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.

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:

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

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

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

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

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.

[latex]\overrightarrow {r_{1}} = \begin{bmatrix}-0.1905\\0\\0\end{bmatrix}\kern 1pc\overrightarrow {r_{2}} = \begin{bmatrix}0.1905\\0\\0\end{bmatrix}[/latex]

[latex]\overrightarrow {F_{1}} = \begin{bmatrix}0\\150\\0\end{bmatrix}\kern 1pc\overrightarrow {F_{2}} = \begin{bmatrix}0\\-200\\0\end{bmatrix}[/latex]

[latex]\overrightarrow{M_1}=\overrightarrow{r_{1}} \times \overrightarrow{F_{1}} \kern 1pc +  \kern 1pc\overrightarrow{r_{2}} \times \overrightarrow{F_{2}}[/latex]

[latex]\overrightarrow{M_{1}}=\:\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\-0.1905 & 0 & 0\\0 &  150 & 0\end{vmatrix}[/latex][latex]\kern 1pc + \kern 1pc  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\0.1905 & 0 & 0\\0 &  -200  & 0\end{vmatrix}[/latex][latex]\:=-28.575\: \hat{k} - 38.1 \hat{k}[/latex]

[latex]\overrightarrow{M_{1}}=\: -66.675 \: \hat{k}[/latex]

b. LEFT TURN

Radius and force in vector form

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

[latex]\overrightarrow {r_{1}} = \begin{bmatrix}-0.1905\\0\\0\end{bmatrix}\kern 1pc\overrightarrow {r_{2}} = \begin{bmatrix}0.1905\\0\\0\end{bmatrix}[/latex]

[latex]\overrightarrow {F_{3}} = \begin{bmatrix}0\\-150\\0\end{bmatrix}\kern 1pc\overrightarrow {F_{4}} = \begin{bmatrix}0\\200\\0\end{bmatrix}[/latex]

[latex]\overrightarrow{M_2}=\overrightarrow{r_{1}} \times \overrightarrow{F_{3}} \kern 1pc +  \kern 1pc\overrightarrow{r_{2}} \times \overrightarrow{F_{4}}[/latex]

[latex]\overrightarrow{M_{2}}=\:\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\-0.1905 & 0 & 0\\0 &  -150 & 0\end{vmatrix}[/latex][latex]\kern 1pc + \kern 1pc  \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\0.1905 & 0 & 0\\0 &  200  & 0\end{vmatrix}[/latex][latex]\:=28.575\: \hat{k} + 38.1 \hat{k}[/latex]

[latex]\overrightarrow{M_{2}}=\: 66.675 \: \hat{k}[/latex]

6. Review 

Because the radius and force values in both instance have same magnitude but different directions the net moments for left and  right turn was expected to have same magnitudes and different direction 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 [latex]\bar{x}=\frac{\sum F_{i}x_i}{\sum F_i}[/latex].
• 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 [latex]F_{A}=F_{B}=F_{C}[/latex].

From the fig above

[latex]F_{R1}=1000 N/m \times 1.5 m = 1500 N[/latex]

[latex]F_{R2} = \frac{1}{2} \times 1.5 m \times 1000 N/m = 750 N[/latex]

[latex]F_{A} = 1500 N + 750 N = 2250N[/latex]

[latex]\bar {x}[/latex] for resultant force of car A

From O [latex]F_{R1}[/latex] locates at [latex]1+ \frac{1}{2}(1.5) = 1.75 m[/latex]

[latex]F_{R2}[/latex] locates at [latex]1+ 1.5+ \frac{1}{3}(1.5) = 3 m[/latex]

now, [latex]\bar{x}=\frac{\sum F_{i}x_i}{\sum F_i}[/latex]

thus [latex]\bar{x_{A}}= \frac {(1500 N/m\times 1.75m+(750 N/m\times 3m)}{(1500N +750N)}[/latex]

[latex]\bar{x_{A}} = 2.17 m[/latex] from left side.

Also for one car the [latex]\bar{x}[/latex] of 2250 N force is [latex]2.17m - 1 m = 1.17 m[/latex]

Using these calculations, the following figure can be drawn.

The total resultant force of all 3 cars is [latex]F_{R3} = 2250N \times 3 = 6750 N[/latex]

[latex]\bar{x}=\frac{\sum F_{i}x_i}{\sum F_i}[/latex]

therefore,

[latex]\bar{x} = \frac{(2.17 \times 2250)+(6.17\times 2250)+(10.17\times 2250)}{6750}[/latex]

[latex]\bar{x} = 6.17 m[/latex]  from left side .

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

[latex]F_{D} = F_{E}=2250 N[/latex]

Total force of the 2 cars [latex]F_{R2} = 2250 \times 2= 4500 N[/latex]

Position of the resultant forces from the 2 cars,

[latex]\bar {x_{2}} = \frac {(4.17 \times 2250) + (9.17 \times 2250)}{4500}=6.67 m[/latex]

6.67 m is from the right side, but from the  left side  [latex]\bar{x_{2}} = 13-6.67 = 6.33 m[/latex]

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

Total force on bridge due to the 5 cars [latex]F_{R5} =F_{R2} +F_{R3} = 4500 N + 6750 N = 11250 N[/latex]

[latex]\bar{x} = \frac{(F_{R2}\times \bar{x_{2}})+(F_{R3}\times\bar{x_{3}})}{F_{R5}}[/latex]

[latex]\bar {x} =\frac{ (6.33 \times 4500)+ (6.17 \times 6750 )}{11250}= 6.234 m[/latex]

[latex]\bar {y} [/latex] for all the cars is half of its width

therefore, [latex]\bar{y_{5}} = \frac{(\bar{y_{2}}\times F_{R3} )+(\bar{}y_{3}\ times F_{R3})}{F_{R5}}[/latex]

[latex]\bar{y_{5}} = \frac{(6m \times 4500 N )+(2m \times 6750 N)}{11250 N}[/latex]

[latex]\bar{y_{5}} = 3.6 m[/latex]

[latex]F_{R5} = 11250 N \quad at \quad (6.234m , 3.6 m )[/latex]

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 moment, M

4. Approach 

Using Equation [latex]M = F\times d[/latex]

5. Analysis 

Couple moment, [latex]M = F\times d = 50N \times 1 m= 50 Nm [/latex]

6. Review  

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