowns:

• 6a,
• 
• 
• 

4. Approach

Use dot product and cross product equations

5. Analysis

Part a:

$$6\underline{a}=6*[6\;\;\;5\;\;\;3]\\6\underline{a}=[36\;\;\;30\;\;\;18]$$

Part b:

$$\underline{a}\cdot\underline{b}=[6\;\;\;5\;\;\;3]\cdot[8\;\;\;1\;\;\;3]\\=6\cdot 8+5\cdot 1+3\cdot3\\=48+5+9\\\underline{a}\cdot\underline{b}=62$$

Part c:

$$\underline{a}\times\underline{b}=\begin{bmatrix}
\underline{\hat{i}} &\underline{\hat{j}} & \underline{\hat{k}} \\
6 & 5 & 3 \\
8 & 1 & 3
\end{bmatrix}\\(5\cdot 3-3\cdot 1)\underline{\hat{i}}-(6\cdot 3-3\cdot 8)\underline{\hat{j}}+(6\cdot 1-5\cdot 8)\underline{\hat{k}}\\\underline{a}\times\underline{b}=12\underline{\hat{i}}+6\underline{\hat{j}}-34\underline{\hat{k}}$$

Part d:

$$ 2\underline{a}=2*[6\;\;\;5\;\;\;3]=[12\;\;\;10\;\;\;6]\\\underline{b}=[8\;\;\;1\;\;\;3]\\2\underline{a}\times\underline{b} = \begin{bmatrix}
\underline{\hat{i}} & \underline{\hat{j}} & \underline{\hat{k}} \\
12 & 10 & 6 \\
8 & 1 & 3
\end{bmatrix} \\=(10\cdot 3-6\cdot 1)\underline{\hat{i}}-(12\cdot 3-6\cdot 8)\underline{\hat{j}}+(12\cdot 1-10\cdot 8)\underline{\hat{k}}\\2\underline{a}\times\underline{b}=24\underline{\hat{i}}+12\underline{\hat{j}}-68\underline{\hat{k}}$$

6. Review

The answer to part d is double the answer for part c, which makes sense. It also makes sense that the answers to a, c, and d have values in three directions, while b only has magnitude.

To start riding her bicycle, Jane must push down on one of her bike’s pedals which are on 16 centimeter long crank arms. Jane can push directly downwards with her legs with a force of 100N. Jane notices that the pedal’s starting position can sometimes make it more or less useful in generating torque.

a) What is the ideal angle that Jane’s bike pedal should be at in order to generate the most torque? Prove this mathematically. (Assume we only care about the very start of her very first push, and choose a reference frame for the angle that makes most sense for you).

b) What angle(s) should the bike pedal be at if Jane wants to generate exactly half of the maximum amount of torque?

c) Is there any position(s) at which the pedal will create zero torque? Where are they and why?

Source: https://commons.wikimedia.org/wiki/File:Girl_on_a_Bike_(Imagicity_116).jpg

 

Source: https://pixabay.com/illustrations/bicycle-cycle-two-wheeler-pedal-3168934/

2. Draw

3. Knowns and Unknowns

Knowns:

Unknowns:

• position of r for maximum torque
• position of r for half of maximum torque
• position of r for zero torque, and why

4. Approach

For part a), I will find a general equation for torque based on the given values in terms of θ, then analyze the function for its maxima

For part b), I will find the magnitude of 50% of maximum torque and then reverse-engineer the equation to determine what angle(s) the pedal needs to be at to satisfy the equation.

For part c), I will look back at my equation and find when the equation equals zero, then try to understand why given the example problem.

5. Analysis

Part a:

Thinking about the shape of the sine function in the first period, the maximum occurs at 90 degrees.

You could say algebraically that the maximum is at 90, 450, 810 etc., but these angles all represent the same position on the wheel. Therefore, we will use 90.

Part b:

Find 50% of the maximum torque:

Rearrange T₂ equation:

Therefore, Jane could push at 30 from vertical, or 150 from vertical to create half the torque.

*Interesting to note is that half the angle does not yield half the torque; the angle is 30, not 45. This is because the sine function is non-linear.*

Part c:

T = 16 sinθ tells us that the angles of 0 and 180 will give us zero torque.

This makes sense given that pushing straight down on a stable pendulum will not cause the pendulum to rotate!

Likewise, if you just stand on your pedals, you’re providing lots of downward force, but creating zero torque since the crank arm and the direction of the force are parallel (or antiparallel)!

6. Review

These answers have the correct units (Nm and degrees) and are within a reasonable order of magnitude based on the given information. See logic/explanations above for more detail.

1. Problem

2. Draw

Sketch:

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• F = 100N
• r₁ = 45cm
• r₂ = 75cm
• α = 45°

Unknowns:

• M₁
• M₂
• Angle when M is zero

4. Approach

Use equation below.

$$ M=|r|\cdot|F|\cdot\sin\theta$$

5. Analysis

Part a)

The angle we were given is not technically the one we should use in the moment equation. The angle should be between r and F. Therefore, we have to find the new angle.

As shown below, the angle we find is also 45°. Now we can continue and solve for M₁ and M₂.

$$\theta=90^{\circ}-45^{\circ}$$

$$\theta=45^{\circ}$$

$$ M_1=|r_1|\cdot|F|\cdot\sin\theta\\M_1=0.45m\cdot 100N\cdot\sin(45^{\circ})\\m_1=31.82Nm\\\\M_2=|r_2|\cdot |F|\cdot\sin\theta\\M_2=0.75m\cdot 100N\cdot\sin(45^{\circ})\\M_2=53.03Nm$$

Part b)

$$M=|r|\cdot|F|\cdot\sin\theta\\if \sin\theta=0, M=0\\\sin\theta=0\\\theta=\sin^{-1}(0)\\\theta=0^{\circ}, 180^{\circ}, 360^{\circ}$$

Answer: the moment is zero when the angle between the force and the moment arm is 0° or 180° (360 would represent the same angle as 0°, as would 540°, etc.)

6. Review

It makes sense that the moment is zero when the door is either closed or wide open, because when we apply a force at those positions, no movement of the door is possible.

1. Problem

George travelled a displacement of d_g = [7   0  8] m from his car. George’s dog, named Sparky, on the other hand, travelled a displacement of ds = [0 6 6] m from George’s car. George called Sparky’s name, and the dog ran to George’s position. It took Sparky four seconds to get there.

a. What is the displacement from George to his dog?
b. What is Sparky’s velocity? (No need to draw.)
c. What is Sparky’s speed? (No need to draw.)

 

Source: https://www.piqsels.com/en/public-domain-photo-oekac

 

2. Draw

3. Knowns and Unknowns

Knowns:

• d_g = [7   0  8] m
• d_s = [0  6  6] m
• t= 4 seconds

Unknowns:

• v_sg= ?
• d_sg= ?

4. Approach

We are going to use vector operations (both subtraction and division), the velocity-displacement relationship, the velocity-speed relationship, and Pythagoras’ theorem to solve this problem.

5. Analysis

Part a:

$$ d_{sg}=d_{g}-d_{s}$$
$$ d_{sg}=[7\;\;\;0\;\;\;8]m-[0\;\;\;6\;\;\;6]m$$
$$ d_{sg}=[7-0\;\;\;0-6\;\;\;8-6]m$$
$$ d_{sg}=[7\;\;\;-6\;\;\;2]m$$

Part b:

$$ v_{sg}=d_{sg}/t$$
$$v_{sg}=[7\;\;\;-6\;\;\;2]m/4s$$
$$v_{sg}=[7/4\;\;\;-6/4\;\;\;2/4]m/s$$
$$v_{sg}=[1.75\;\;\;-1.5\;\;\;0.5]m/s$$

Part c:

$$V_{sg}=\sqrt{V_{sgx}^2+V_{sgy}^2+V_{sgz}^2}$$
$$V_{sg}=\sqrt{1.75^2+(-1.5)^2+0.5^2}$$
$$ V_{sg}=2.36m/s$$

6. Review

Part a:

One way to review the question is to walk through the solution verbally. Our solution shows that for Sparky to get to George, he must walk 7 m in the positive x-direction (almost out of the page), 6 m in the negative y-direction (left), and finally 2 m in the positive z-direction (up).

Firstly, since the dog initially did not go in the x-direction, it makes sense that Sparky would have to copy George’s exact x movement. Secondly, since George did not move in the y-direction, it would make sense that Sparky would just have to retrace his steps, and if he initially went 6 m right, he would have to go 6 m left. Thirdly, George and Sparky both went upwards, but George went 2 m higher with an altitude of 8 m compared to Sparky’s 6 m, correlating to Sparky having to go positive 2 m in the z-direction to meet George.

Therefore, since all the movements make sense for Sparky to meet George (using logic), the answer is proven to be right.

Parts b and c:

Since B and C correlate to the same magnitude, they can be reviewed together. From a quick search, an average dog tops out at a speed of 19 miles per hour. We can convert this to SI units:

The top speed of an average dog is 8.49 m/s. So 2.36m/s is approximately a quarter of the top speed of an average dog. Sparky probably was not sprinting at full speed and he could be a slower dog breed, making  2.36m/s a reasonable answer.

Example 1.8.7: Cross Product, Submitted By Victoria Keefe

2: Sketch

N/A

3. Knowns and Unknowns

Knowns:

• F = (5 i – 2 j + k)N
• l = 5m
• w = 1m
• h = 0.25m

Unknown:

• M

4: Approach

Use the equation: M = r x F

5. Analysis

$$ r = (-1 \hat{i} + 0.25 \hat{j} + 5 \hat{k})m \\ \bar{F} = (5 \hat{i} – 2 \hat{j} + \hat{k})N \\  \bar{M} =[ [(0.25 \cdot 1) – (-2 \cdot 5)] \hat{i} – [(-1 \cdot 1) – (5 \cdot 5)] \hat{j} + [(-1 \cdot -2) – (5 \cdot 0.25)] \hat{k}]Nm \\ \bar{M} = (10.25 \hat{i} + 26 \hat{j} +0.75 \hat{k})Nm $$

6.0 Review

The resulting moment has reasonable values and can be verified with a cross product calculator.

Particles are bodies where all the mass is concentrated at a single point in space. Particle analysis will only have to take into account the forces acting on the body and translational motion because rotation is not considered for particles.

Rigid bodies, on the other hand, have mass that is distributed throughout a finite volume. Rigid body analysis is more complex and also has to take into account moments and rotational motions. In actuality, no bodies are truly particles, but some bodies can be approximated as particles to simplify analysis. Bodies are often assumed to be particles if the rotational motions are negligible when compared to the translational motions, or in systems where there is no moment exerted on the body, such as a concurrent force system.

The rotation of this comet and the moments exerted on the comet are unimportant in modeling its trajectory through space, therefore we would treat it as a particle. Public Domain image by Buddy Nath.

The gravitational forces and the tension forces on the skycam all act through a single point, making this a concurrent force system that can be analyzed as a particle. Image by Despeaux CC-BY-SA 3.0.

Rotation and moments will be key to the analysis of the crowbar in this system, therefore the crowbar needs to be analyzed as an extended body. Public Domain image by Pearson Scott Foresman.

Source: Engineering Mechanics, Jacob Moore, et al. http://www.oercommons.org/courses/mechanics-map-open-mechanics-textbook/view

As you draw a free body diagram, there are a couple of things you need to keep in mind:

(1) Include only those forces acting ON the object whose free body diagram you are drawing. Any force exerted BY the object on some other object belongs on the free body diagram of the other object.
(2) All forces are contact forces and every force has an agent. The agent is “that which is exerting the force.” In other words, the agent is the life form or thing that is doing the pushing or pulling on the object. No agent can exert a force on an object without being in contact with the object.

We are going to introduce the various kinds of forces by means of examples. Here is the first
example:

There is no “force of motion” acting on an object. Once you have the force or forces
exerted on the object by everything that is touching the object, you have all the forces. Do not add a “force of motion” to your free body diagram. It is especially tempting to add this force when there are no actual forces in the direction in which an object is going. Keep in mind, however, that an object does not need a force on it to keep going in the direction in which it is going; moving along at a constant velocity is what an object does when there is no net force on it.

Source: Calculus-Based Physics 1, Jeffery W. Schnick. https://openlibrary.ecampusontario.ca/catalogue/item/?id=ce74a181-ccde-491c-848d-05489ed182e7 page 86

Objects in static equilibrium are objects that are not accelerating (either linear acceleration or angular acceleration). These objects may be stationary, or they may have a constant velocity.

Newton’s Second Law states that the force exerted on an object is equal to the mass of the object times the acceleration it experiences. Therefore, if we know that the acceleration of an object is equal to zero, then we can assume that the sum of all forces acting on the object is zero. Individual forces acting on the object, represented by force vectors, may not have zero magnitude but the sum of all the force vectors will always be equal to zero for objects in equilibrium.

The equations used when dealing with particles in equilibrium are:

$$\sum\vec F=0$$

Which leads to:

$$\sum F_x=0\\\sum F_y=0\\\sum F_z=0$$

Since it is a particle, there are no moments involved like there is when it comes to rigid bodies.

Source: Engineering Mechanics, Jacob Moore, et al. http://mechanicsmap.psu.edu/websites/2_equilibrium_concurrent/2-1_static_equilibrium/staticequilibrium.html

The first step in finding the equilibrium equations is to draw a free body diagram of the body being analyzed. This diagram should show all the known and unknown force vectors acting on the body. In the free body diagram, provide values for any of the know magnitudes or directions for the force vectors and provide variable names for any unknowns (either magnitudes or directions).

Next, you will need to choose the x, y, and z axes. These axes do need to be perpendicular to one another, but they do not necessarily have to be horizontal or vertical. If you choose coordinate axes that line up with some of your force vectors you will simplify later analysis.

Once you have chosen axes, you need to break down all of the force vectors into components along the x, y and z directions (see the vectors page in Appendix 1 if you need more guidance on this). Your first equation will be the sum of the magnitudes of the components in the x direction being equal to zero, the second equation will be the sum of the magnitudes of the components in the y direction being equal to zero, and the third (if you have a 3D problem) will be the sum of the magnitudes in the z direction being equal to zero. Collectively, these are known as the equilibrium equations.

Once you have your equilibrium equations, you can solve them for unknowns using algebra. The number of unknowns that you will be able to solve for will be the number of equilibrium equations that you have. In instances where you have more unknowns than equations, the problem is known as a statically indeterminate problem, and you will need additional information to solve for the given unknowns.

Example:

1. Problem

2. Draw

3. Knowns and Unknowns

Known:

• M_g = 6kg

Unknown:

• T₁
• T₂

4. Approach

Derive equilibrium equations and solve simultaneously to find T₂ and T₁

5. Analysis

$$F_g=(9.8)(6)\\F_g=58.8N\\\sum F_x=-T_1+T_2\cos(15^{\circ})=0\\\sum F_y=T_2\sin(15^{\circ})-58.8=0\\T_2=\frac{58.8}{\sin(15^{\circ})}=227.2N\\-T_1+227.2\cos(15^{\circ})=0\\T_1=227.2\cos(15^{\circ})=219.4N\\T_1=219.4N\\T_2=227.2N$$

6. Review

It makes sense that the tension in T₂ is greater than T₁ since it is at an angle.

Source: Engineering Mechanics, Jacob Moore, et al. Mechanics Map – Equilibrium Analysis for Concurrent Force Systems. Many more examples are available at this site.

See additional examples: P3.pdf, P5.pdf, P8.pdf

1. Problem

2. Draw 

3. Knowns and Unknowns

Knowns:

• Desmond’s mass = 80kg
• Mass of bag caught = (y)kg
• Weight of Desmond + bag, W = 9.81(80 + y)
• Weight of the block = x
• Wall A = First wall
• Wall B = Second wall
• g = 9.81m/s²

Unknowns:

• x
• y

4. Approach 

Apply equilibrium equations and solve simultaneously to find x and y

5. Analysis 

Taking moment about wall B,

Summing up vertical forces

Solving simultaneously;

In kilograms; (Divide by 9.81m/s²)

6. Review 

The block would need to weigh 97.86 kg at 2m from Wall A, and Desmond should catch the bag weighing 6.81kg while he hangs 7m from Wall A.

