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:

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 ( [latex]\sum\underline{F}=0[/latex] , [latex]\sum\underline{M}=0[/latex] ) . 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

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 ( [latex]\sum\underline{F}=0[/latex] , [latex]\sum\underline{M}=0[/latex] ), 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 ( [latex]\sum\underline{F}=0[/latex] , [latex]\sum\underline{M}=0[/latex] ), 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 ( [latex]\sum\underline{F}=0[/latex] , [latex]\sum\underline{M}=0[/latex] ) 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 [latex]\sum{F_{y}} = 0[/latex]
• Calculate F_F using [latex]\sum{F_{x}} = 0[/latex]
• Calculate µ using [latex]\mu= \dfrac{F_{F}}{N}[/latex]
• Compare µ to µ_s

5.  Analysis

Calculating w

[latex]w = m\cdot g[/latex]

[latex]\:\:\:=500kg \cdot 9.81 m/s^2 = 4905 N[/latex]

w = 4905 N

Calculating N 

[latex]\sum {F_y} = N-w =0[/latex]

[latex]N=w[/latex]

N = 4905 N

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

Finding F_F

[latex]\sum {F_x}= P-F_{F} = 0[/latex]

[latex]P = F_{F}[/latex]

F_F  = 5000 N

Finding µ

[latex]\mu = \dfrac {F_{F}}{N} = \dfrac{5000 N }{4905 N}[/latex]

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 

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, [latex]F_{g }= m\times g=19,620 N[/latex]

Since the tension applied by Spider-Man is a force with both x and y components, we would need angle [latex]\theta[/latex]

[latex]\theta = \tan^{-1} \left( \frac{h_c - h_s}{20} \right) = 22.3^\circ[/latex]

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.

[latex]M_{B} =\: -h_{c}\times T_{x}+\frac{w}{2}\times F_{g} \:=0\\ \kern    1pc =10m \times T_{x}+\frac{2.5}{2}\times 19,620 N[/latex]

thus,   [latex]T_{x} = 2,452.5 N[/latex]

Also

[latex]T_{x} = T\cos{\theta}[/latex]

[latex]\lvert T\rvert = \frac{T_{x}}{\cos{\theta}}=2,651 N[/latex]

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.

[latex]\begin{aligned} F_y &= -T_y + F_n - F_g \\ 0 &= -T \sin{\theta} + F_n - F_g \\ 0 &= -2651 \cdot \sin{22.3^\circ} + F_n - 19620 \\ F_n &= 20626\,N \end{aligned}[/latex]

now [latex]F_{f}=\mu \times F_{n} \\ F_{f} = 4,125 N[/latex]

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

[latex]F_{g} = m \cdot g = 70 kg \cdot 9.81m/s^2 = 686.7 N[/latex]

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.

[latex]F_{gx} =686.7\cdot \sin{30}[/latex]

[latex]F_{gx} =343.35N[/latex]

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.