Question

A 2 kg and 2 meter long stick is held horizontally. A 4 kg mass is placed at 30 cm, and a 5 kg mass is placed at 75 cm. Determine the torque on the meterstick if it is received and allowed

Answer 1

Answer:

τ = 32.8635 N-m (counterclockwise)

Explanation:

Given

M = 2 kg

L = 2 m

r = 0.10 m

m₁ = 4 kg

r₁ = (1.00-0.30)m = 0.70 m

m₂ = 5 Kg

r₂ = (0.90-0.75)m = 0.15 m

In order to determine the torque on the meterstick if it is received and allowed to pivot about the 90 cm mark (ypu can see the pic to understand the question), we apply:

τ = r₁*m₁*g + r₂*m₂*g - r*M*g = g*(r₁*m₁+r₂*m₂-r*M)

⇒  τ = (9.81 m/s²)(0.70 m*4 kg + 0.15 m*5 Kg - 0.10 m*2 kg)

⇒  τ = 32.8635 N-m (counterclockwise)