Chapter 8, Problem 1STP

To determine

To Choose: The correct option.

Answer to Problem 1STP

C. $1.1\text{m}$

Explanation of Solution

Given:

The mass of left box is $m_1=25\text{kg}$ .

The mass of right box is $m_2=15\text{kg}$ .

The length of the board, $l=3.0\text{m}$

Formula used:

The torque is given by,

  $\tau =Fr\sin \theta$

Where F is the applied force, r is the distance of the point of application of force from the axis of rotation, and $\theta$ is the angle between F and r .

Calculation:

If the fulcrum is at a distance $x$ from left side of the board, then the weight of the boxes will act in vertical downward direction which generates torque.

Now, torque due to boxes will be balanced about the fulcrum. Then,

  $\begin{array}{c}{\tau }_1={\tau }_2\\ W_1{r}_1\sin 90°=W_2{r}_2\sin 90°\\ \left(m_{1}g\right)\left(x\right)=\left(m_{2}g\right)\left(l-x\right)\\ 25x=15\left(3-x\right)\\ 25x=45-15x\\ x=1.1\text{m}\end{array}$

Conclusion:

Thus, thedistance of the fulcrum from the left side of the board in order to balance the masses horizontally is $1.1\text{m}$ .