Rotational Equilibrium and Newton's First Law in Rotational Form

AP Physics 1· difficulty 3/5

W rope (30°)

A uniform 5 kg5~\text{kg}, 2.0 m2.0~\text{m} horizontal beam is hinged to a wall and supported by a rope at 3030^\circ above the beam at its far end. A 10 kg10~\text{kg} sign hangs at the far end. Using g10 m/s2g \approx 10~\text{m/s}^2, find the rope tension.

  • A

    150 N150~\text{N}

  • B

    300 N300~\text{N}

  • C

    250 N250~\text{N}

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  • D

    200 N200~\text{N}

Explanation

Torque about hinge: Tsin302.0=(5)(10)(1.0)+(10)(10)(2.0)T(1.0)=50+200=250T=250 NT \sin 30^\circ \cdot 2.0 = (5)(10)(1.0) + (10)(10)(2.0) \Rightarrow T(1.0) = 50 + 200 = 250 \Rightarrow T = 250~\text{N}.

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