Question

A uniform meter stick of mass $M$ has a half-filled can of fruit juice of mass $m$ attached to one end. The meter stick and the can balance at a point 21.2 cm from the end of the stick where the can is attached. When the balanced stick-can system is suspended from a scale, the reading on the scale is 2.58 N. Calculate the mass of the meter stick.

Answer 1

111.6 g

Step-by-step explanation:

Given

m + M = 2.58 / 9.8

= 0.2632 kg

When the can of fruit juice is balance from scale , we get following relation

M x ( 50 - 21.2 ) = m x 21.2 ( balancing the torque due to weight of scale and can about the balancing point )

M x 28.8 = 21.2 m

= 21.2 ( 0.2632 - M )

= 5.58 - 21.2 M

M ( 28.8 + 21.2 ) = 5.58

M = .1116 kg

= 111.6 g