1. Problem

2. Draw

Free Body Diagram:

3. Knowns and Unknowns

Knowns:

• m=5000 kg
• ∠A = 40°
• ∠B = 60°

Unknowns:

• F₁
• F₂ 

4. Approach

Considering the load is evenly distributed, the upper pair of string have equal lengths, and so does the lower pair, FBD can be drawn for each points. Use equilibrium equations for x and y direction to find tension on each string.

5. Analysis

Weight,

Point A

Point B

Point C

Thus both the upper strings are at a tension of 19070.76 N and lower strings are at a tension of 14159.51 N.

6. Review

Note that in this problem the vehicle’s weight was evenly distributed which is convenient in symmetrical situations like this. The tension on the strings can be verified as follows.

1. Problem

2. Draw 

3. Knowns and Unknowns

Knowns:

Unknowns:

• F_A
• F_B

4. Approach 

Analyzing the dimensions of the situation and applying the equilibrium equation.

5. Analysis 

b. When the length of the strings is equal, the weight on point A is equal to the weight on point B. F_A = F_B

No Force in the x-direction

c. When one chain of the swing is longer, the tension on the chains varies such that ; in turn, chain B will experience more tension than chain A. To solve for tension in the chains, we would require more information about the distribution of the weights on points A and B.

6. Review 

In questions involving free-body diagrams, it is convenient to have perfectly symmetrical situations. However, when faced with non-symmetrical or distributed loads, different techniques need to be applied, which will be explained in the following chapters.

The direction of the cross product vector A x B is given by the right-hand rule for the cross product of two vectors. To apply this right-hand rule, extend the fingers of your right hand so that they are pointing directly away from your right elbow. Extend your thumb so that it is at
right angles to your fingers.

Keeping your fingers aligned with your forearm, point your fingers in the direction of the first vector (the one that appears before the “×” in the mathematical expression for the cross product; e.g. the A in A x B ).

Now rotate your hand, as necessary, about an imaginary axis extending along your forearm and along your middle finger, until your hand is oriented such that, if you were to close your fingers, they would point in the direction of the second vector.

Your thumb is now pointing in the direction of the cross product vector. C = A x B. The cross product vector C is always perpendicular to both of the vectors that are in the cross product (the A and the B in the case at hand). Hence, if you draw them so that both of the vectors that are in the cross product are in the plane of the page, the cross product vector will always be perpendicular to the page, either straight into the page, or straight out of the page. In the case at hand, it is straight out of the page.

When we use the cross product to calculate the torque due to a force F whose point of application has a position vector r, relative to the point about which we are calculating the torque, we get an axial torque vector τ. To determine the sense of rotation that such a torque vector would correspond to, about the axis defined by the torque vector itself, we use the Right Hand Rule for Something Curly Something Straight. Note that we are calculating the torque with respect to a point rather than an axis—the axis about which the torque acts comes out in the answer.

Source: Jeffrey W. Schnick https://openlibrary.ecampusontario.ca/catalogue/item/?id=ce74a181-ccde-491c-848d-05489ed182e7 pages 135–137

A couple is a set of equal and opposite forces that exerts a net moment on an object but no net force. Because the couple exerts a net moment without exerting a net force, couples are also sometimes called pure moments.

The two equal and opposite forces exerted on this lug wrench are a couple. They exert a moment on the lug nut on this wheel without exerting any net force on the wheel. Adapted from image by Steffen Heinz Caronna CC-BY-SA 3.0.

The moment exerted by a couple also differs from the moment exerted by a single force in that it is independent of the location you are taking the moment about. In the example below we have a couple acting on a beam. Each force has a magnitude F and the distance between the two forces is d.

The moment exerted by this couple is independent of the distance x.

Now we have some point A, which is distance x from the first of the two forces. If we take the moment of each force about point A, and then add these moments together for the net moment about point A we are left with the following formula.

$$M=-(F\ast x)+(F\ast(x+d))$$

If we rearrange and simplify the formula above, we can see that the variable x actually disappears from the equation, leaving the net moment equal to the magnitude of the forces (F) times the distance between the two forces (d).

$$M=-(F\ast x)+(F\ast x)+(F\ast d)\\\\M=(F\ast d)$$

This means that no matter what value of x we have, the magnitude of the moment exerted by the couple will be the same. The magnitude of the moment due to the couple is independent of the location we are taking the moment about. This will also work in two or three dimensions as well. The magnitude of the moment due to a couple will always be equal to the magnitude of the forces times the perpendicular distance between the two forces.

Source: Engineering Mechanics, Jacob Moore et al., http://mechanicsmap.psu.edu/websites/3_equilibrium_rigid_body/3-3_couples/couples.html

A distributed load is any force where the point of application of the force is an area or a volume. This means that the “point of application” is not really a point at all. Though distributed loads are more difficult to analyze than point forces, distributed loads are quite common in real-world systems, so it is important to understand how to model them.

Distributed loads can be broken down into surface forces and body forces. Surface forces are distributed forces where the point of application is an area (a surface on the body). Body forces are forces where the point of application is a volume (the force is exerted on all molecules throughout the body). Below are some examples of surface and body forces.

The water pressure pushing on the surface of this dam is an example of a surface force. Image by Curimedia CC-BY-SA 2.0

The gravitational force on this apple is distributed over the entire volume of the fruit. Gravitational forces are an example of body forces. Image by Zátonyi Sándor CC-BY 3.0.

Distributed loads are represented as a field of vectors. This is drawn as a number of discrete vectors along a line, over a surface, or over a volume, that are connected with a line or a surface, as shown below.

This is a representation of a surface force in a 2D problem (A force distributed over a line). The magnitude is given in units of force per unit distance.

This is a representation of a surface force in a 3D problem (a force distributed over an area). The magnitude is given in units of force per unit area (also called a pressure).

Though these representations show a discrete number of individual vectors, there is actually a magnitude and direction at all points along the line, surface, or body. The individual vectors represent a sampling of these magnitudes and directions.

It is also important to realize that the magnitudes of distributed forces are given in force per unit distance, area, or volume. We must integrate the distributed load over its entire range to convert the force into the usual units of force.

Analyzing Distributed Load:

For analysis purposes in statics and dynamics, we will usually substitute in a single point force that is statically equivalent to the distributed load in the problem. This single point force is called the equivalent point load and it will cause the same accelerations or reaction forces as the distributed load while simplifying the math.

Source: Engineering Mechanics, Jacob Moore et al., http://mechanicsmap.psu.edu/websites/4_statically_equivalent_systems/4-4_distributed_forces/distributedforces.html

This is a more complex example of a distributed load. This is a cartoon of an airplane with its wing covered in a combination of snow and ice. In a real-world situation, loads will not accommodate people for ease of calculation; you get what you get. In this case, we could approximate this shape with two semi-circles on each end of the wing with a triangle in the middle. For more accuracy, we could use a system similar to the trapezoidal rule.

Source: ” Statics” by LibreTexts is licensed under CC BY-NC-SA. https://eng.libretexts.org/Bookshelves/Introduction_to_Engineering/EGR_1010%3A_Introduction_to_Engineering_for_Engineers_and_Scientists/14%3A_Fundamentals_of_Engineering/14.11%3A_Mechanics/14.11.01%3A_Statics

An equivalent point load is a single point force that will have the same effect on a body as the original loading condition, which is usually a distributed load. The equivalent point load should always cause the same linear acceleration and angular acceleration as the original load it is equivalent to (or cause the same reaction forces if the body is constrained). Finding the equivalent point load for a distributed load often helps simplify the analysis of a system by removing the integrals from the equations of equilibrium or equations of motion in later analysis.

If the body is unconstrained as shown on the left, the equivalent point load (shown as a solid vector) will cause the same linear and angular acceleration as the original distributed load (shown with dashed vectors). If the body is constrained as shown on the right, the equivalent point load (shown as a solid vector) will cause the same reaction forces as the original distributed force (shown with dashed vectors).

Finding the Equivalent Point Load

When finding the equivalent point load we need to find the magnitude, direction, and point of application of a single force that is equivalent to the distributed load we are given. In this course we will only deal with distributed loads with a uniform direction, in which case the direction of the equivalent point load will match the uniform direction of the distributed load. This leaves the magnitude and the point of application to be found. There are two options available to find these values:

1. We can find the magnitude and the point of application of the equivalent point load via integration of the force functions.
2. We can use the area/volume and the centroid/center of volume of the area or volume under the force function.

The first method is more flexible, allowing us to find the equivalent point load for any force function that we can make a mathematical formula for (assuming we have the skill in calculus to integrate that function). The second method is usually faster, assuming that we can look up the values for the area or volume under the force curve and the values for the centroid or center of volume for the area under the curve.

Using Integration in 2D Surface Force Problems:

Finding the equivalent point load via integration always begins by determining the mathematical formula that is the force function. The force function mathematically relates the magnitude of the force (F) to the position (x). In this case the force is acting along a single line, so the position can be entirely determined by knowing the x coordinate, but in later problems we may also need to relate the magnitude of the force to the y and z coordinates. In our example below, we can relate magnitude of the force to the position by stating that the magnitude of the force at any point in Newtons per meter is equal to the x position in meters plus one.

The block shown above has a distributed force acting on it. The force function relates the magnitude of the force to the x-position along the top of the box.

The magnitude of the equivalent point load will be equal to the area under the force function. This will be the integral of the force function over its entire length (in this case, from x = 0 to x = 2).

$$F_{eq}=\int_{xmin}^{xmax}F(x)dx$$

Now that we have the magnitude of the equivalent point load such that it matches the magnitude of the original force, we need to adjust the position (x_eq) such that it would cause the same moment as the original distributed force. The moment of the distributed force will be the integral of the force function (F(x)) times the moment arm about the origin (x). The moment of the equivalent point load will be equal to the magnitude of the equivalent point load that we just found times the moment arm for the equivalent point load (x_eq). If we set these two things equal to one another and then solve for the position of the equivalent point load (x_eq) we are left with the following equation:

$$x_{eq}=\frac{\int_{xmin}^{xmax}(F(x)\ast x)dx}{F_{eq}}$$

Now that we have the magnitude, direction, and position of the equivalent point load, we can draw the point load in our original diagram. This point force can be used in place of the distributed force in further analysis.

The values for F_eq and x_eq that we have solved for are the magnitude and position of the equivalent point load.

Using the Area and Centroid in 2D Surface Force Problems:

As an alternative to using integration, we can use the area under the force curve and the centroid of the area under the force curve to find the equivalent point load’s magnitude and point of application, respectively.

The magnitude of the equivalent point load is equal to the area under the force function. Also, the equivalent point load will travel through the centroid of the area under the force function.

The magnitude(F_eq) of the equivalent point load will be equal to the area under the force function. We can find this area using calculus, but there are often easier geometry-based ways of finding the area under the force function.

The equivalent point load will also travel through the centroid of the area under the force function. This allows us to find the value for x_eq. The centroid for many common shapes can be looked up in tables, and the parallel axis theorem can be used to determine the centroid of more complex shapes (see the centroid page for more details).

Source: Engineering Mechanics, Jacob Moore et al., Mechanics Map – Equivalent Point Load via Integration

Example 2 (note: 1 kip = 1000 lb):

Example 3:

Example 4:

Source: ” Equilibrium Structures, Support Reactions, Determinacy and Stability of Beams and Frames” by LibreTexts is licensed under CC BY-NC-ND . https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.03%3A_Equilibrium_Structures_Support_Reactions_Determinacy_and_Stability_of_Beams_and_Frames

3.4.1 Pin or Hinge Support

A pin support allows rotation about any axis but prevents movement in the horizontal and vertical directions. Its idealized representation and reactions are shown in Table 3.1

Table 3.1 Reaction Representations

3.4.2 Roller Support

A roller support allows rotation about any axis and translation (horizontal movement) in any direction parallel to the surface on which it rests. It restrains the structure from movement in a vertical direction. The idealized representation of a roller and its reaction are also shown in Table 3.1.

3.4.3 Rocker Support

The characteristics of a rocker support are like those of the roller support. Its idealized form is depicted in Table 3.1.

3.4.4 Link

A link has two hinges, one at each end. It permits movement in all direction, except in a direction parallel to its longitudinal axis, which passes through the two hinges. In other words, the reaction force of a link is in the direction of the link, along its longitudinal axis.

3.4.5 Fixed Support

A fixed support offers a constraint against rotation in any direction, and it prevents movement in both horizontal and vertical directions.

Example 1:

Example 2 (A_x added even though it turns out to be 0):

Example 3:

Example 4:

Source: ” Equilibrium Structures, Support Reactions, Determinacy and Stability of Beams and Frames” by LibreTexts is licensed under CC BY-NC-ND . https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.03%3A_Equilibrium_Structures_Support_Reactions_Determinacy_and_Stability_of_Beams_and_Frames

Source: ” Equilibrium Structures, Support Reactions, Determinacy and Stability of Beams and Frames” by LibreTexts is licensed under CC BY-NC-ND . https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.03%3A_Equilibrium_Structures_Support_Reactions_Determinacy_and_Stability_of_Beams_and_Frames

1. Problem

A family is sitting watching TV on their couch. The couch is 5 m long and weighs 120 N. The child is sat 1 m away from one end and has a mass of 30 kg. The mother is sat 0.5 m away from the child and has a mass of 60 kg. The father is 3 m away from the mother and has a mass of 70 kg.

a) Draw a free-body diagram of the couch

b) Calculate the reaction force on each of the two legs.

Assume the couch is supported by two rollers.

Source: https://www.maxpixel.net/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817

2. Draw

Sketch:

3. Knowns and Unknowns

Knowns:

• g = 9.81 m/s²
• m_c = 30 kg
• m_m = 60 kg
• m_f = 70 kg
• F_g = 120 N
• r_c = 1 m
• r_M = 1.5 m
• r_f = 4.5 m
• r_B = 5 m
• r_g = 2.5 m

Unknowns:

• N_A
• N_B

4. Approach

Use equilibrium equations

5. Analysis

Part a:

Part b:

$$\sum M_A=0=N_B\times r_B-F_c\times r_c-F_M\times r_M-F_f\times r_f-F_g\times r_g\\\\N_B(5m)=(30kg\times 9.81m/s^2)(1m)+(60kg\times 9.81m/s^2)(1.5m)\\+(70kg\times 9.81m/s^2)(4.5m)+(120N)(2.5m)$$$$\\N_B(5m)=294.3Nm+882.9Nm +3090.15Nm+300Nm\\\\N_B(5m)=4567.35N m\\\\N_B\frac{4567.35Nm}{5m}\\\\N_B=913.47N\\\\N_B=913N$$

$$\sum F_y=0=N_A+N_B-F_C-F_M-F_f-F_g\\\\N_A=F_C+F_M+F_f+F_g-N_B\\\\N_A=(30kg\times 9.81m/s^2)+(60kg\times 9.81m/s^2)\\+(70kg\times 9.81m/s^2)+120N-913.47N$$$$\\N_A=294.3N+588.6N+686.7N+120N-913.47N$$$$\\N_A=776.13N\\\\\\\underline{N_A=776N}$$

6. Review

It is interesting that N_B is larger than N_A, because the weight of the mother and child combined (80 kg) is larger than that of the father (70 kg). However, when you sum the moments at point B instead of A, you get the same answer. The distance between the reaction forces and the nearest forces is important, as well as the magnitude of the forces themselves. The distance between A and F_c is 1 m, while the distance between B and F_f is only 0.5 m.

Additionally, it makes sense that both N_A and N_B are positive, i.e. are in the positive y direction.

1. Problem

A water valve is opened by a wheel with a diameter of 10 inches. It takes 7.5 lb of force to open the valve. What is the moment it takes to open the valve?

Real-life scenario:

Source: https://www.pxfuel.com/en/free-photo-ekahu

2. Draw

Sketch:

Free-body diagram:

3. Knowns and Unknowns

Known:

• d = 10 in
• F = 7.5 lb

Unknown:

• M

4. Approach

Determine the moment by finding the cross product of r_A and F_A, then r_B and F_B, then add.

