Question

a 3.0 m uniform beam of mass 15 kg is pivoted 1.0 m from the end as shown below. a 35 kg child sits 0.60 m from the pivot. how far, d, from the pivot, must a 20 kg child sit in order for the beam to be in equilibrium?

Answer 1

The distance in which the 2 kg child must sit in order for the beam to be in equilibrium is 1.1 m.

How to calculate the position of the 2 kg child?

The distance in which the 2 kg child must sit in order for the beam to be in equilibrium is calculated as follows;

taking moment about the pivot;

sum of clockwise moment = sum of anti clockwise moment

35 kg x (0.6 m) = 20 kg x (d)

21 = 20d

d = 21 / 20

d = 1.05 m

d ≈ 1.1 m

Thus, the distance in which the 2 kg child must sit in order for the beam to be in equilibrium is 1.1 m.