Question

A woman of mass m = 53.4 kg sits on the left end of a seesaw—a plank of length L = 4.25 m—pivoted in the middle as shown in the figure.

(a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 67.7 kg sit if the system (seesaw plus man and woman) is to be balanced?

(b) Find the normal force exerted by the pivot if the plank has a mass of m_pl = 12.8 kg.

(c) Repeat part (a), but this time compute the torques about an axis through the left end of the plank (m).

EXERCISE

Suppose a 32.0 kg child sits 1.87 m to the left of center on the same seesaw as the problem you just solved. A second child sits at the end on the opposite side, and the system is balanced.

(a) Find the mass of the second child (kg).
m_{child 2} =

(b) Find the normal force acting at the pivot point (N).
F_n =   

Answer 1

Problem - 1

Given is:-

Mass of the woman m = 53.4 kg

Length of the plank L = 4.25 m

Mass of the man M = 67.7 kg

Now,

part - a

For equilibrium condition the net torque around the pivot point must be zero, therefore

or

by plugging the values we get

(We took clockwise torque as negative)

thus

by plugging all the values we get

which gives us

hence, man should sit 1.676m towards the right of pivot point for equilibrium condition.

Part - b

Mass of the plank = 12.8 kg

Thus for equilibrium condition the normal force exerted by the pivot point is

by plugging all the values we get

which gives us

Part - c

Again summing the torques and equating it to zero

(Taking clockwise torques negative)

by plugging all the values we get

which gives us