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Torque

AP Physics 1· difficulty 2/5

A 60 N60~\text{N} force is applied at the end of a 0.50 m0.50~\text{m} lever at 30∘30^\circ from the lever's direction. What torque is produced about the pivot?

  • A

    15 N⋅m15~\text{N·m}

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  • B

    30 N⋅m30~\text{N·m}

  • C

    26 N⋅m26~\text{N·m}

  • D

    7.5 N⋅m7.5~\text{N·m}

Explanation

τ=rFsin⁡θ=(0.50)(60)(sin⁡30∘)=(0.50)(60)(0.5)=15 N⋅m\tau = r F \sin\theta = (0.50)(60)(\sin 30^\circ) = (0.50)(60)(0.5) = 15~\text{N·m}.

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