5. Analysis

Find radius:

$$r=\frac{d}{2}\\r=\frac{10in}{2}\\r=5in\\5in\times\frac{1ft}{12in}=0.42ft$$

Find r_A, F_A, r_B, and F_B in vector form:

$$ \underline{r}_A= \begin{bmatrix}
0.42 \\
0
\end{bmatrix}ft\:\; \underline{F}_A=\begin{bmatrix}
0 \\
7.5
\end{bmatrix}lb \\\underline{r}_B=\begin{bmatrix}
-0.42 \\
0
\end{bmatrix}ft\:\; \underline{F}_B=\begin{bmatrix}
0 \\
-7.5
\end{bmatrix}lb $$

Find M_A:

$$\underline{M}_A=\underline{r}_A\times \underline{F}_A=\begin{bmatrix}
\underline{\hat{i}} & \underline{\hat{j}} & \underline{\hat{k}} \\
0.42 & 0 & 0 \\
0 & 7.5 & 0
\end{bmatrix}$$ $$\underline{M}_A=\hat{i} \begin{bmatrix}
0 & 0 \\
7.5 & 0
\end{bmatrix} -\underline{\hat{j}} \begin{bmatrix}
0.42 & 0 \\
0 & 0
\end{bmatrix}+\underline{\hat{k}} \begin{bmatrix}
0.42 & 0 \\
0 & 7.5
\end{bmatrix}\\\underline{M}_A=(\underline{\hat{i}}(0)-\underline{\hat{j}}(0)+\underline{\hat{k}}(0.42\cdot 7.5-0\cdot 0))ft\cdot lb\\\underline{M}_A=3.15\underline{\hat{k}} ft\cdot lb$$

Find M_B:

$$\underline{M}_B=\underline{r}_B\times \underline{F}_B=\begin{bmatrix}
\underline{\hat{i}} & \underline{\hat{j}} & \underline{\hat{k}} \\
-0.42 & 0 & 0 \\
0 & -7.5 & 0
\end{bmatrix}$$$$\underline{M}_B=\hat{i} \begin{bmatrix}
0 & 0 \\
-7.5 & 0
\end{bmatrix} -\underline{\hat{j}} \begin{bmatrix}
-0.42 & 0 \\
0 & 0
\end{bmatrix}+\underline{\hat{k}} \begin{bmatrix}
-0.42 & 0 \\
0 & -7.5
\end{bmatrix}\\\underline{M}_B=(\underline{\hat{i}}(0)-\hat{j}(0)+\underline{\hat{k}}(-0.42\cdot -7.5-0\cdot 0))ft\cdot lb\\\underline{M}_B=3.15\underline{\hat{k}} ft\cdot lb$$

Add M_A and M_B to get M:

$$\underline{M}=\underline{M}_A+\underline{M}_B\\\underline{M}=3.15ft\cdot lb+3.15ft\cdot lb\\\underline{M}=6.3\underline{\hat{k}}ft\cdot lb$$

6. Review

Notice that for the x-coordinates, the positive x-direction was taken to the left, and this was consistently followed in the calculation. It’s important to stay consistent with your chosen direction, as mixing it up can lead to sign errors.

This answer makes sense because there is only moment acting in the k direction.

Note: We could have come to the same answer using the formula M = F*d, which would have been faster.

$$M=f\cdot d\\M=7.5lb\cdot 10in(\frac{1ft}{12in})\\M=7.5lb\cdot \frac{5}{6}ft\\\\M=6.25ft\cdot lb$$

This answer is slightly more accurate because we didn’t round when converting between inches and feet (in the original solution, we rounded 0.416667 to 0.42).

1. Problem

A shelf on the wall is 1.5 meters away from the floor. The shelf has a length of 100 cm. A person starts putting different objects on it to create a distributed load. The load created a curve described by:

w = 4x⁴ +2 N/m.

Calculate the resultant force and how far it is acting from the wall (Fixed end).

Source: https://pixabay.com/vectors/shelf-floating-bathroom-glass-576088/

2. Draw

Sketch:

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• w = 4x⁴ + 2 N/m
• L = 100 cm = 1m
• xmin = 0m
• xmax = 1m

Unknowns:

• x_r
• F_r

4. Approach

Use distributed load equations:

$$F_r=\int^{xmax}_{xmin} wdx\\X_r=\frac{\int^{xmax}_{xmin} x*w(x)*dx}{\int^{xmax}_{xmin}wdx}$$

5. Analysis

Solve for F_r:

$$ F_r=\int^1_0 (4x^4+2)dx\;\; N\\F_r=(\frac{4x^5}{5}+2x)\vert^1_0\;\;N\\F_r=(\frac{4}{5}+2)N\\F_r=2.8N$$

Solve for x_r:

$$X_r=\frac{(\int^1_0x(4x^4+2)dx)N/m}{\int^1_0(4x^4+2)dx)N}\\X_r=\frac{\int^1_0(4x^5+2x)dxN/m}{2.8N}\\X_r=\frac{(\frac{4x^6}{6}+\frac{2x^2}{2})\vert^1_0N/m}{2.8N}\\X_r=\frac{(\frac{2}{3}+1)N/m}{2.8N}\\X_r=0.59m$$

6. Review

The function shows an increasing curve on the interval, so it makes sense that the resultant force would be applied closer to the right end of the beam than the left end.

1. Problem

A mason jar lid with a diameter of 11cm is firmly closed and requires a moment of 5.5 Nm to be opened. If you are to open the lid with your finger and thumb, assuming they are applying equal force, determine the force required by your thumb and finger.

Source: https://commons.wikimedia.org/wiki/File:Fingertip_tightening_lid_on_pickled_beet_jar.jpg

2. Draw

3. Knowns and Unknowns

Knowns:

• M = 5.5 Nm
• d = 11 cm = 0.11 m

Unknowns:

• F – Force exerted by thumb and finger

4. Approach

Using couple to find F

5. Analysis

Since the forces exerted by the finger and thumb are equal and opposite, it is possible to find the force exerted by the finger and thumb using the couple equation.

$$5.5Nm=F\times0.11m$$

$$F=\frac{5.5Nm}{0.11m}\\F =50N$$

Thus, your thumb and finger are applying a force of 50N each.

6. Review

The moment equation is applicable since the force of the thumb and finger is equal. The equation produces a reasonable value of 50N

1. Problem

You are driving home from work, and you need to turn right to get into your driveway. To turn the steering, you are applying 115 N with each hand. Consider your steering wheel has a diameter of 40 cm.

a. Determine the couple moment produced on the steering wheel.

b. Compare the above result with a situation where you use single hand to navigate the wheel. How much force does it take to turn the steering wheel in this case?

Source:https://commons.wikimedia.org/wiki/File:Woman_hand_steering_wheel_%28Unsplash%29.jpg

 

2. Draw

3. Knowns and Unknowns

Knowns:

• F = 115 N
• d = 40 cm = 0.40m

Unknowns:

• Couple moment, M
• Force applied by hand in the second scenario, F₂

4. Approach

Use the couple Moment equation to find the moment, and then use the moment to find the force required to turn the handle single-handedly.

5. Analysis

a.  Both the arms are applying a 115 N force each. The diameter of the wheel is 40 cm.

Therefore,

$$\overrightarrow{M}= \overrightarrow{F}\times\overrightarrow{d}=115N\times0.4m$$

$$\overrightarrow{M}=46Nm$$

b. Consider 46Nm as the required moment to turn the steering wheel.

When just one hand applies force on the wheel it forms a moment which is equal to 46 Nm, with distance and angle given we can find the force it takes to turn the steering wheel.

$$\overrightarrow{M} = F_{2}\times\frac{d}{2}\times\sin{\theta}\\46Nm =F_{2}\cdot\frac{0.4m}{2}\cdot\sin{90}\\F_{2}=230 N$$

6. Review 

As expected, the force it took to turn the steering wheel is twice the force it took with one hand when both hands are used.

1. Problem

Ellen is taking her driving test and is asked to perform a right turn. The diameter of the steering wheel is 15 inches. Ellen’s right hand exerts a force of 200 N, and her left exerts 150 N.

a. Calculate the net moment in the above-given situation using the cross product.

b. If Ellen were to perform a left turn with the same forces given, what would be the net moment?

Source:https://www.wikihow.com/Drive-Safely-in-Fog

The above image is a close illustration of the problem. In the problem, the hands of the driver are along the diameter.

2. Draw 

3. Knowns and Unknowns

Knowns:

• Diameter of the wheel = 15 in = 0.381m
• Right hand force = 200 N
• left hand force = 150 N

Unknowns:

• M₁ (for right hand)
• M₂ (for left hand)

4. Approach 

Convert inches to meters and add the cross product of distance and force to find the net moment.

5. Analysis 

a.RIGHT TURN

Express the radius of the steering wheel and forces in vector form for tight turn.

b. LEFT TURN

Radius and force in vector form

Radius is the same as it is for a right turn, but the forces have a different direction.

6. Review 

Because the radius and force values in both instances have the same magnitude but different directions, the net moments for left and right turns were expected to have the same magnitudes and different directions; thus, the answers make sense.

1. Problem

2. Sketch:

3. Knowns and Unknowns:

Knowns:

• m₁ = 35kg
• m₂ = 50kg
• g = 9.81 m/s²

Unknown:

• R₁
• R₂

4. Approach:

Convert the masses to forces using gravity.
Equate the moments caused by each force.
Solve for the ratio of lengths.

5. Analysis:
For the children to apply an equal amount of force while playing on the seesaw, they must create an equal moment.
Therefore:
$$ M_1 = M_2$$
These moments can be rewritten in terms of F and R.
$$ F_1 \cdot R_1 = F_2 \cdot R_2$$
The F values can be solved by multiplying the child’s mass by gravity to find the force they exert on the seesaw.
$$ (m_1 \cdot g) \cdot R_1 = (m_2 \cdot g) \cdot R_2$$
Gravity can be cancelled out as it exists on both sides of the equation, leaving:
$$m_1  \cdot R_1 = m_2 \cdot R_2$$
Putting values into this equation gives:
$$35kg  \cdot R_1 = 50kg \cdot R_2$$
The equation can now be adjusted to solve for R₁/R₂.
$$ \frac{R_1}{R_2} = \frac{50kg}{35kg}$$
This can be simplified to give the ratio:
$$ \frac{R_1}{R_2} = \frac{10}{7} \\ R_1 : R_2 = 10 : 7$$

6. Review

It makes sense that the lighter child needs to be further away to produce an equal moment to that of the heavier child. 10:7 is a reasonable ratio that matches the ratio of their weights.

1. Problem

2. Draw

3. Knowns and Unknowns

Knowns:

• Dimensions of the cars and the bridge
• Weight of one car, w = 1000 N/m

Unknowns:

• The resultant force on either side of the bridge.
• The location of the resultant force of all 5 cars.

4.  Approach

• The weight of the cars is distributed along the dimensions. We figure out he resultant forces and the location of the resultant forces along the x direction using the equation .
• Note that the shape of the car consists of a rectangle and an adjacent triangle. Hence, the distributed load is found for both shapes, using equations appropriate to find the area under the force, and then added to form the resultant force of one car.

5. Analysis

a. Finding the location along the x-axis and the magnitude of the resultant force of 3 cars

Since all 3 cars are identical .

From the fig above

for resultant force of car A

From O locates at

locates at

now,

thus

from left side.

Also for one car the of 2250 N force is

Using these calculations, the following figure can be drawn.

The total resultant force of all 3 cars is

therefore,

  from left side .

b. Position and magnitude of the resultant force of the 2 cars on the lane. 

Total force of the 2 cars

Position of the resultant forces from the 2 cars,

6.67 m is from the right side, but from the  left side 

c. Position and magnitude of the total resultant force on the bridge.

Total force on bridge due to the 5 cars

for all the cars is half of its width

therefore,

6. Review

This problem is less complex because of the symmetry of the cars and the bridge. The coordinates of the resultant force are just as expected. x is closer to the side with more weight (due to the distribution of weight of the vehicles), and y is less than halfway across the bridge because that side has more vehicles

1. Problem

2. Draw

3. Known and Unknowns 

Knowns:

• F =50 N
• Diameter, d = 1m

Unknown:

• Couple moments, M

4. Approach 

Using Equation

5. Analysis 

Couple moment,

6. Review  

Note that in this question, the direction of the rotation of the tap is not mentioned. But the sign of the couple moment changes with respect to the direction of rotation and the coordinates which you choose to be positive.

Forces are given many names, such as push, pull, thrust, and weight. Traditionally, forces have been grouped into several categories and given names relating to their source, how they are transmitted, or their effects. Several of these categories are discussed in this section.

Normal Force

Weight (also called the force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling. You must support the weight of a heavy object by pushing up on it when you hold it stationary. But how do inanimate objects like a table support the weight of a mass placed on them, such as shown in the figure below? When the bag of dog food is placed on the table, the table sags slightly under the load. This would be noticeable if the load were placed on a card table, but even a sturdy oak table deforms when a force is applied to it. Unless an object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or a trampoline or diving board). The greater the deformation, the greater the restoring force. Thus, when the load is placed on the table, the table sags until the restoring force becomes as large as the weight of the load. At this point, the net external force on the load is zero. That is the situation when the load is stationary on the table. The table sags quickly, and the sag is slight, so we do not notice it. But it is similar to the sagging of a trampoline when you climb onto it.

Figure: (a) The person holding the bag of dog food must supply an upward force F̅_hand equal in magnitude and opposite in direction to the weight of the food w̅ so that it doesn’t drop to the ground. (b) The card table sags when the dog food is placed on it, much like a stiff trampoline. Elastic restoring forces in the table grow as it sags until they supply a force N̅ equal in magnitude and opposite in direction to the weight of the load

We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting the weight of an object, or a load, is perpendicular to the surface of contact between the load and its support, this force is defined as a normal force and here is given by the symbol $\stackrel{}{}$(This is not the newton unit for force, N.) The word normal means perpendicular to a surface. This means that the normal force experienced by an object resting on a horizontal surface can be expressed in vector form as follows:

$$\vec N=-m\vec g$$

In scalar form, this becomes:

$$N=mg$$

The normal force can be less than the object’s weight if the object is on an incline.

When an object rests on an incline that makes an angle θ with the horizontal, the force of gravity acting on the object is divided into two components: a force acting perpendicular to the plane, w_y, and a force acting parallel to the plane, w_x. The normal force is typically equal in magnitude and opposite in direction to the perpendicular component of the weight w_y. The force acting parallel to the plane, w_x, causes the object to accelerate down the incline.

Figure: An object rests on an incline that makes an angle theta with the horizontal

Be careful when resolving the weight of the object into components. If the incline is at an angle θ to the horizontal, then the magnitudes of the weight components are:

$$w_x=w\sin\theta=mg\sin\theta$$

and

$$w_y=w\cos\theta=mg\cos\theta$$

We use the second equation to write the normal force experienced by an object resting on an inclined plane:

$$N=mg\cos\theta$$

Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, we draw the right angle formed by the three weight vectors. The angle θ of the incline is the same as the angle formed between w and w_y. Knowing this property, we can use trigonometry to determine the magnitude of the weight components:

$$\cos\theta=\frac{w_y}{w},\:w_y=w\cos\theta=mg\cos\theta\\\sin\theta=\frac{w_x}{w},\:w_x=w\sin\theta=mg\sin\theta$$

Tension

A tension is a force along the length of a medium; in particular, it is a pulling force that acts along a stretched flexible connector, such as a rope or cable. The word “tension” comes from a Latin word meaning “to stretch.” Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons.

Any flexible connector, such as a string, rope, chain, wire, or cable, can only exert a pull parallel to its length; thus, a force carried by a flexible connector is a tension with a direction parallel to the connector. Tension is a pull in a connector. Consider the phrase: “You can’t push a rope.” Instead, tension force pulls outward along the two ends of a rope.

Consider a person holding a mass on a rope. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero and the net force is zero. The only external forces acting on the mass are its weight and the tension supplied by the rope. Thus,

If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope.

When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force $\vec{T}$, that force must be parallel to the length of the rope, as shown. By Newton’s third law, the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the weight of the rope). The rope is the medium that carries the equal and opposite forces between the two objects. The tension anywhere in the rope between the hand and the mass is equal. Once you have determined the tension in one location, you have determined the tension at all locations along the rope.

Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a tendon, or a bicycle brake cable. If there is no friction, the tension transmission is undiminished; only its direction changes, and it is always parallel to the flexible connector, as shown below:

If we wish to create a large tension, all we have to do is exert a force perpendicular to a taut flexible connector. We can see that the tension in the rope is related to the force acting perpendicularly in the following way:

$$T=\frac{w}{2\sin\theta}$$

We can extend this expression to describe the tension T created when a perpendicular force (F⊥) is exerted at the middle of a flexible connector:

$$T=\frac{F\perp}{2\sin\theta}$$

The angle between the horizontal and the bent connector is represented by θ. In this case, T becomes large as θ approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., θ=0 and sin θ=0). For example, the image below shows a situation where we wish to pull a car out of the mud when no tow truck is available. Each time the car moves forward, the chain is tightened to keep it as straight as possible. The tension in the chain is given by and since θ is small, T is large. This situation is analogous to the tightrope walker, except that the tensions shown here are those transmitted to the car and the tree rather than those acting at the point where F⊥ is applied.

Figure: We can create a large tension in the chain, and potentially a big mess, by pushing on it perpendicular to its length, as shown.

Friction

Friction is a resistive force opposing motion or its tendency. Imagine an object at rest on a horizontal surface. The net force acting on the object must be zero, leading to equality of the weight and the normal force, which act in opposite directions. If the surface is tilted, the normal force balances the component of the weight perpendicular to the surface. If the object does not slide downward, the component of the weight parallel to the inclined plane is balanced by friction. Friction is discussed in greater detail in the next chapter.

Spring Force

A spring is a special medium with a specific atomic structure that has the ability to restore its shape, if deformed. To restore its shape, a spring exerts a restoring force that is proportional to and in the opposite direction in which it is stretched or compressed. This is the statement of a law known as Hooke’s law, which has the mathematical form

$$\vec F=-k\vec x$$

The constant of proportionality k is a measure of the spring’s stiffness. The line of action of this force is parallel to the spring axis, and the sense of the force is in the opposite direction of the displacement vector. The displacement must be measured from the relaxed position; x=0 when the spring is relaxed.

A spring exerts its force proportional to a displacement, whether it is compressed or stretched. (a) The spring is in a relaxed position and exerts no force on the block. (b) The spring is compressed by displacement [latex]\Delta \vec{x}_1[/latex] of the object and exerts restoring force [latex]-k \Delta \vec{x}_1[/latex]. (c) The spring is stretched by displacement [latex]\Delta \vec{x}_2[/latex] of the object and exerts restoring force [latex]-k \Delta \vec{x}_2[/latex].

A free body diagram is a tool used to solve engineering mechanics problems. As the name suggests, the purpose of the diagram is to “free” the body from all other objects and surfaces around it so that it can be studied in isolation. We will also draw in any forces or moments acting on the body, including those forces and moments exerted by the surrounding bodies and surfaces that we removed.

The diagram below shows a ladder supporting a person and the free body diagram of that ladder. As you can see, the ladder is separated from all other objects and all forces acting on the ladder are drawn in with key dimensions and angles shown.

A ladder with a man standing on it is shown on the left. Assuming friction only at the base, a free body diagram of the ladder is shown on the right.

The first step in solving most mechanics problems will be to construct a free-body diagram. This simplified diagram will allow us to more easily write out the equilibrium equations for statics or strengths of materials problems, or the equations of motion for dynamics problems.

To construct the diagram, we will use the following process.

1. First, draw the body being analyzed, separated from all other surrounding bodies and surfaces. Pay close attention to the boundary, identifying what is part of the body and what is part of the surroundings.
2. Second, draw in all external forces and moments acting directly on the body. Do not include any forces or moments that do not directly act on the body being analyzed. Do not include any forces that are internal to the body being analyzed. Some common types of forces seen in mechanics problems are:
   • Gravitational Forces: Unless otherwise noted, the mass of an object will result in a gravitational weight force applied to that body. This weight is usually given in pounds in the English system, and is modelled as 9.81 (g) times the mass of the body in kilograms for the metric system (resulting in a weight in Newtons). This force will always point down towards the center of the Earth and act on the center of mass of the body.
     
     

     Gravitational forces always act downward on the center of mass.
     • Normal Forces (or Reaction Forces): Every object in direct contact with the body will exert a normal force on that body, which prevents the two objects from occupying the same space at the same time. Note that only objects in direct contact can exert normal forces on the body.
       • An object in contact with another object or surface will experience a normal force that is perpendicular (hence normal) to the surfaces in contact.
         
         

         Normal forces always act perpendicular to the surfaces in contact. The barrel in the hand truck shown on the left has a normal force at each contact point.
       • Joints or connections between bodies can also cause reaction forces or moments, and we will have one force or moment for each type of motion or rotation the connection prevents.
         
         

         The roller on the left allows for rotation and movement along the surface, but a normal force in the y direction prevents motion vertically. The pin joint in the center allows for rotation, but normal forces in the x and y directions prevent motion in all directions. The fixed connection on the right has a normal force preventing motion in all directions and a reaction moment preventing rotation.
     • Friction Forces: Objects in direct contact with the body can also exert friction forces on the body, which will resist the two bodies sliding against one another. These forces will always be perpendicular to the surfaces in contact. Friction is the subject of an entire chapter in this book, but for simple scenarios, we usually assume rough or smooth surfaces.
       • For smooth surfaces, we assume that there is no friction force.
       • For rough surfaces, we assume that the bodies will not slide relative to one another, no matter what. In this case, the friction force is always just large enough to prevent this sliding.
         
         

         For a smooth surface, we assume only a normal force perpendicular to the surface. For a rough surface, we assume normal and friction forces are present.
     • Tension in Cables: Cables, wires or ropes attached to the body will exert a tension force on the body in the direction of the cable. These forces will always pull on the body, as ropes, cables and other flexible tethers cannot be used for pushing.
       
       

       The tension force in cables always acts along the direction of the cable and will always be a pulling force.
     • The above forces are the most common, but other forces such as pressure from fluids, spring forces and magnetic forces may exist and may act on the body.
3. Once the forces are identified and added to the free body diagram, the last step is to label any key dimensions and angles on the diagram.

Source: Engineering Mechanics, Jacob Moore, et al. http://wwhttp://mechanicsmap.psu.edu/websites/1_mechanics_basics/1-6_free_body_diagrams/free_body_diagrams.htmlw.oercommons.org/courses/mechanics-map-open-mechanics-textbook/view

The car shown below is moving and then slams on the brakes, locking up all four wheels. The distance between the two wheels is 8 feet, and the center of mass is 3 feet behind and 2.5 feet above the point of contact between the front wheel and the ground. Draw a free-body diagram of the car as it comes to a stop.

Source: Engineering Mechanics, Jacob Moore et al., http://mechanicsmap.psu.edu/websites/1_mechanics_basics/1-6_free_body_diagrams/pdf/P3.pdf

Image adapted. Source: Engineering Mechanics, Jacob Moore, et al. http://mechanicsmap.psu.edu/websites/4_statically_equivalent_systems/4-1_statically_equivalent_systems/images/equivalentexample.png

Two equally sized barrels are being transported in a hand truck as shown below. Draw a free-body diagram of each of the two barrels.

Source: Engineering Mechanics, Jacob Moore et al., http://mechanicsmap.psu.edu/websites/1_mechanics_basics/1-6_free_body_diagrams/pdf/P2.pdf

For a rigid body in static equilibrium, that is, a non-deformable body where forces are not concurrent, the sum of both the forces and the moments acting on the body must be equal to zero. The addition of moments (as opposed to particles, where we only looked at the forces) adds another set of possible equilibrium equations, allowing us to solve for more unknowns as compared to particle problems.

Moments, like forces, are vectors. This means that our vector equation needs to be broken down into scalar components before we can solve the equilibrium equations. In a two-dimensional problem, the body can only have clockwise or counterclockwise rotation (corresponding to rotations about the z-axis). This means that a rigid body in a two-dimensional problem has three possible equilibrium equations; that is, the sum of force components in the x and y directions, and the moments about the z axis. The sum of each of these will be equal to zero.

For a two-dimensional problem, we break our one vector force equation into two scalar component equations.

$$\sum\vec F=0\\\sum F_x=0\:\sum F_y=0$$

The one moment vector equation becomes a single moment scalar equation.

$$\sum\vec M=0\\\sum M_z=0$$

If we look at a three-dimensional problem, we will increase the number of possible equilibrium equations to six. There are three equilibrium equations for force, where the sum of the components in the x, y, and z directions must be equal to zero. The body may also have moments about each of the three axes. The second set of three equilibrium equations states that the sum of the moment components about the x, y, and z axes must also be equal to zero.

We break the forces into three component equations

$$\sum\vec F=0\\\sum F_x=0\:\sum F_y=0\:\sum F_z=0$$

We break the moments into three component equations

$$\sum\vec M=0\\\sum M_x=0\:\sum M_y=0\:\sum M_z=0$$

Finding the Equilibrium Equations:

As with particles, the first step in finding the equilibrium equations is to draw a free-body diagram of the body being analyzed. This diagram should show all the force vectors acting on the body. In the free body diagram, provide values for any of the known magnitudes, directions, and points of application for the force vectors and provide variable names for any unknowns (either magnitudes, directions, or distances).

Next, you will need to choose the x, y, and z axes. These axes do need to be perpendicular to one another, but they do not necessarily have to be horizontal or vertical. If you choose coordinate axes that line up with some of your force vectors, you will simplify later analysis.

Once you have chosen axes, you need to break down all of the force vectors into components along the x, y and z directions. Your first equation will be the sum of the magnitudes of the components in the x direction being equal to zero, the second equation will be the sum of the magnitudes of the components in the y direction being equal to zero, and the third (if you have a 3D problem) will be the sum of the magnitudes in the z direction being equal to zero.

Next, you will need to come up with the moment equations. To do this, you will need to choose a point to take the moments about. Any point should work, but it is usually advantageous to choose a point that will decrease the number of unknowns in the equation. Remember that any force vector that travels through a given point will exert no moment about that point. To write out the moment equations, simply sum the moments exerted by each force (adding in pure moments shown in the diagram) about the given point and the given axis (x, y, or z) and set that sum equal to zero. All moments will be about the z-axis for two-dimensional problems, though moments can be about x, y and z axes for three-dimensional problems.

Once you have your equilibrium equations, you can solve these equations for the unknowns. The number of unknowns that you will be able to solve for will again be the number of equations that you have.

Source: Engineering Mechanics, Jacob Moore, et al. Mechanics Map – Equilibrium Analysis for a Rigid Body

The car below has a mass of 1500 lbs with the center of mass 4 ft behind the front wheels of the car. What are the normal forces on the front and the back wheels of the car?

Source: Engineering Mechanics, Jacob Moore et al., P1.pdf

While sitting in a chair, a person exerts the forces in the diagram below. Determine all forces acting on the chair at points A and B. (Assume A is frictionless and B is a rough surface).

Source: Engineering Mechanics, Jacob Moore et al., P5.pdf

Dry friction is the force that opposes one solid surface sliding across another solid surface. Dry friction always opposes the surfaces sliding relative to one another and can have the effect of either opposing motion or causing motion in bodies.

The most commonly used model for dry friction is coulomb friction. This type of friction can further be broken down into static friction and kinetic friction. These two types of friction are illustrated in the diagram below. First imagine a box sitting on a surface. A pushing force is applied parallel to the surface and is constantly being increased. A gravitational force, a normal force, and a frictional force are also acting on the box.

As the pushing force increases, the static friction force will be equal in magnitude and opposite in direction until the point of impending motion. Beyond this point, the box will begin to slip as the pushing force is greater in magnitude than the kinetic friction force opposing motion.

Static friction occurs prior to the box slipping and moving. In this region, the friction force will be equal in magnitude and opposite in direction to the pushing force itself. As the magnitude of the pushing force increases, so does the magnitude of the friction force.

If the magnitude of the pushing force continues to rise, eventually the box will begin to slip. As the box begins to slip, the type of friction opposing the motion of the box changes from static friction to what is called kinetic friction. The point just before the box slips is known as impending motion. This can also be thought of as the maximum static friction force before slipping. The magnitude of the maximum static friction force is equal to the static coefficient of friction times the normal force existing between the box and the surface. This coefficient of friction is a property that depends on both materials and can usually be looked up in tables.

Kinetic friction occurs beyond the point of impending motion when the box is sliding. With kinetic friction, the magnitude of the friction force opposing motion will be equal to the kinetic coefficient of friction times the normal force between the box and the surface. The kinetic coefficient of friction also depends upon the two materials in contact, but will almost always be less than the static coefficient of friction.

Source: Engineering Mechanics, Jacob Moore, et al. http://mechanicsmap.psu.edu/websites/7_friction/7-1_dry_friction/dryfriction.html

Imagine a box sitting on a rough surface as shown in the figure below. Now, imagine that we start pushing on the side of the box. Initially, the friction force will resist the pushing force, and the box will sit still. As we increase the force pushing the box, however, one of two things will occur.

1. The pushing force will exceed the maximum static friction force, and the box will begin to slide across the surface (slipping).
2. Or, the pushing force and the friction force will create a strong enough couple that the box will rotate and fall on its side (tipping).

As the pushing force increases on the box, it will either begin to slide along the surface (slipping) or it will begin to rotate (tipping).

When we look at cases where either slipping or tipping may occur, we are usually interested in finding which of the two options will occur first. To determine this, we usually determine both the pushing force necessary to make the body move and the pushing force necessary to make the body tip over. Whichever option requires less force is the option that will occur first.

Determining the Force Required to Make an Object “Slip”:

A body will slide across a surface if the pushing force exceeds the maximum static friction force that can exist between the two surfaces in contact. As in all dry friction problems, this limit to the friction force is equal to the static coefficient of friction times the normal force between the body. If the pushing force exceeds this value, then the body will slip.

If the pushing force exceeds the maximum static friction force (μₛ * Fₙ) then the body will begin to slide.

Determining the Force Required to Make an Object “Tip”:

The normal forces supporting bodies are distributed forces. These forces will not only prevent the body from accelerating into the ground due to gravitational forces, but they can also redistribute themselves to prevent a body from rotating when forces cause a moment to act on the body. This redistribution will result in the equivalent point load for the normal force shifting to one side or the other. A body will tip over when the normal force can no longer redistribute itself any further to resist the moment exerted by other forces (such as the pushing force and the friction force).

At rest (A), the normal force is a uniformly distributed force on the bottom of the body. As a pushing force is applied (B), the distributed normal force is redistributed, moving the equivalent point load to the right. This creates a couple between the gravity force and the normal force that will counter the couple exerted by the pushing force and the friction force. If the pushing force becomes large enough (C), the couple exerted by the gravitational force and the normal force will be unable to counter the couple exerted by the pushing force and the friction force.

The easiest way to think about the shifting normal force and tipping is to imagine the equivalent point load of the distributed normal force. As we push or pull on the body, the normal force will shift to the left or right. This normal force and the gravitational force create a couple that exerts a moment. This moment will be countering the moment exerted by the couple formed by the pushing force and the friction force.

Because the normal force is the direct result of physical contact, we cannot shift the normal force beyond the surfaces in contact (aka. the edge of the box). If countering the moment exerted by the pushing force and the friction force requires shifting the normal force beyond the edge of the box, then the normal force and the gravity force will not be able to counter the moment, and as a result, the box will begin to rotate (aka tip over).

The body will tip when the moment exerted by the pushing and friction forces exceeds the moment exerted by the gravity and normal forces. For an impending motion, the normal force will be acting at the very edge of the body.

Source: Engineering Mechanics, Jacob Moore, et al. http://mechanicsmap.psu.edu/websites/7_friction/7-2_slipping_vs_tipping/slippingvstipping.html

A 500 lb box is sitting on a concrete floor. If the static coefficient of friction is .7 and the kinetic coefficient of friction is .6:

• What is the friction force if the pulling force is 150 lbs?
• What pulling force would be required to get the box moving?
• What is the minimum force required to keep the box moving once it has started moving?

