Question

A block of mass 10 kg is suspended at a distance of 20 cm from the center of a uniform bar 1 m long. What force is required to balance it at its center of gravity by applying the force at the other end of the bar? (g = 10m/s² )

Answer 1

Given,

Let the Force applied to balance be x,

According to Principle of Moments,

Sum of Clockwise Moments = Sum of Anti-Clockwise Moments

= (10*10*0.2) = (x*0.8)

= 20 = 0.8x

= 25 N = x

The Force needs to be applied to balance is 25 N at the other end of the bar.

I hope this helps.