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:

$$\underline{M}_A=\underline{r}_A\times \underline{F}_A=\begin{bmatrix}
\underline{\hat{i}} & \underline{\hat{j}} & \underline{\hat{k}} \\
0.42 & 0 & 0 \\
0 & 7.5 & 0
\end{bmatrix}$$ $$\underline{M}_A=\hat{i} \begin{bmatrix}
0 & 0 \\
7.5 & 0
\end{bmatrix} -\underline{\hat{j}} \begin{bmatrix}
0.42 & 0 \\
0 & 0
\end{bmatrix}+\underline{\hat{k}} \begin{bmatrix}
0.42 & 0 \\
0 & 7.5
\end{bmatrix}\\\underline{M}_A=(\underline{\hat{i}}(0)-\underline{\hat{j}}(0)+\underline{\hat{k}}(o.42\cdot 7.5-0\cdot 0))ft\cdot lb\\\underline{M}_A=3.15\underline{\hat{k}} ft\cdot lb$$

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

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.

2. Draw

Sketch:

Free-body diagram:

3. Knowns and Unknowns

Knowns:

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

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.