Source: Engineering Mechanics, Jacob Moore, et al. http://mechanicsmap.psu.edu/websites/7_friction/7-1_dry_friction/pdf/DryFriction_WorkedExample1.pdf

A 30 lb sled is being pulled up an icy incline of 25 degrees. If the static coefficient of friction between the ice and the sled is .4 and the kinetic coefficient of friction is .3, what is the required pulling force needed to keep the sled moving at a constant rate?

Source: Engineering Mechanics, Jacob Moore, et al. http://mechanicsmap.psu.edu/websites/7_friction/7-1_dry_friction/pdf/DryFriction_WorkedExample2.pdf

A plastic box is sitting on a steel beam. One end of the steel beam is slowly raised, increasing the angle of the surface until the box begins to slip. If the box begins to slip when the beam is at an angle of 41 degrees, what is the static coefficient of friction between the steel beam and the plastic box?

Source: Engineering Mechanics, Jacob Moore, et al. http://mechanicsmap.psu.edu/websites/7_friction/7-1_dry_friction/pdf/DryFriction_WorkedExample3.pdf

The box shown below is pushed as shown. If we keep increasing the pushing force, will the box first begin to slide or will it tip over?

Source: Engineering Mechanics, Jacob Moore, et al. http://mechanicsmap.psu.edu/websites/7_friction/7-2_slipping_vs_tipping/pdf/TippingVsSlipping_WorkedExample1.pdf

1. Problem

Billy (160 lbs), Bobby (180 lbs), and Joe (145 lbs) are walking across a small bridge with a length of 11 feet. Both sides of the bridge are supported by rollers. Billy is 2 feet along the bridge whereas Joe is 9 feet along the bridge. If the maximum force that the left side of the bridge can withstand without failing is 225 lbs, where along the bridge can Bobby stand?

Real-life scenario:

Source: https://www.flickr.com/photos/chumlee/48306801162/

2.Draw

Sketch:

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• r_Bi = 2 ft
• r_J = 9 ft
• r_B = 11 ft
• F_Bi = 160 lb
• F_J = 145 lb
• F_Bo = 180 lb
• A_y = 225 lb (since this is the maximum force without failure)

Note: Since the mass of the bridge was not given, we assume it is negligible and ignore it for this question.

Unknown:

• r_Bo

4. Approach

Use equilibrium equations ( , ) . Use sum of forces in y to find By; use sum of moments to find where Bobby can stand. Solve for x.

5. Analysis

$$ \sum F_y=0=-F_{Bi}-F_{Bo}-F_J+A_y+B_y\\\\B_y=F_{Bi}+F_{Bo}+F_J-A_y\\\\B_y=160 lb+180 lb+145 lb-225 lb\\\\\\B_y=260 lb$$

$$\sum M_A=0=-(F_{Bi})(r_{Bi})-(F_{Bo})(r_{Bo})-(F_{J})(r_{J})+(B_{y})(r_{B})\\r_{bo}=\frac{-(F_{Bi})(r_{Bi})-(F_{J})(r_{J})+(B_{y})(r_{B})}{F_{Bo}}\\r_{Bo}=\frac{-(160 lb)(2 ft)-(145 lb)(9 ft)+(260 lb)(11 ft)}{180 lb}$$

$$\underline{r_{Bo}=6.86 ft}$$

6. Review

Bobby can stand anywhere from 6.8611 ft – 11 ft from A with no problems. If Bobby were to stand between 0 ft and 6.811 ft, the left side of the bridge would fail.

1. Problem

A box is sitting on an inclined plane (θ = 15°) and is being pushed down the plane with a force of 20 N. Draw the free-body diagram for the box while it is in static equilibrium.Source:https://www.omnicalculator.com/physics/normal-force

2. Draw

3. Knowns and Unknowns

Knowns:

• θ = 15°
• F_A

Unknown:

• Free-body diagram of box

4. Approach

Draw the box, then draw all forces acting on it

5. Analysis

6. Review

All forces acting upon the box are drawn, including weight/gravitational force, normal force, friction, and applied forces.

1. Problem

Source: https://static.thenounproject.com/png/2745226-200.png

A box is being pushed along level ground with a force of 150 N at an angle of 30° with the horizontal. The mass of the box is 12 kg.

a) What is the normal force between the box and the floor?

b) What is the coefficient of friction between the box and the floor?

2. Draw

Sketch:

Free Body Diagram:

3. Knowns and Unknowns

Knowns:

• F_A = 150 N
• θ = 30°
• m = 12 kg

Unknowns:

• F_N
• μ

4. Approach

Use equilibrium equations ( , ), SOH CAH TOA, friction equation

5. Analysis

Part a:

Find F_g:

$$Fg=m\cdot g\\Fg=(12kg)(9.81m/s^2)\\\\Fg=117.72N$$

Find F_N using equilibrium equations:

$$\sum Fy=0=F_N-F_g-F_A\sin 30^{\circ}\\0=F_N-117.72N-150N\cdot \sin 30^{\circ}\\F_N=117.72+150N\cdot\sin 30^{\circ}\\\\\underline{F_N=192.7 N}$$

Part b:

Find two equations for F_f, set them equal to each other, and solve for μ

$$F_f=150N\cdot\cos 30^{\circ}\\F_f=\mu\cdot F_N\\150N\cdot\cos 30^{\circ}=\mu\cdot F_N$$

$$\mu=\frac{150N\cdot\cos 30^{\circ}}{F_N}\\\mu=\frac{150N\cdot\cos 30^{\circ}}{192.72N}=0.67405$$

$$\underline{\mu=0.67}$$

6. Review

F_N, F_g, and the y component of F_A are the only forces in the y direction so it makes sense that they need to equal zero for the equilibrium equations. F_N is the only positive force in the y direction, so it makes sense that it equals the magnitude of the other two put together.

The coefficient found between the box and the floor is reasonable as it is less than 1, and it’s reasonable for a box on the floor. For example, if the box was wood and the floor was wood, the coefficient of static friction would be anywhere from 0.5-0.7, so having a coefficient of friction being equal to 0.67 makes sense.

1. Problem

2. Draw

Sketch:

Free-body diagram (box):

Free-body diagram (distributed load):

3. Knowns and Unknowns

Knowns:

• Mass of brick (m) = 5kg
• Coefficient of friction (μ₁) = 0.49
• Acceleration due to gravity (g) = 9.81m/s2
• Length of the hand (L) = 16 cm

Unknowns:

• Applied force (F_A)
• Intensity (w)

4. Approach

Use equilibrium equations ( , ), equations for F_g and F_f (see below).

$$F_g=m g\\F_f=\mu F_N$$

5. Analysis

Part a:

$$F_g=m\cdot g\\F_g=5kg\cdot 9.81m/s^2\\F_g=49.05N$$

$$\sum F_y=0=-F_g+F_f\\F_f=F_g\\F_f=49.05N$$

$$F_f=\mu_1  F_N\\F_N=\frac{F_f}{\mu_1}\\F_N=\frac{49.05N}{0.49}$$

$$\underline{F_N=100.1N}$$

Part b:

$$\sum F_x=0=F_N-F_A\\F_A=F_N\\F_A=100.1N$$

$$w=\frac{F_A}{L}\\w=\frac{100.1N}{(16cm\times\frac{1m}{100cm})}$$

$$\underline{w=625.64N/m}$$

6. Review

It makes sense that the applied force is larger than the gravitational force. It also makes sense that the normal and applied forces are equal, since they are the only forces in the x direction (same goes for the friction and gravitational forces).

1. Problem

2. Draw

Sketch:

3. Knowns and Unknowns

• θ_A = 20°
• θ_B = 60°
• θ_C = 15°
• m = 20 kg
• F_A = 200 N
• F_B = 150 N

Unknown:

• F_f

4. Approach

Draw the ball, then add forces. Use equilibrium equations ( , ) to find the friction force.

5. Analysis

Part a:

Part b:

Step 1: Find F_G

$$F_G=m g\\F_G=20kg\cdot 9.81m/s^2\\F_G=196.2N$$

Step 2: Find the x-component of F_G

$$F_{GX}=F_G\sin(15^{\circ})\\F_{GX}=196.2N\cdot\sin(15^{\circ})\\F_{GX}=50.78N$$

Step 3: Find the x-component of F_A

$$\cos(20^{\circ})=\frac{F_AX}{F_{A}}\\F_{AX}=F_A\cdot(cos(20^{\circ}))\\F_{AX}=200N\cdot(\cos(20^{\circ}))\\F_{AX}=187.938N$$

Step 4: Find the x-component of F_B

$$F_{BX}=F_B\cdot(\cos(60^{\circ}))\\F_{BX}=150N\cdot(\cos(60^{\circ}))\\F_{BX}=75N$$

Step 5: Sum forces in the x-direction to find the frictional force

$$\sum F_x=0=-F_f+F_{BX}-F_{AX}+F_{GX}\\F_f=-F_{AX}+F_{BX}+F_{GX}\\F_f=-187.938N+75N+50.78N$$

$$\underline{F_f=-62.158N}$$

Because the frictional force is negative, that means the frictional force actually acts in the opposite direction, so the friction is keeping the ball from going up the plane.

6. Review

The units of F_f are newtons, which makes sense because it is a force. It also makes sense that F_Ax is larger than F_Gx and F_Bx.

1. Problem

2. Draw

3. Knowns and Unknowns

Knowns:

• m = 500 kg
• P = 5000 N
• 9 = 9.81 m/s^2

Unknowns:

• w
• N
• F_F
• µ

4. Approach

• Calculate w using the formula w = mg
• Calculate N using
• Calculate F_F using
• Calculate µ using
• Compare µ to µ_s

5.  Analysis

Calculating w

w = 4905 N

Calculating N 

N = 4905 N

Therefore, N_A = N_B = 4905N/2 = 2452.5N

Finding F_F

F_F  = 5000 N

Finding µ

µ = 1.019 

Comparing µ and µ_s

1.019> 0.72

µ > µ_s

Thus, the trailer is is in motion, i.e. sliding along the surface.

6. Review

From initial inspection of the FBD, it is clear that w (acting downwards) is opposite to the normal forces (acting upwards). The sum of the normal forces would not exceed w.

F_F should be equal and opposite to P.

1. Problem 

 

1.8 metres tall Spiderman with strong spider-like webs is attempting to tip over a 10 m tall and 2.5 m wide shipping container sitting upright. Given that the coefficient of static friction between the bottom of the shipping container and the ground is 0.2, and the mass of the container is 2000 kg, how much tension must Spiderman impart on his web in order to barely tip over the container towards him if he is 20 metres away from the bottom of the container?

 

Source:https://commons.wikimedia.org/wiki/File:Container_%E3%80%90_22G1_%E3%80%91_GVTU_201551(1)_%E3%80%90_Container_pictures_taken_in_Japan_%E3%80%91.jpg

2. Draw

3. Knowns and Unknowns 

Knowns:

• m = 2000 kg
• h_c = 10 m
• h_s = 1.8 m
• d = 20 m
• w =2.5
• μ  = 0.2 

Unknowns:

• T_x
• T_y 

4. Approach

Componentizing forces in the x-y direction and using the summation of moments and forces to be equal to zero.

5. Analysis

Because mass is given force by gravity,

Since the tension applied by Spider-Man is a force with both x and y components, we would need angle

Analyzing the force diagram:

Looking at the force diagram for the shipping container, it is understood that if the container were to tip towards Spiderman, the sum of moments on point B would equal zero. It is noticeable that among the forces on the container( frictional force, normal force, force of gravity, and tension) only force of gravity and x component of tension contribute to the moment at point B. Therefore, the following equation for the moment at B is derived.

thus,   

Also

The required tension to tip the container is 2,651 N

6. Review

We can ensure that the container will tip, not slip, by examining the role of frictional force in the net force. If the maximum frictional force is greater than the pulling force, the container would tip. To verify this, we analyze the below given force diagram to find the frictional force.

now

Because F_f > T_x, the container will tip.

1. Problem 

2. Draw 

3. Knowns and Unknowns 

Knowns:

• m= 70 kg
• P = 300N
• θ = 30°

Unknowns:

• F_g
• F_F

4. Approach 

• Find F_g by using F_g = m⨯g.
• Figuring out forces in x and y components
• Apply equilibrium equation to figure out the rest of the forces.

5. Analysis 

Finding F_g

Forces in x and y components

Since P, F_F, and N are in the direction of either x or y, only F_g  has a x and y components.

Equilibrium equation in x-axis  

∑F_x = 300N – F_F− 686.7N sin(30°) = 0

F_F=-43.35N

Since this value is negative it just means that the drawing has the force of friction acting in the opposite direction.

6. Review 

The answer obtained makes sense as the force of friction should be small and be working against the cart sliding down the incline. The force of gravity in the x component is only slightly larger than that of the person pulling the cart so the force of friction being only 43.35N makes sense.

1. Problem

2. Sketch

3. Knows and Unknowns:

Knowns:

• F_A = 10N
• D₂ = 0.2m
• D₃ = 0.1m
• L = 0.8m
• H = 0.15m
• m = 4kg

Unknowns:

• D₁ = ?
• F_g = ?
• F_fa = ?
• F_fb = ?
• N_A = ?
• N_B = ?
• μ = ?

4. Approach:

Define an origin point to measure all forces from to simplify the problem.
Determine D₁ and the coordinates of all forces measured from the origin O.
Find F_g.
Use sum of moment around A to find N_B.
Use sum of forces in the Y direction to find N_A.
Use sum of forces in the X direction to find μ.

5. Analysis:
Defining the origin as the bottom left corner of the gaming platform:
$$ N_A\ is\ measured\ D_3 + \frac{1}{2} D_2\ from\ the\ origin. \\ N_B\ is\ measured\ D_3 + D_2 + D_1 + \frac{1}{2} D_2 \ from\ the\ origin. \\ F_g\ is\ measured\ D_3 + D_2 + \frac{1}{2} D_1 \ from\ the\ origin.$$
To find D₁:
$$ L = 2D_3 + 2D_2 + D_1 \\ D_1 = L – 2D_3 – 2D_2 \\ D_1 = 0.8m – 2 \cdot 0.1m – 2 \cdot 0.2m \\ D_1 = 0.2m$$
To find F_g:
$$ F_g = mg = 4kg \cdot 9.81 m/s^2 = 39.24N$$
The coordinates of these forces can now be shown as:
$$N_A\ is\ at\ [0.2, 0]m \\ N_B\ is\ at\ [0.6, 0]m \\ F_g\ is\ at\ [0.4, 0]m \\ F_A\ is\ at\ [0, 0.15]m$$

The sum of moments around point A can now be done to find N_B. The distance between each force and point A is the difference of their coordinates.
$$\sum M_A = 0 = -F_g  \cdot (0.4m – 0.2m) + N_B \cdot (0.6m – 0.2m) – F_A (0.15m – 0m) \\ N_B = \frac{F_g \cdot 0.2m + F_A \cdot 0.15m}{0.4m} \\ N_B = \frac{39.24N \cdot 0.2m + 10N \cdot 0.15m}{0.4m} \\ N_B = 23.37 N$$
The Sum of forces in the Y direction can be used to find N_A now:
$$\sum F_y = 0 = -F_g + N_A + N_B \\ N_A = F_g – N_B = 39.24N – 23.37N \\ N_A = 15.87N$$
Before the sum of forces in the X is done, it must be noted that F_fa and F_fb are equivalent to μN_A and μN_B , respectively.
$$\sum F_x = 0 = F_A – F_{fa} – F_{fb} = F_A – \mu \cdot N_A – \mu \cdot N_B \\ -F_A = -\mu \cdot N_A – \mu \cdot N_B \\ \frac{-F_A}{\mu} = -N_A – N_B \\ \frac{1}{\mu} = \frac{-N_A – N_B}{-F_A} \\ \mu = \frac{-F_A}{-N_A – N_B} \\ \mu = \frac{-10N}{-15.87N – 23.37N} \\ \mu = 0.25$$

6. Review:

To review this solution, the value for N_A can be solved for by taking the moment around point B.
$$ \sum M_B = 0 = F_g \cdot (0.6m – 0.4m) – N_A \cdot (0.6m – 0.2m) – F_A \cdot (0.15m) \\ N_A = \frac{(F_g \cdot 0.2m ) – (F_A \cdot 0.15m)}{0.4m} \\ N_A = \frac{(39.24N \cdot 0.2m) – (10N \cdot 0.15)}{0.15m}{0.4m} \\ N_A = 15.87N$$
The value found is the same value originally found, providing evidence that there was no error in the calculation.

