Question

on a see saw two children of masses 25 kg and 50 kg are sitting on one side of it a distance 2 m and 1.5 m respectively from its middle. Where should a man of mass 75 kg sit to balance it? ​

Answer 1

Answer:

5/3 m

Explanation:

The toque of the system should be zero.

Torque due to two children's= F1r1 + F2r2

=25×2 + 50×1•5

=50+75

=125

Let the distance be= x m

Torque due to man = Fr

=75x

Now,

ATQ, 75x = 125

=> x= 125/75= 5/3

The required distance is 5/3 m.