Question

A student measured a meter stick to be 150 gm. The student then placed a knife edge 

on 30-cm mark of the stick. If the student placed a 500-gm weight on 5-cm mark and 

a 300-gm weight on somewhere on the meter stick, the meter stick then was 

balanced. Where did the student place the 300-gram weight?

Answer 1

Distance of 500 gm weight from pivot point,

$x_1 = 30 - 5 = 25 cm$

Distance of center of mass point,

$x_2 = 50 - 30 = 20 cm$

suppose 300 gm weight is put at distance r from 30cm mark,

$(500 \times 25) - (150 \times 20) - (300 \times r) = 0$

$r = \dfrac{9500}{300} = 31.7 cm$

$x = 30 + 31.7 = 61.7 \text{ cm mark }$