AP Physics 1· difficulty 4/5
A uniform 100 N beam of length 4 m rests on a pivot at its center. A 60 N weight hangs from one end. To keep the beam horizontal, you must apply a downward force at the opposite end of magnitude
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Explanation
Beam's CG is over the pivot, contributing zero torque. The 60 N hanging weight produces 60·2 = 120 N·m on its side. To balance, apply 60 N at the far end (also 2 m from pivot): 60·2 = 120 N·m.
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