A two-force member is a body that has forces (and only forces, no moments) acting on it in only two locations. In order to have a two-force member in static equilibrium, the net force at each location must be equal, opposite, and collinear. This will result in both force members being in either tension or compression as shown in the diagram below.

The forces acting on two force members need to be equal, opposite, and collinear for the body to be in equilibrium.

Imagine a beam where forces are only exerted at each end of the beam (a two-force member). The body has some non-zero force acting at one end of the beam, which we can draw as a force vector. If this body is in equilibrium, then we know two things: 1) the sum of the forces must be equal to zero, and 2) the sum of the moments must be equal to zero.

In order to have the sum of the forces equal to zero, the force vector on the other side of the beam must be equal in magnitude and opposite in direction. This is the only way to ensure that the sum of the forces is equal to zero with only two forces.

In order to have the sum of the moments equal to zero, the forces must be collinear. If the forces were not collinear, then the two equal and opposite forces would form a couple. This couple would exert a moment on the beam when there are no other moments to counteract the couple. Because the moment exerted by the two forces must be equal to zero, the perpendicular distance between the forces (d) must be equal to zero. The only way to achieve this is to have the forces be collinear.

In order to have the sum of the moments be equal to zero, the forces acting on two force members must always be collinear, acting along the line connecting the two points where forces are applied.

Source: Engineering Mechanics, Jacob Moore, et al. http://mechanicsmap.psu.edu/websites/5_structures/5-1_structures/structures.html

Adapted from image by ToddC4176 CC-BY-SA 3.0

A truss is an engineering structure that is made entirely of two force members. In addition, statically determinate trusses (trusses that can be analyzed completely using the equilibrium equations), must be independently rigid. This means that if the truss was separated from its connection points, no one part would be able to move independently with respect to the rest of the truss.

When we talk about analyzing a truss, we are usually looking to identify not only the external forces acting on the truss structure, but also the forces acting on each member internally in the truss. Because each member of the truss is a two-force member, we simply need to identify the magnitude of the force on each member and determine if each member is in tension or compression.

To determine these unknowns, we have two methods available: the method of joints and the method of sections. Both will give the same results, but each through a different process.

The method of joints focuses on the joints, or the connection points where the members come together. We assume we have a pin at each of these points that we model as a particle, we draw out the free body diagram for each pin, and then write out the equilibrium equations for each pin. This will result in a large number of equilibrium equations that we can use to solve for a large number of unknown forces.

The method of sections involves pretending to split the truss into two or more different sections and then analyzing each section as a separate rigid body in equilibrium. In this method, we determine the appropriate sections, draw free-body diagrams for each section, and then write out the equilibrium equations for each section.

The method of joints is usually the easiest and fastest method for solving for all the unknown forces in a truss. The method of sections, on the other hand, is better suited to targeting and solving for the forces in just a few members without having to solve for all the unknowns. In addition, these methods can be combined if needed to best suit the goals of the problem solver.

Source: Engineering Mechanics, Jacob Moore, et al. https://mechanicsmap.psu.edu/websites/5_structures/5-1_structures/structures.html

Treat the entire truss as a rigid body and solve for the reaction forces supporting the truss structure.

Source: Engineering Mechanics, Jacob Moore, et al. https://mechanicsmap.psu.edu/websites/5_structures/5-3_method_of_joints/methodofjoints.html 

Source: Internal Forces in Beams and Frames, Libretexts. https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.05%3A_Internal_Forces_in_Plane_Trusses

The method of joints is a process used to solve for the unknown forces acting on members of a truss. The method centers on the joints or connection points between the members, and it is usually the fastest and easiest way to solve for all the unknown forces in a truss structure.

Using This Method:

The process used in the method of joints is outlined below:

1. In the beginning, it is usually useful to label the members and the joints in your truss. This will help you keep everything organized and consistent in later analysis. In this book, the members will be labelled with letters, and the joints will be labelled with numbers.
   
   

   The first step in the method of joints is to label each joint and each member.
2. Treating the entire truss structure as a rigid body, draw a free body diagram, write out the equilibrium equations, and solve for the external reacting forces acting on the truss structure. This analysis should not differ from the analysis of a single rigid body.
   
   

   Treat the entire truss as a rigid body and solve for the reaction forces supporting the truss structure.
3. Assume there is a pin or some other small amount of material at each of the connection points between the members. Next, you will draw a free-body diagram for each connection point. Remember to include:
   • Any external reaction or load forces that may be acting at that joint.
   • A normal force for each of the two force members connected to that joint. Remember that for a two-force member, the force will be acting along the line between the two connection points on the member. We will also need to guess if it will be a tensile or a compressive force. An incorrect guess now, though, will simply lead to a negative solution later on. A common strategy then is to assume all forces are tensile, then later in the solution, any positive forces will be tensile forces, and any negative forces will be compressive forces.
   • Label each force in the diagram. Include any known magnitudes and directions, and provide variable names for each unknown.
     
     

     Drawing a free body diagram of each joint, we draw in the known forces as well as tensile forces from each two force member.
4. Write out the equilibrium equations for each of the joints. You should treat the joints as particles, so there will be force equations but no moment equations. This should give you a large number of equations.
   • The sum of the forces in the x direction will be zero, and the sum of the forces in the y direction will be zero for each of the joints.$$\sum\vec F=0\\\sum F_x=0\:\sum F_y=0$$
5. Finally, solve the equilibrium equations for the unknowns. You can do this algebraically, solving for one variable at a time, or you can use matrix equations to solve for everything at once. If you assumed that all forces were tensile earlier, remember that negative answers indicate compressive forces in the members.

Source: Engineering Mechanics, Jacob Moore, et al. https://mechanicsmap.psu.edu/websites/5_structures/5-3_method_of_joints/methodofjoints.html 

Find the force acting in each of the members in the truss bridge shown below. Remember to specify if each member is in tension or compression.

Solution:

Source: Engineering Mechanics, Jacob Moore, et al. MethodOfJoints_WorkedExample1.pdf

Source: Internal Forces in Beams and Frames, Libretexts. https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.05%3A_Internal_Forces_in_Plane_Trusses

The method of sections is a process used to solve for the unknown forces acting on members of a truss. The method involves breaking the truss down into individual sections and analyzing each section as a separate rigid body. The method of sections is usually the fastest and easiest way to determine the unknown forces acting in a specific member of the truss.

Using This Method:

The process used in the method of sections is outlined below:

1. In the beginning, it is usually useful to label the members in your truss. This will help you keep everything organized and consistent in later analysis. In this book, the members will be labelled with letters.
   
   

   The first step in the method of sections is to label each member.
2. Treating the entire truss structure as a rigid body, draw a free body diagram, write out the equilibrium equations, and solve for the external reacting forces acting on the truss structure. This analysis should not differ from the analysis of a single rigid body.
   
   

   Treat the entire truss as a rigid body and solve for the reaction forces supporting the truss structure.
3. Next, you will imagine cutting your truss into two separate sections. The cut should travel through the member that you are trying to solve for the forces in, and should cut through as few members as possible (The cut does not need to be a straight line).
   
   

   Next, you will imagine cutting the truss into two parts. If you want to find the forces in a specific member, be sure to cut through that member. It also makes things easier if you cut through as few members as possible.
4. Next, you will draw a free-body diagram for either one or both sections that you created. Be sure to include all the forces acting on each section.
   • Any external reaction or load forces that may be acting on the section.
   • An internal force in each member that was cut when splitting the truss into sections. Remember that for a two-force member, the force will be acting along the line between the two connection points on the member. We will also need to guess if it will be a tensile or a compressive force. An incorrect guess now, though, will simply lead to a negative solution later on. A common strategy then is to assume all forces are tensile, then later in the solution, any positive forces will be tensile forces, and any negative forces will be compressive forces.
   • Label each force in the diagram. Include any known magnitudes and directions, and provide variable names for each unknown.
     
     

     Next, draw a free-body diagram of one or both halves of the truss. Add the known forces, as well as the unknown tensile forces for each member that you cut.
5. Write out the equilibrium equations for each section for which you drew a free body diagram. These will be extended bodies, so you will need to write out the force and the moment equations.
   • You will have three possible equations for each section: two force equations and one moment equation.$$\sum\vec F=0\; \; \sum\vec M=0\\\sum F_x=0\; \; \sum F_y=0\; \; \sum M_z=0$$
6. Finally, solve the equilibrium equations for the unknowns. You can do this algebraically, solving for one variable at a time, or you can use matrix equations to solve for everything at once. If you assumed that all forces were tensile earlier, remember that negative answers indicate compressive forces in the members.

Source: Engineering Mechanics, Jacob Moore, et al. https://mechanicsmap.psu.edu/websites/5_structures/5-4_method_of_sections/methodofsections.html 

Find the forces acting on members BD and CE. Be sure to indicate if the forces are tensile or compressive.

Source: Engineering Mechanics, Jacob Moore, et al. https://mechanicsmap.psu.edu/websites/5_structures/5-4_method_of_sections/pdf/MethodOfSections_WorkedExample1.pdf

Find the forces acting on members AC, BC, and BD of the truss. Be sure to indicate if the forces are tensile or compressive.

If we make a cut in the top section, we don’t need to solve for the reaction forces.

Source: Engineering Mechanics, Jacob Moore, et al.  https://mechanicsmap.psu.edu/websites/5_structures/5-4_method_of_sections/pdf/MethodOfSections_WorkedExample2.pdf

1. Problem

A flower cart at a local garden center is being pushed with a force of 500N at joint G. Its back wheels (A) are locked, so it is not moving. There is 1 meter of space between each of the four shelves in height, and each shelf is four meters long.

a) Calculate the reaction forces of the locked wheels and the unlocked wheels.

b) Calculate the load carried by F_CG, and whether it is in tension or compression

Which method did you use, joints or sections? Which is faster for the style of questions? How would your strategy change if you were calculating the load in each member?

Source: https://flic.kr/p/txjSpP

2. Draw

Sketch:

Free-body Diagram:

3. Knowns and Unknowns

Knowns:

• P = 500 N
• Width = 4 m
• Total height = 3 m

Unknowns:

• R_Ax
• R_Ay
• R_By
• F_CG

4. Approach

Part a: Determine reaction forces using equilibrium equations

Part b: Calculate F_CG using the method of sections. Make a cut, then solve internal forces using equilibrium equations

5. Analysis

Part a:

Solving for R_Ax:

$$\sum F_x=0=P+R_{Ax}\\R_{Ax}=-P\\R_{Ax}=-500N$$

Solving for R_By:

$$\sum M_A=0=(r_{BA}\cdot R_{By})-(r_{GA}\cdot P)\\r_{BA}\cdot R_{By}=r_{GA}\cdot P\\R_{By}=\frac{r_{GA}\cdot P}{r_{BA}}\\R_{By}=\frac{2m\cdot 500N}{4m}\\R_{By}=250N$$

Solving for R_Ay:

$$\sum F_y=0=R_{By}+R_{Ay}\\R_{Ay}=-R_{By}\\R_{Ay}=-250N$$

The answers we got that were negative numbers mean that the direction of the vector is drawn wrong on our original diagram (in reference to our coordinate frame). This makes sense as R_Ay and R_By are the only external forces in the y direction, so they have to cancel each other for the equilibrium equations to be true (Same for the vectors in the x direction.) Therefore, one of them should have a negative direction. We will leave this answer as is for now, but the next time we draw the system, we will change the direction of the arrow.

$$ \underline{R_{Ax}=-500N,\; R_{Ay}=-250N, \; R_{By}=250N}$$

Part b:

Firstly, we redraw the diagram, changing the direction of the R_Ay and R_Ax. Then, since we are using the method of sections, we make a cut so that the member F_CG (the one we want to find) is cut.

Now we redraw, choosing one of the pieces from the cut. Here, the top half is chosen, but you could also choose the bottom half and get the right answer. The only reason the top half was chosen here is because there are fewer external forces to consider for the top.

Solve for the member we are looking for:

 $$\sum F_x=0=P+\frac{4}{\sqrt{17}}F_{CG}\\\frac{4}{\sqrt{17}}F_{CG}=-P\\F_{CG}=-P(\frac{\sqrt{17}}{4})\\F_{CG}=-500N(\frac{\sqrt{17}}{4})\\F_{CG}=-515.388N$$

Again, the number we get is negative. The way we drew F_CG originally was as if the member was in tension. The negative number just means that it is actually compression, not tension.

$$ \underline{F_{CG}=515 \text{N (Compression)}}$$

Part C:

For part b, I used the method of sections, as it would be the fastest method. The method of joints would require the lower joints to be solved first, which would be a much slower process, whereas with this method, a simple cut can be made and the member’s load can be quickly solved using equilibrium equations. Had the question asked for all member loads to be solved, however, the method of joints would have been the faster approach.

6. Review

Part a:

R_Ax is equal and opposite to P so we know it is correct, and the value of R_By should also be correct as its moment about A (250 * 4m = 1000 Nm), is equal and opposite to the moment of P about A, (500 N * 2 m = 1000 Nm). as R_Ay is equal and opposite to R_By it is also correct.

Part b:

The x component of the calculated value of F_CG is equal in magnitude to P (see equation below), and it is the only cut member acting in the x direction. Therefore, it must be correct.

$$\frac{4}{\sqrt{17}}(515 N)=500 N$$

Part C:

The method of sections allows you to solve a very specific area of the system’s internal forces (the members that are cut), whereas the method of joints usually requires you to solve most, if not all, of the internal forces of the system. Therefore, the method of sections is the most efficient for finding the internal forces of specific parts of the system, whereas the method of joints is more efficient for solving the whole system.

1. Problem

2. Draw

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• All of the known values do not mean anything – we only need to know where the forces exist

Unknown:

• Which members are zero-force

4. Approach

Look at each joint and determine how many forces are in each direction. If there is only one force in a direction, that member is zero-force.

5. Analysis

Part a:

Let’s start with joint C. If we think of the forces acting in the x and y directions as shown below by the coordinate frame, we see that there are two forces acting in the y direction, and only one in the x direction. Therefore, assuming the joint is in static equilibrium, member CE is a zero-force.

If we do the same type of analysis for the other joints and remove the zero-force members, the structure now looks like this:

After one more analysis of the joints, we find one more zero-force member, as shown below.

Answer: CE, DG, and FG are zero-force members.

Part b:

Zero-force members exist to provide stability to the truss, to keep the shape rigid.

6. Review

Although the new truss (without zero-force members) looks strange, there are no joints where there’s only one force in one direction; therefore, there are no more zero-force members.

1. Problem 

A small truss bridge is over a river, and it has a bucket full of water hanging off the middle. The mass of the water bucket is 15 kg.

a. Solve for the reaction forces.

b. Use the method of joints to solve each member.

c. Draw forces and how they act.

In the real world, the truss bridge might be similar to this,

Source:https://commons.wikimedia.org/wiki/File:Railroad_Truss_bridge_over_trinity_river_near_Goodrich,_Texas.jpg.

2. Draw

3. Knowns and Unknowns

Knowns:

• F_g = 15 kg . 9.81 m/s² = 147.15 N
• r_AF = r_BD = r_CE =2 m
• r_DF = r_AB = 6 m

Unknowns:

• R_Ax
• R_Ay
• R_By
• F_EF
• F_ED
• F_BD
• F_BE
• F_CE
• F_AE
• F_AC
• F_BC
• F_AF

4. Approach 

First, use the equilibrium equations to find the reaction forces. Apply these reaction forces in the method of joints to find the force of each member.

