Question

A horizontal uniform meter stick supported at the 50-cm mark has a mass of 0.50 kg hanging from it at the 20-cm mark and a 0.30 kg mass hanging from it at the 60-cm mark. Determine the position on the meter stick at which one would hang a third mass of 0.60 kg to keep the meter stick balanced.

Answer 1

Answer:

70 cm

Explanation:

0.5 kg at 20 cm

0.3 kg at 60 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$0.5(50-20)=0.3(60-50)+0.6x\\\Rightarrow x=\dfrac{0.5(50-20)-0.3(60-50)}{0.6}\\\Rightarrow x=20\ cm$

The position of the third mass of 0.6 kg is at 20+50 = 70 cm

Answer 2

Answer:

70 cm mark

Explanation:

m1 = 0.5 kg

m2 = 0.3 kg

m3 = 0.6 kg

let the third mass is at d cm from 50 cm mark. take moments about the 50 cm mark.

Anticlockwise torque = clock wise torque

0.5 x ( 50 - 20) + 0.6 x d = 0.3 (60 - 50)

0.5 x 30 + 0.6 d = 0.3 x 10

15 + 0.6 d = 3

0.6 d = - 12

d = - 20 cm

So, it means third mass is at 20 cm right to the 50 cm mark. So, it is at 70 cm mark.