Question

A non-uniform lever of mass 6.2 kg and length 4 m is pivoted at its mid-point. The lever will be in
equilibrium when a vertically downward force of 5.6 Nis acting at its right end and a vertically upward force of 4.7 N is acting on its left end. Calculate the location of its center of gravity?

Answer 1

Because the upward force is added at the left end, we know the center of gravity [c.g.]
is somewhat to the left of the pivot or fulcrum. Call the c.g.'s distance from the pivot x.

clockwise torque = (4.7N)(2m) + (5.6N)(2m)=20.6N-m

counter-clockwise torque = (6.2kg)(9.8N/kg) x = 60.8 x

set them equal for equilibrium and x = 20.6/60.8 = 0.34 m

which seems reasonable, the c.g. being a bit off center compared to the 4 m length..

Answer 2

question: how about pivot at its mid-point?

Answer 3

The question states that the pivot is at the mid-point. We are to find where the center of gravity is. Done by setting net torque = 0.

Answer 4

Then we place the lever's mass at the center of gravity.