Chapter 8.2, Problem 16PP

To determine

The distance from the center of a seesaw where S should sit.

Answer to Problem 16PP

  $1.48\text{m}$

Explanation of Solution

Given:

Mass of A is $m_A=43\text{kg}$ .

Mass of S is $m_S=52\text{kg}$ .

Ashok sits at a distance of $1.8\text{m}$ from the center of seesaw.

Formula used:

The torque can be obtained 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:

When A and S sit on the seesaw, their weight will act in vertical downwards direction which generates torque.

The torque will be balanced about the pivot point that is the center of seesaw.

Let S sit at a distance $r_S$ from the center of seesaw, then balance of torque can be written as,

  $\begin{array}{l}{\tau }_S={\tau }_A\\ W_S{r}_S\sin 90°=W_A{r}_A\sin 90°\\ r_S=\frac{m_{A}g{r}_A}{m_{S}g}\\ =\frac{m_A{r}_A}{m_S}\end{array}$

Substitute given values in above

  $\begin{array}{c}{r}_S=\frac{m_A{r}_A}{m_S}\\ =\frac{43\times 1.8}{52}\\ =1.48\text{m}\end{array}$

Conclusion:

Thus, S should sit $1.48\text{m}$ from the center of seesaw.