Question

A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.

pls answer quickly. Thanks​

Answer 1

The mass of the rule is 56.41 g

Step-by-step explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0 22cm 100cm

-------------------------Δ------------------------------------

↓ ↓

200g m₂

Apply principle of moment

(200 g)(22 cm - 0) = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ = (200 g)(22 cm) / (78 cm)

m₂ = 56.41 g

Therefore, the mass of the rule is 56.41 g