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Rotational Equilibrium and Newton's First Law in Rotational Form

AP Physics 1· difficulty 3/5

W A B

A uniform 20 kg20~\text{kg} horizontal beam is 3.0 m3.0~\text{m} long and supported at each end (A on the left, B on the right). A 40 kg40~\text{kg} sign hangs 2.5 m2.5~\text{m} from A. Using g≈10 m/s2g \approx 10~\text{m/s}^2, what is the upward force at support B?

  • A

    270 N270~\text{N}

  • B

    430 N430~\text{N}

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  • C

    600 N600~\text{N}

  • D

    333 N333~\text{N}

Explanation

Take torques about A: FB(3.0)=(20)(10)(1.5)+(40)(10)(2.5)=300+1000=1300F_B (3.0) = (20)(10)(1.5) + (40)(10)(2.5) = 300 + 1000 = 1300. So FB=1300/3.0≈430 NF_B = 1300/3.0 \approx 430~\text{N}.

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