Question

A pivot clip is positioned at the center of a meter stick with two hanging clips placed on the left side (each clip is 5 g) one at 10 cm from the center and other at 30 cm from the center. Each added mass is 30 g. What single mass would need to be positioned at 40 cm to the right of center to balance the system

Answer 1

let m1 = 5 + 30 = 35 grams
x1 = 10 cm
m2 = 5 + 30 = 35 grams
x2 = 30 cm

let m3 is the third mass to be added at x3=40 cm distance on the right side

As the meter stick is in equilibrium, net torque acting on it must be zero.

m1*x1 + m2*x2 - m3*x3 = 0

m3 = (m1*x1 + m2*x2)/x3

= (35*10 + 35*30)/40

= 35 g <<<<<<<<----------------Answer