Answer to Question #87462 in Mechanics | Relativity for Naznin

Question

A uniform metre rule is pivoted at 30 cm Mark.it is 100 long. The rule balances when a weight of 6N is placed in 0cm and a weight of 2N is placed on 70cm Mark. What is the weight of the rule

Answer 1

For the equilibrium:

$$6(0-30)+2(70-30)+W(50-30)=0$$

$$20W=100$$

$$W= \frac{100}{20}=5N$$