A 0.117 kg meterstick is supported at its 37.7 cm mark by a string attached to t

Question

A 0.117 kg meterstick is supported at its 37.7 cm mark by a string attached to the ceiling. A 0.624 kg mass hangs vertically from the 4.04 cm mark. A mass is attached somewhere on the meterstick to keep it horizontal and in both rotational and translational equilibrium. The force applied by the string attaching the meter stick to the ceiling is 23.9 N. Find the value of the unknown mass. The acceleration of gravity is 9.81 m/s^2 . Answer in units of kg. Find the point where the mass attaches to the stick. Answer in units of cm

Explanation / Answer

24.8 = 0.116 . 9.81 + 0.688 . 9.81 + M . 9.81 (Upward force = downward force ) M = 1.724 kg 0.116 . 12.3 + 1.724 . X = 0.668 . 33.45 (Torques cancel) X = 12.13329 So the mass is at the 49.8 cm mark

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