Rotational Equilibrium and Newton's First Law in Rotational Form

AP Physics 1· difficulty 4/5

A uniform 4.0 m beam of mass 10 kg is supported by a pivot 1.0 m from one end. To balance, what mass should be hung at the closer end? (g=10g = 10 m/s²; treat beam center at 2.0 m from short end)

  • A

    40 kg

  • B

    5 kg

  • C

    20 kg

  • D

    10 kg

    check_circle

Explanation

Beam center of mass is 2.0 m from each end, so it sits 1.0 m on the FAR side of the pivot. Beam weight produces 10g1.0=10010 \cdot g \cdot 1.0 = 100 N·m of torque about the pivot. The mass MM at the short end sits 1.0 m on the opposite side. Balance: Mg1.0=100M g \cdot 1.0 = 100 N·m, so M=10M = 10 kg.

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