Question

A boy and girl sit on a seesaw of length 4m, balanced at its center. The girl sits at the far end and has a mass of 50kg. The boy is twice as heavy as the girl, and therefore sits midway between his end and the center. Then the girl is given a bag of oranges weighing 5kg, and the seesaw rotates out of balance. When the boy is given a bag of apples, balance is still not restored. He has to place the apples 0.5m behind him for them to be balanced again. What is the weight of the bag of apples?

Answer 1

Let x = mass of Bag of apples {weight of Bag of apples = xg}
Static Equilibrium about Pivot
¦²CCW Torque = ¦²CW Torque: {CW & CCW directions depend on the diagram}
Torque of Girl + bag of oranges {at Right side of pivot} = {50 + 5)(9.8)(2) = 1078 m-N {CW}
Torque of Boy {Left side of pivot} = 100(9.8)(1) = 980 m-N {CCW}
Torque of Bag of apples {Left side of pivot} = x(9.8)(1.5) = 14.7x m-N {CCW}
980 + 14.7x = 1078
14.7x = 98
x = 6.67 kg
ANS = xg = 6.67(9.8) ¡Ö 65 N