Question

A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced when a 60-kg person sits on one end and a 78-kg person sits on the other end.

Required:
Find a displacement of the center of mass of the system relatively to the seesaw's midpoint.

Answer 1

x = 0.9 m

Step-by-step explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

∑ τ = 0

60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

90 - 117 + 30 x = 0

x = 27/30

x = 0.9 m

In summary the center of mass is on the side of the lightest weight x = 0.9 m