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. Determin
e the torque on the meterstick if it is received and allowed
1 answer:
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)
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