5. Analysis 

a. Calculating the reaction forces using the equilibrium equations and the diagram.

$$\sum F_x=0=R_{Ax}\\R_{Ax}=0N$$

Solving for R_By:

$$\sum M_A=0= -(r_{AC}\cdot F_g)+(r_{AB}\cdot R_{By})\\r_{AB}\cdot R_{By}=r_{AC}\cdot  F_g $$

$$R_{By}=\frac{r_{AC}\cdot F_g}{r_{AB}}\\R_{By}=\frac{3m\cdot 147.15 N}{6m} $$

$$R_{By}=73.575 N$$

Solving for R_Ay:

$$\sum F_y=0=R_{By}+R_{Ay} -F_g\\R_{Ay}=F_g-R_{By}\\R_{By}=73.575N$$

b. F_AF, F _EF, F_DE, F_BD are zero force members because for joints F and D, there are only 2 members each, and there is no external load.

Finding the angle between AC and AE:

tan θ = r_CE/r_AC

therefore, θ = 33.7 °

Analyzing joint A:

$$\sum F_y=0 = R_{Ay}+ F_{AE}\cdot sin 33.7 \\ =\frac{-73.575 N}{sin 33.7} $$

$$ F_{AE} = -132.6 N $$

$$\sum F_x=0 = F_{AC} + F_{AE} cos 33.7\\ F_{AC}=132.6N\cdot \cos {33.7}$$

$$ F_{AC} = 110.32 N $$

Analyzing Joint C:

From the FBD,

F_AC= F_BC = 110.32 N

Also, F_g  =F_CE = 147.15 N

Analyzing Joint B

$$\sum F_y=0 = R_{By} +F_{BE}\cdot sin 33.7  $$

$$ F_{BE} = -132.6 N $$

$$\sum F_x=0 = F_{BC} -F_{BE} cos 33.7\\ F_{BC}=132.6N\cdot \cos {33.7}$$

$$ F_{BC} = 110.32 N $$

c.

6. Review 

The answers make sense because F_CEis in tension as imagined due to the weight of the bucket. Members AC and BC are under reaction forces, are equally under tension, and members AE and BE are balanced by compression. The identified zero-force members also make sense due to the rules being met.

1. Problem

2. Draw

3. Knowns and Unknowns

Knowns:

•
•
•
•
•
•
•
•

Unknowns:

•
•
•
•
•
•
•

4. Approach

Use trigonometry and equilibrium equations, as well as the method of joints, to calculate unknowns

5. Analysis

Finding total mass:

There are four 75cm rods, two rods that are cm, and one rod that is 1.5m long, all of which have a diameter of 2cm. The total mass is therefore $$ 8g/cm^3 \cdot ((4 \pi \cdot (1cm)^2\cdot (75cm) + 2\pi\cdot (1cm)^2 \cdot \\ (75 \sqrt(2)cm) + \pi  (1cm)^2 \cdot (150cm)) \\ \approx 16,641g = 16.641kg = m_{tot}$$

F_g can then be found using $$ F_g = g \cdot m_{tot} = -9.81m/s^2 \cdot 16.641kg \approx -163.25N$$

The sum of forces and moments can now be used to determine reaction forces

$$ \sum F_x = P_x – R_{Ax} \\ R_{Ax} = P \cdot cos(\theta) = 70N cos(15^\circ) \\ R_{Ax} \approx 67.6N$$

$$\sum F_y = -F_g – P_y + R_{Ay} + R_{By} = 0$$

With two unknowns, sum of moments must be considered. Summing moments around point A finds:

$$ \sum M_A = 0 = -(0.75m \cdot Pcos(\theta)) – (0.75m \cdot F_g) + (1.5m \cdot R_{By}) $$

$$ \sum M_A = 0 = -(0.75m \cdot 70N cos(15^\circ)) – (0.75m \cdot 163.25N) + (1.5m \cdot R_{By})$$

$$ R_{By} \approx 115.43N$$

Sum of Forces in Y can now be used to find R_Ay

$$ F_y = -F_g – P_y + R_{Ay} +R_{By} = 0$$

$$ R_{Ay} \approx 65.94N $$

The method of joints can now be used to determine if any zero-force members exist in the shopping cart.

Looking at Point A, it can be found that in compression and in tension. Continuing through each joint, it can be found that member CB is a zero-force member as point B only has one component in the X direction, which, in a static system, implies that the force must be zero.

6. Review

Given the high weight of the cart creating a large F_g and the applied force of 70N, the reaction forces found and the internal forces all seem to be of appropriate magnitude and direction.

1. Problem

You are designing a shelf for a friend, and it can be analyzed as a truss shown below. If you know the forces in the member CD can take a tensile force of 25 N, and GH can take a compressive force of 15 N before breaking. If w = 10 cm and h = 20 cm, what mass can be supported if it is situated exactly on point E?

2. Sketch
Draw a free-body diagram of the truss with a slice through CD, GH, and IH.

3. Knowns and Unknowns

Knowns:

• w = 0.1 m
• h = 0.2 m
• F_CD = -25 N
• F_GH = 15 N

Unknowns:

• F_HI =?
• d_HI = ?
• P =?
• m = ?

4. Approach
Use the method of sections to show internal forces and analyze as a rigid body.

5. Analysis
First, the distance d_HI can be found:
$$ d_{HI} = \sqrt{(0.1 m)^2 + (0.2 m)^2} \\ d_{HI} = 0.2236 m$$
This value can be used to find the equivalents to cosθ and sinθ:
$$ \cos{\theta} = \frac{0.1 m}{0.2236 m} \\ \sin{\theta} = \frac{0.2 m}{0.2236 m} $$
The sum of forces in the X direction can be used to determine the value of F_HI:
$$ \sum F_x = 0 = -F_{CD} + F_{GH} + F_{HI} \cdot \cos{\theta} \\ F_{HI} = \frac{(F_{CD} – F_{GH})}{\cos{\theta}} \\ F_{HI} = \frac{0.2236 m}{0.1 m} \cdot (25 N – 15 N) \\ F_{HI} = 22.36N$$
The sum of forces in the Y direction can now be used to determine the value of P:
$$ \sum F_y = 0 = -P + F_{HI} \cdot \sin{\theta} \\ P = F_{HI} \cdot \sin{\theta} \\ P = (22.36 N) \cdot \frac{0.2 m}{0.2236 m} \\ P = 20 N$$
The mass can now be simply found by dividing the force by gravity:
$$ P = mg \\ m = \frac{P}{g} \\ m = \frac{20 N}{9.81 \frac{m}{s^2}} \\ m = 2.04 kg$$

6. Review
A brief review of this question is the sign and magnitude of the result. The force generated by the mass pushes downwards with gravity, which is expected. The magnitude of the mass, while defining this shelf as one that is particularly flimsy, is a reasonable mass.

When a beam or frame is subjected to transverse loadings, the three possible internal forces that are developed are the normal or axial force, the shearing force, and the bending moment, as shown in section k of the cantilever of the figure below. To predict the behaviour of structures, the magnitudes of these forces must be known. In this chapter, the student will learn how to determine the magnitude of the shearing force and bending moment at any section of a beam or frame and how to present the computed values in a graphical form, which is referred to as the “shearing force” and the “bending moment diagrams.” Bending moment and shearing force diagrams aid immeasurably during design, as they show the maximum bending moments and shearing forces needed for sizing structural members.

Normal Force

The normal force at any section of a structure is defined as the algebraic sum of the axial forces acting on either side of the section.

Shearing Force

The shearing force (SF) is defined as the algebraic sum of all the transverse forces acting on either side of the section of a beam or a frame. The phrase “on either side” is important, as it implies that at any particular instance, the shearing force can be obtained by summing up the transverse forces on the left side of the section or on the right side of the section.

Bending Moment

The bending moment (BM) is defined as the algebraic sum of all the forces’ moments acting on either side of the section of a beam or a frame.

Source: Internal Forces in Beams and Frames, Libretexts. https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.04%3A_Internal_Forces_in_Beams_and_Frames

Axial (Normal) Force

An axial force is regarded as positive if it tends to tier the member at the section under consideration. Such a force is regarded as tensile, while the member is said to be subjected to axial tension. On the other hand, an axial force is considered negative if it tends to crush the member at the section being considered. Such force is regarded as compressive, while the member is said to be in axial compression.

Shear Force

A shear force that tends to move the left of the section upward or the right side of the section downward will be regarded as positive. Similarly, a shear force that has the tendency to move the left side of the section downward or the right side upward will be considered a negative shear force.

Bending Moment

A bending moment is considered positive if it tends to cause concavity upward (sagging). If the bending moment tends to cause concavity downward (hogging), it will be considered a negative bending moment.

Source: Internal Forces in Beams and Frames, Libretexts. https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.04%3A_Internal_Forces_in_Beams_and_Frames

Shearing Force Diagram

This is a graphical representation of the variation of the shearing force on a portion or the entire length of a beam or frame. As a convention, the shearing force diagram can be drawn above or below the x-centroidal axis of the structure, but it must be indicated if it is a positive or negative shear force.

Bending Moment Diagram

This is a graphical representation of the variation of the bending moment on a segment or the entire length of a beam or frame. As a convention, the positive bending moments are drawn above the x-centroidal axis of the structure, while the negative bending moments are drawn below the axis.

Below is a simple example of what shear and moment diagrams look like. Afterwards, the relation between the load on the beam and the diagrams will be discussed.

Source: Internal Forces in Beams and Frames, LibreTexts. https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.04%3A_Internal_Forces_in_Beams_and_Frames

For the derivation of the relations among w, V, and M, consider a simply supported beam subjected to a uniformly distributed load throughout its length, as shown in the figure below. Let the shear force and bending moment at a section located at a distance of x from the left support be V and M, respectively, and at a section x + dx be V + dV and M + dM, respectively. The total load acting through the center of the infinitesimal length is wdx.

To compute the bending moment at section x + dx, use the following:

or

  (Equation 6.1)

Equation 6.1 implies that the first derivative of the bending moment with respect to the distance is equal to the shearing force. The equation also suggests that the slope of the moment diagram at a particular point is equal to the shear force at that same point. Equation 6.1 suggests the following expression:

(Equation 6.2)

Equation 6.2 states that the change in moment equals the area under the shear diagram. Similarly, the shearing force at section x + dx is as follows:

or

  (Equation 6.3)

Equation 6.3 implies that the first derivative of the shearing force with respect to the distance is equal to the intensity of the distributed load. Equation 6.3 suggests the following expression:

(Equation 6.4)

Equation 6.4 states that the change in the shear force is equal to the area under the load diagram. Equations 6.1 and 6.3 suggest the following:

(Equation 6.5)

Equation 6.5 implies that the second derivative of the bending moment with respect to the distance is equal to the intensity of the distributed load.

Source: Internal Forces in Beams and Frames, LibreTexts. https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.04%3A_Internal_Forces_in_Beams_and_Frames

Shear Diagram

To create the shear force diagram, we will use the following process.

1. Solve for all external forces acting on the body.
2. Draw out a free body diagram of the body horizontally. Leave all distributed forces as distributed forces and do not replace them with the equivalent point load.
3. Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal shear force.
4. Starting at zero at the right side of the plot, you will move to the right, pay attention to forces in the free body diagram above. As you move right in your plot, keep steady except…
   • Jump upwards by the magnitude of the force for any point forces going up.
   • Jump downwards by the magnitude of the force for any point forces going down.
   • For any uniform distributed forces you will have a linear slope where the magnitude of the distributed force is the slope of the line (positive slopes for upwards distributed forces, negative slopes for downwards distributed forces).
   • For non-uniform distributed forces, the shape of the shear diagram plot will be the integral of the force function.
   • You can ignore any moments or horizontal forces applied to the body.

   By the time you get to the left end of the plot, you should always wind up coming back to zero. If you don’t wind up back at zero, go back and check your previous work.

To read the plot, you simply need to find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot. Positive numbers represent an upward internal shearing force to the right of the cross section and a downward force on the left, and negative numbers indicate a downward internal shearing force to the right of the cross section and an upward force on the left. A visual of these forces can be seen in the diagram to the right.

Moment Diagram

The moment diagram will plot out the internal bending moment within a horizontal beam that is subjected to multiple forces and moments perpendicular to the length of the beam. For practical purposes, this diagram is often used in the same circumstances as the shear diagram, and generally, both diagrams will be created for analysis in these scenarios.

To create the moment diagram for a shaft, we will use the following process.

1. Solve for all external forces and moments, create a free body diagram, and create the shear diagram.
2. Lined up below the shear diagram, draw a set of axes. The x-axis will represent the location (lined up with the shear diagram and free body diagram above), and the y-axis will represent the internal bending moment.
3. Starting at zero on the right side of the plot, you will move to the right, pay attention to the shear diagram and the moments in the free body diagram above. As you move right in your plot, the moment diagram will primarily be the integral of the shear diagram, except…
   • Jump upwards by the magnitude of the moment for any negative (clockwise) moments.
   • Jump downwards by the magnitude of the moment for any positive (counter-clockwise) moments.
   • You can ignore any forces in the free body diagram.

   By the time you get to the left end of the plot, you should always wind up coming back to zero. If you don’t wind up back at zero, go back and check your previous work.

To read the plot, you simply need to take the find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot. Positive internal moments would cause the beam to bow downwards (think a smile shape) negative internal moments will cause the beam to bow upwards (think a frown shape). You can also see the positive and negative internal moments in the figure to the right.

Source: Engineering Mechanics, Jacob Moore, et al.  http://mechanicsmap.psu.edu/websites/6_internal_forces/6-4_shear_moment_diagrams/shear_moment_diagrams.html

Draw the shearing force and bending moment diagrams for the cantilever beam subjected to a uniformly distributed load in its entire length, as shown in Figure 4.5a.

Answer:

Support reactions.

First, compute the reactions at the support. Since the support at B is fixed, there will possibly be three reactions at that support, namely B_y, B_x, and M_B, as shown in the free-body diagram in Figure 4.4b. Applying the conditions of equilibrium suggests the following:

Shear Force Function

Let x be the distance of an arbitrary section from the free end of the cantilever beam, as shown in Figure 4.5b. The shearing force of all the forces acting on the segment of the beam to the left of the section, as shown in Figure 4.5e, is determined as follows:

The obtained expression is valid for the entire beam. The negative sign indicates a negative shearing force, which was established from the sign convention for a shearing force. The expression also shows that the shearing force varies linearly with the length of the beam.

Shearing force diagram. Note that because the expression for the shearing force is linear, its diagram will consist of straight lines. The shearing forces at x = 0 m and x = 5 m were determined and used for plotting the shearing force diagram, as shown in Figure 4.5c. As shown in the diagram, the shearing force varies from zero at the free end of the beam to 100 kN at the fixed end. The computed vertical reaction of B_y at the support can be regarded as a check for the accuracy of the analysis and diagram.

Bending Moment Function

The expression for the bending moment at a section of a distance x from the free end of the cantilever beam is as follows:

The negative sign indicates a negative moment, which was established from the sign convention for moment. As seen in Figure 4.5f, the moment due to the distributed load tends to cause the segment of the beam on the left side of the section to exhibit an upward concavity, and that corresponds to a negative bending moment, according to the sign convention for bending moment.

Bending moment diagram. Since the function for the bending moment is parabolic, the bending moment diagram is a curve. In addition to the two principal values of bending moment at x = 0 m and at x = 5 m, the moments at other intermediate points should be determined to correctly draw the bending moment diagram. The bending moment diagram of the beam is shown in Figure 4.5d.

Source: Internal Forces in Beams and Frames, Libretexts. https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.04%3A_Internal_Forces_in_Beams_and_Frames

Example 3

Example 4

Example 5

Example 6

Source: Internal Forces in Beams and Frames, Libretexts. https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.04%3A_Internal_Forces_in_Beams_and_Frames

1. Problem

2. Draw

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• w = 220 N/m
• OA = 0.5 m
• AC = 0.2 m
• AB = 0.4 m
• L = 1.9 m

Unknowns:

• N_c
• V_c
• M_c

4. Approach

Use equilibrium equations. First, solve for reaction forces, then make a cut at C and solve for the internal forces.

