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If a 5.0 N force is applied at an angle of 30 degrees at the end of a 2.0 m. lever, what will be the magnitude of the torque (with solution please)

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KizunaHighgo

Calculating the magnitude of a torque.

Given

  • Force (F) = 5.0 N
  • Length (r) = 2.0 m
  • Axis of rotation (θ) = 30°
  • Magnitude (τ) = ?

τ = rF sinθ

τ = (2.0m)(5.0N)(sin 30)

τ = (10)(sin 30)

τ = 5 Nm

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