Question

The center of gravity of a 50 g meter stick is at 51 cm and the stick is supported at 70 cm. Where must an 80 g object be hung in order to have equilibrium?

Answer 1

`0.81875~~0.82m`

Explanation:

`\Sigma"Clockwise moments"=\Sigma"Anticlockwise moments"`

`\Sigma"Anticlockwise moments"=\SigmaFx=(0.05*9.81)(0.70-0.51)=(0.4905)(0.19)=0.093195Nm`

`\SigmaClockwise moments"=0.093195=\SigmaFx=(0.08*9.81)x`

`0.093195=0.7848x`

`x=0.093195/0.7848=0.11875~~0.12m`

However, we want the actual potion, which would be `x`m away from the pivot point, which is `0.70+0.11875=0.81875~~0.82m`