5. Analysis

$$w=\frac{F}{L}\\F=wL\\F_R=220N/m\cdot 1.9m\\F_R=418N\\\sum F_X=0=B_X$$

Find reaction forces:

$$\sum M_A=0=B_y(0.4m)-F_R(0.45m)\\(0.4m)B_y=418N(0.45m)\\B_y=\frac{188.1 N\cdot m}{0.4m}\\B_y=470.25N$$

$$\sum F_y=0=-F_R+A_y+B_y\\A_y=F_R-B_y\\A_y=418N-470.25N\\A_y=-52.25N$$

The answer we got for A_y is negative, which means that the arrow should be drawn in the other direction. We will change it for our next sketch.

Make a cut at C:

Now solve for the internal forces:

$$\sum F_x=0\:\:;\:\:N_c=0\\\sum F_y=0=-A_y-V_c-(w\cdot L_A)\\V_c=-52.25N-(220N/m\cdot 0.7m)\\V_c=-206.25 N\\\sum M_c=A_y(0.2m)+M_c+(F_{Rc}\cdot 0.35m)\\M_c=52.25 N (0.2m)+(220N/m\cdot 0.7m\cdot 0.35m)\\M_c=64.35N\cdot m$$

Final FBD, showing the arrows in the correct directions:

6. Review

It makes sense that A_y and B_y are in different directions, because the resultant force Fr of the solar panel on the beam is not between A and B. It also makes sense that the moment at C is in the counterclockwise direction rather than the clockwise direction, when you think about the direction of the forces applied to the beam.

1. Problem

2. Draw

Sketch:

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• F1 = 500 N
• F2 = 300 N

Unknowns:

• A_y
• A_x
• C_y
• V_B
• M_B
• N_B

4. Approach

Shear/moment equations, EOM equations

5. Analysis

Solve for reaction forces:

(Ax, Cy)

\begin{aligned}
\sum F_{x}=0=A_{x}=0 \\
\sum M_{A}=0 &=-F_{1} \cdot 2 m-F_{2} \cdot 5.5 m+C_{y} \cdot 8 m \\
C_{y}=&\frac{F_{1} \cdot 2 m+F_{2} \cdot 5.5 m}{8m} \\
C_{y} &=\frac{500 N \cdot 2 m+300 N \cdot 5.5 m}{8 m} \\
C_{y} &=331.25 \mathrm{~N}
\end{aligned}

(Ay)

\begin{aligned}
\sum F_{y}=0 &=A_{y}+C_{y}-F_{1}-F_{2} \\
A_{y} &=F_{1}+F_{2}-C_{y} \\
A_{y} &=500 \mathrm{~N}+300 \mathrm{~N} – 331.25 \mathrm{~N} \\
A_{y} &=468.75 \mathrm{~N}
\end{aligned}

Cut 1: at B

\begin{aligned}
\sum F_{X}=0=A_{X} &+N_{B}=0 \\
& N_{B}=0 \\
\sum F_{y}=0 &=A_{y}-V_{B}-F_{1} \\
V_{B} &=A_{y}-F_{1} \\
V_{B} &=468.75 N – 500 N \\
V_{B}=-31.25 N
\end{aligned}

\begin{aligned}
\sum M_{B}=& 0=-A_{y}(4 m)+F_{1}(2 m)+M_{B} \\
& M_{B}=A_{y}(4 m)-F_{1}(2 m) \\
& M_{B}=468.75 N(4 m)-500 N(2 m) \\
M_{B} &=875 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}

Cut 2: At the point where F₁ is applied

\begin{aligned}
\sum M_{1}=0 &=-A_{y}(2 m)+M_{1}=0 \\
M_{1} &=A_{y}(2 m) \\
M_{1} &=468.75 N(2 m) \\
M_{1} &=937.5 \mathrm{~N} \cdot m
\end{aligned}

Cut 3: At the point where F₂ is applied

\begin{aligned}
\sum M_{2}=0 =-A_{y}(5.5 \mathrm{~m})+F_{1}(3.5 \mathrm{~m})+M_{2} \\
M_{2} =A_{y}(5.5 \mathrm{~m})-F_{1}(3.5 \mathrm{~m}) \\
M_{2} =468.75 \mathrm{~N}(5.5 \mathrm{~m})-500 \mathrm{~N}(3.5 \mathrm{~m}) \\
M_{2} =828.125 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}

Answer: N_B = 0, V_B = -31.25 N, M_B = 875 Nm

6. Review

The reaction forces make sense as they offset the applied forces. The shear/moment diagrams returned to zero, so they are correct too. The moment found at B is in the moment diagram; it is smaller than the maximum.

1. Problem 

2. Draw 

3. Knowns and Unknowns 

Knowns:

• F_g = 30N
• r _AB =  r _BC =  0.04 m

Unknowns:

• Reaction forces = R_Ax, R_Ay, R_Cy
• For the segment of beam AB nd BC,
• Shear = V, V₂
• Bending Moment = M, M₂
• Normal force = N, N₂

4. Approach 

Find reaction forces from equilibrium equations. Use them to find the shear and bending moment of segment AB and BC. Draw the V-M diagram from the answers.

5. Analysis 

Finding reaction forces:

$$\sum F_x=0=R_{Ax}\\R_{Ax}=0N$$

Solving for RC_y:

$$\sum M_A=0= -(r_{AB}\cdot F_g)+(r_{AC}\cdot R_{Cy})\\r_{AB}\cdot F_{g}=r_{AC}\cdot R_{Cy} $$

$$R_{Cy}=\frac{r_{AB}\cdot F_g}{r_{AC}}\\R_{By}=\frac{0.04m\cdot 30 N}{0.08 m} $$

$$R_{Cy}=15 N$$

Solving for RA_y:

$$\sum F_y=0=R_{Cy}+R_{Ay} -F_g\\R_{Ay}=F_g-R_{Cy}\\R_{Ay}= 15 N$$

Finding shear and Moment from A to B :

$$\sum F_x=0\\N=0N$$

Solving for V

$$\sum F_y=0=R_{Ay}+-V\\ V= 15 N$$

Solving for M :

$$\sum M_A=0= -( V\cdot 0.04) +M\\M = 15\cdot 0.04\\M=0.6N $$

Finding shear and Moment from B to C:

$$\sum F_x=0\\N=0N$$

Solving for V

$$\sum F_y=0=R_{Cy}+V_2\\ V_2= -15 N$$

Solving for M :

$$\sum M_C=0= -( V_2\cdot(0.04) -M_2\\M_2 = 15\cdot 0.04\\M=0.6N $$

V-M diagram:

For V-x graph, from A to B there constant shear that is 15 N then from B to C the shear has same magnitude but different direction.  To plot the M-x graph, substitute x in the equation for M and M₂ to find the peak bending moment. Also, because M is the integral of shear force, horizontal plots in shear correspond to a linear plot in the M-x graph.

6. Review 

The answers make sense. The moment, shear, and reaction forces are equal in both segments of the beam. This is as expected because of the symmetry of the beam.

1. Problem

Roughly sketch the shear and moment diagram for the following structure. Measure from point A.

2. Sketch

N/A, Sketch Provided.

3. Knows and Unknowns:

Knowns:

• Distributed Load: w
• Dimensions and shown

Unknown:

• Shear/Moment Diagram

4. Approach:

Break the beam into sections as forces change.

Determine the plotted shape of the shear and moment of each section.

Draw the V/M diagrams using these shape diagrams.

5. Analysis:

A distributed load creates a linear section across it, with the magnitude of the shear force increasing across the load. The maximum shear will be at the end of the distributed load. With no other forces being applied except for reactions, section B-C should be constant. Since the beam is static, the reactions must bring the shear back to zero.

The moment can be drawn using the knowledge that dM = V. Section A-B in the shear diagram is linear; integrating a linear function gives a quadratic function, thus, section A-B in the moment diagram must be curved. Section B-C is constant in shear. Integrating a constant function gives a linear function; therefore, the moment across this section must be linear, and since the shear is negative, the slope of the moment must be negative. Finally, because the beam is static, the moment must return to zero due to the reaction forces.

6. Review

Using principles of shear force (V) and moments and their integral/derivative relationship, the shear and moment diagrams make sense for their general shape.

1. Problem

Find the reaction forces and create the shear and moment diagrams for the beam.

2. Draw

$$ 4N/m \cdot \frac{5.5m}{2} = 11N$$

$$ \bar{x} = 1/3 \cdot 5.5m = 1.83m$$

$$ \bar{x} \ from  A = 4m + 4.5m + 1.83m = 10.3m$$

3. Knowns and Unknowns

Knowns:

• Applied Forces:
• F_B = -20N
• F_C = -400N
• F_D = 500N
• F_Load = -4N/m
• Dimensions as shown

Unknowns:

• Reaction Forces: R_AX, R_AY, M_A
• Shear/Moment Diagram

4. Approach

1. Use equilibrium equation to find the reaction forces
2. Use beam analysis/cutting between forces to find equations for the shear/moment diagram.

5. Analysis

First, use the sum of forces and the sum of moments to determine R_Ay and M_A.

$$ \sum F_x = R_{Ax} = 0$$

$$ \sum F_y = -R_{Ay} – 20N – 400N – 11N +500N = 0$$

$$ R_{Ay} = 69N$$

$$ \sum M_A = – M_A – (20N \cdot 4m) – (400N \cdot 8.5m) – (11N \cdot 10.3m) + (500 N \cdot 14m) = 0 $$

$$ M_A = 3406.7Nm$$

Next, cut the beam between A and B and analyze the internal forces.

$$ \sum F_x = N + 0 = 0 \\ N = 0$$

$$ \sum F_y = -69N -V = 0 \\ V= -69N$$

$$ \sum M_{cut} =  M + (69N \cdot x) – (3406.7Nm) = 0 \\ M = -69x + 3406.7$$

Repeat the analysis of internal forces from A to C.

$$ \sum F_x = N + 0 = 0 \\ N = 0$$

$$ \sum F_y = -69N – 20N -V = 0 \\ V = -89N$$

$$ \sum M_{cut} = M + (69N \cdot x) + (20N \cdot{(x-4m)}) – 3406.7Nm = 0 \\ M+89x-3326.7Nm = 0 \\ M = -89x + 3326.7$$

Finally,  analyze the beam in the area from C to D. Ensure that the value of X used is now taken from point D.

$$ \sum F_x = -N = 0 \\ N = 0$$

Since the section does not reach the end of the distributed load, the slope of the load must be calculated:

$$\frac{4}{5.5} = \frac{m}{14-x} \\ m = \frac{56-4x}{5.5}$$

This can be used to find the equivalent point load in terms of x.

$$ F_{eq} = \frac{\frac{56-4x}{5.5} \cdot (14-x)}{2} \\ F_{eq} = \frac{4x^2}{11} – \frac{112x}{11} + \frac{784}{11}$$

This point load can be used when summing forces to find v.

$$\sum F_y = 0 = V – F_{eq} + 500 \\ V = F_{eq} – 500 \\ V = \frac{4x^2}{11} – \frac{112x}{11} + \frac{784}{11} – 500 \\ V = \frac{4x^2}{11} – \frac{112x}{11} – 428.7$$

To save time, rather than use the sum of moments to determine the equation for moments in this section, the moment can be found by the integral of V.

$$ M = \int \frac{4x^2}{11} – \frac{112x}{11} – 428.7 dx  \\ M = \frac{4x^3}{33} – \frac{112x^2}{22} – 428.7x +C$$

We know that for this beam to be static, the moment at x=14m must be 0. This can be used to determine C.

$$ M(14) = 0 =  \frac{4(14)^3}{33} – \frac{112(14)^2}{22} – 428.7(14) +C \\ C = 6667 \\ Therefore: M_{cd}(x) = \frac{4x^3}{33} – \frac{112x^2}{22} – 428.7x + 6667$$

With the values for internal forces found between each change in force, the internal forces can be plotted.

6. Review

Using principles of shear force (V) and moments and their integral/derivative relationship, the shear and moment diagrams make sense for their general shape. The numerical values calculated align with the variables provided.

1. Problem

A person who weighs 140lbs is standing on the edge of a 10ft long diving board. The board is secured by a pin at 0ft and supported by a roller at 4ft.

a. Calculate the reaction forces on the board.

b. Calculate the internal forces at the midpoint between A and B.

c. Calculate the internal forces at the midpoint between B and C.

d. Draw the shear and moment diagram.

source: https://www.flickr.com/photos/lac-bac/40229806264

2. Draw 

3. Knowns and Unknowns 

Knowns:

• F_g = 140 lbs
• X_A= 0 ft
• X_B= 4 ft
• X_C= 10 ft

Unknowns:

• R_Ax
• R_Bx
• R_Ay
• Internal forces N, V, M between B and C, & A and B

4. Approach 

Calculate the reaction forces using equilibrium equations. Cut the diving board between A & B, and use equilibrium equations to find internal forces. Do the same to find internal forces between B and C. Use the values to draw the shear and moment diagram.

5. Analysis

a.

now,

The negative sign for means that the direction of   is not +y but in the -y direction.

Thus, the reation forces are

b. Finding internal forces between A and B 

To find the internal forces between A and B, cut in the middle of A and B, giving

Internal forces between A and B are

c.  Finding internal forces between B and C

Cut at the midpoint between B and C to find the internal forces.

Finding x

To apply the equilibrium equation, you need to find x in this scenario.

now,

Internal forces between C and B are

d. Shear and Moment diagram 

6. Review 

There is no force in the x direction; therefore, it makes sense that the reaction force, R_Ax, is zero. Because the moment at A is generated by the person’s weight, R_By should be considerably large. As R_By is larger than the person’s weight, R_Ay should act downwards and would be of the same magnitude as the difference between R_By and the person’s weight to make the system stable.

For section b, the shear force between A and B is equal and opposite to the reaction force R_Ay. Similarly, with the concept of equilibrium, shear is equal and opposite to R_Ax, and the moment in this section is the negative of the moment generated at A.

It also makes sense that the shear between B and C is equal to the difference between R_By and R_Ay, and as expected, acting downward, thus it equals the weight of the person.

1. Problem

A bolt attached to a stair rail is to be removed by a technician. A solid wrench of length 30cm, arched horizontally to the bolt and assumed to be rigidly connected. The technician applies a downward vertical force of 200N at the free end of the wrench. Determine the internal normal force, shear force and bending moment at a point 0.15m along the wrench. Draw the full shear and moment diagram.

Source: https://www.alamy.com/bolt-and-wrench-isolated-on-white-3d-rendering-image211658127.html

2. Draw

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• F=200N
• L= 30cm => 0.3m

Unknowns:

• Internal normal force, shear force and bending moment at point B
• Shear + moment diagram

4. Approach

Solve for reaction forces at A, then take arbitrary sections between A and B and between B and C to derive V and M equations. Plot results.

5. Analysis

Solve for reaction forces:

\begin{aligned}
\sum M_{A} &= 0 = -200\,\mathrm{N} \cdot 0.3\,\mathrm{m} – M_A \\
M_A &= -60\,\mathrm{Nm}
\end{aligned}

\begin{aligned}
\sum F_{y} &= 0 = A_y – 200\,\mathrm{N} \\
A_y &= 200\,\mathrm{N}
\end{aligned}

\begin{aligned}
\sum F_{x} &= 0 = N \\
N &= 0
\end{aligned}

No force in the x-direction

Cut 1: With reaction forces known, cut a section between A and B

0m<x<0.15m

\begin{aligned}
\sum F_{y} &= 0 = A_y – V \\
V &= A_y = 200\,\mathrm{N}
\end{aligned}

\begin{aligned}
\sum M_x &= 0 = -M_A – A_y \cdot x +M(x) \\
M(x) &= 200x-60\,\mathrm{Nm}
\end{aligned}

You can also cut the section between B and C

0.15m<x<0.3m

\begin{aligned}
\sum F_{y} &= 0 = A_y – V \\
V &= A_y = 200\,\mathrm{N}
\end{aligned}

\begin{aligned}
\sum M_x &= 0 = M_x – M_A- A_y \cdot x \\
M(x) &= 200x – 60\,\mathrm{Nm}
\end{aligned}

\begin{aligned}
\text{At } x = 0, \quad &M = 0 \\
\text{At } x = 0.15\,\mathrm{m}, \quad &M = -30\,\mathrm{Nm} \\
\text{At } x = 0.3\,\mathrm{m}, \quad &M = 0
\end{aligned}