Chapter 11, Problem 25A

a. Calculate the individual torques produced by the weights of the girl and boy on the seesaw in the figure. What is the net torque?

b. Calculate the distance a 600-N boy should sit from the fulcrum.

c. Calculate the distance a 300-N girl should sit when the boy weighs 400 N.

To determine

The individual torques produced by the weights of the girl and the boy and the net torque.

Answer to Problem 25A

The torque produced by the weights of the girl is $900\text{ N}\cdot \text{m}$ .

The torque produced by the weights of the boy is $900\text{ N}\cdot \text{m}$ .

The net torque produced is $0$ .

Explanation of Solution

Given info:

The weight of the girl is $300\text{ N}$ .

The weight of the boy is $600\text{ N}$ .

Formula used:

The expression for the torque as follows:

  $\tau =r\times F$

Here, $r$ is the lever arm and $F$ is the applied force.

Calculation:

Sketch the free body diagram of the system as shown below.

  

Refer to Figure.

The torque produced by the $300\text{ N}$ weight as follows:

  $\begin{array}{l}{\tau }_{\text{300 N}}=r\times F\\ {\tau }_{\text{300 N}}=\left(3\text{ m}\right)\times \left(300\text{ N}\right)\\ {\tau }_{\text{300 N}}=900\text{ N}\cdot \text{m}\end{array}$

Hence, the torque produced by the weights of the girl is $900\text{ N}\cdot \text{m}$ .

The distance x as follows:

  $\begin{array}{l}{\displaystyle \sum M_{\text{@center}}}=0\\ \left(300\text{ N}\times 3\text{ m}\right)-\left(600\text{ N}\times x\right)=0\\ x=1.5\text{ m}\end{array}$

The torque produced by the $300\text{ N}$ weight as follows:

  $\begin{array}{l}{\tau }_{\text{600 N}}=r\times F\\ {\tau }_{\text{600 N}}=\left(\text{1}\text{.5 m}\right)\times \left(600\text{ N}\right)\\ {\tau }_{\text{600 N}}=900\text{ N}\cdot \text{m}\end{array}$

Hence, the torque produced by the weights of the boy is $900\text{ N}\cdot \text{m}$ .

The net torque as follows:

  $\begin{array}{l}\tau ={\tau }_{300\text{ N}}-{\tau }_{600\text{ N}}\\ \tau =\left(900\text{ N}\cdot \text{m}\right)-\left(900\text{ N}\cdot \text{m}\right)\\ \tau =0\end{array}$

Hence, the net torque is $\tau =0$ .

Conclusion:

Thus, the torque produced by the weights of the girl is $900\text{ N}\cdot \text{m}$ .

Thus, the torque produced by the weights of the boy is $900\text{ N}\cdot \text{m}$ .

Thus, the net torque produced is $0$ .

To determine

The distance of the $600\text{ N}$ boy from the fulcrum.

Answer to Problem 25A

The distance of the $600\text{ N}$ boy from the fulcrum is $1.5\text{ m}$ .

Explanation of Solution

Given info:

The weight of the girl is $300\text{ N}$ .

The weight of the boy is $600\text{ N}$ .

Calculation:

The distance x is as follows:

  $\begin{array}{l}{\displaystyle \sum M_{\text{@center}}}=0\\ \left(300\text{ N}\times 3\text{ m}\right)-\left(600\text{ N}\times x\right)=0\\ x=1.5\text{ m}\end{array}$

Conclusion:

Thus, the distance of the $600\text{ N}$ boy from the fulcrum is $1.5\text{m}$ .

To determine

The distance of the $300\text{ N}$ girl should sit when the boy weighs $400\text{ N}$ .

Answer to Problem 25A

The distance of the $300\text{ N}$ girl should sit when the boy weighs $400\text{ N}$ is $2\text{ m}$ .

Explanation of Solution

Given info:

The weight of the girl is $300\text{ N}$ .

The weight of the boy is $600\text{ N}$ .

Calculation:

Consider the girl sits from the fulcrum at a distance d.

Find the distance x as follows:

  $\begin{array}{l}{\displaystyle \sum \tau }=0\\ \left(300\text{ N}\times d\right)-\left(400\text{ N}\times 1.5\text{ m}\right)=0\\ d=2\text{ m}\end{array}$

Conclusion:

Thus, the distance of the $300\text{ N}$ girl should sit when the boy weighs $400\text{ N}$ is $2\text{ m}$ .