Answer to Question #128432 in Mechanics | Relativity for Alyana

Question

loads of 50g 80g and 100g are placed at the 12cm 22cm and 62cm marks of a uniform meter stick. If the stick is supported at its midpoint. What load must be suspended at the 100cm mark to maintain equilibrium?

Answer 1

The moment of force rotating stick clockwise must be equal to the moment of force rotating it counterclockwise. The moment of force is equal to the product of the force per shoulder.

First shoulder = 50 - 12 = 38 cm

Second shoulder = 50 - 22 = 28 cm

Trird shoulder = 62 - 50 = 12 cm

Fourth shoulder = 100 - 50 = 50 cm

$50\times38+80\times28=100\times12+m\times50$

m=58.8 g

Answer: load of 58.8 g