Two people of different masses sit on a seesaw (Figure 11.103). ${\mathit{M}}_{\mathbf{1}}\mathbf{,}$the mass of personis$\mathbf{90}\mathit{k}\mathit{g}\mathbf{,}{\mathit{M}}_{\mathbf{2}}$is$\mathbf{42}\mathit{k}\mathit{g}\mathbf{,}{\mathit{d}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}\mathit{m}\mathbf{,}$and${\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3}\mathit{m}\mathbf{.}$The people are initially at rest. The mass of the board is negligible.

(a) What are the magnitude and direction of the torque about the pivot due to the gravitational force on person(b) What are the magnitude and direction of the torque about the pivot due to the gravitational force on person(c) Since at this instant the linear momentum of the system may be changing, we don’t known the magnitude of the “normal” force exerted by the pivot. Nonetheless, it is possible to calculate the torque due to this force. What are the magnitude and direction of the torque about the pivot due to the force exerted by the pivot on the board? (d) What are the magnitude and direction of the net torque on the system (board + people)? (e) Because of this net torque, what will happen? (A) The seesaw will begin to rotate clockwise. (B) The seesaw will begin to rotate counterclockwise. (C) The seesaw will not move. (f) Person 2 moves to a new position, in which the magnitude of the net torque about the pivot is now$\mathbf{0}\mathbf{,}$and the seesaw is balanced. What is the new value of${\mathit{d}}_{\mathbf{2}}$in this situation?

Torque is the measure of the force that can cause an object to rotate about an axis. Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes an angular acceleration. Hence, torque can be defined as the rotational equivalent of linear force.

The torque $\left(\tau \right)$due to the force about a point is defined as the defined as the product of the magnitude of the force $\left(F\right)$and the perpendicular distance $\left(d\right).$It is given by the equation $\tau =Fd$

The seesaw with two persons and the forces are shown in the below figure.

Mass of the first person, $M_1=90kg$

Mass of the second person, $M_2=42kg$

Distance of the first person from the pivot point, $d_1=0.8m$

Distance of the second person from the pivot point, $d_2=1.3m$

Acceleration due to gravity, $g=9.8m/s^2$

(a)The gravitational force acting on the first person is defined as the product of the mass of the first person $\left(M_1\right)$and acceleration due to gravity$\left(g\right).$

$F_{g1}=M_{1}g$

Therefore, the magnitude of the torque $\left({\tau }_1\right)$about the pivot point due to gravitational force on the first person is the product of the force $\left(F_{1g}\right)$and the perpendicular distance from the pivot $\left(d_1\right)$

$$\begin{gathered} {\tau }_1=F_{g1}{d}_1 \\ =\left(M_{1}g\right)d_1 \\ =\left(90kg\right)(9.8m/s^2)(0.8m) \\ =705.6N.m \end{gathered}$$

The torque due to gravitational force about the pivot point rotate the sea saw in counter clockwise direction. Hence the direction of this torque is taken as positive.

(b)The gravitational force acting on the second person is defined as the product of the mass of the second person $\left(M_2\right)$and acceleration due to gravity $\left(g\right).$

$F_{g2}=M_{2}g$

Therefore, the magnitude of the torque $\left({\tau }_2\right)$about the pivot point due to gravitational force on the second person is the product of the force $\left(F_{2g}\right)$and the perpendicular distance from the pivot $\left(d_2\right).$

$$\begin{gathered} {\tau }_2=F_{g2}{d}_2 \\ =\left(M_{2}g\right)d_2 \\ =\left(42kg\right)(9.8m/s^2)(1.3m) \\ =535.08N.m \end{gathered}$$

The torque due to the gravitational force about the point rotate the sea saw in clockwise direction. Hence the direction of this torque is taken as negative.

(c)Let $F_n$be the normal force on the seesaw board. Then the perpendicular distance from the pivot is zero. Hence, the torque due to the normal force is

$$\begin{gathered} {\tau }_3=F_n\left(0m\right) \\ =0N.m \end{gathered}$$

The magnitude of the torque is zero. Hence the direction of the torque is indeterminate.

(d)The net torque on the system is the sum of the torque due to the two gravitational forces of the persons and the normal force on the board. The net torque acting on the system is

$$\begin{gathered} {\tau }_{net}={\tau }_1+{\tau }_2+{\tau }_3 \\ =705.6N.m-535.08N.m+0N.m \\ =170.52N.m \end{gathered}$$

(e)The magnitude of the net torque is positive. Hence the direction of the net torque is counter clockwise direction.

Therefore, the seesaw will begin to rotate in counter clockwise direction.

Hence, option (B) is correct.

(f)If the new position of the second person is $d_2\text{'}$then the net torque about the pivot point should be zero. Now, we get the new value of $d_2$as,

$$\begin{gathered} \Sigma {\tau }_{net}=0 \\ {\tau }_1-{\tau }_2=0 \\ \left(M_{1}g\right)d_1-\left(M_{2}g\right)d_2\text{'}=0 \\ {d}_2\text{'}=\frac{M_1}{M_2}{d}_1 \\ =\left(\frac{90kg}{42kg}\right)(0.8m) \\ =1.71m \end{gathered}$$