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 ceil- ing. 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. The acceleration of gravity is 9.81 m/s2 . -Find the point where the mass attaches to the stick. Answer in units of cm

Explanation / Answer

force up = 23.9N weights down =(0.111kg+0.622kg+x)*9.81m/s/s total weight down = 7.18N + 9.81x where x is the unknown mass so we have 23.9=7.18+9.81x => x=1.70kg find the location of this mass by employing rotational equilibrium; the sum of all torques must be zero let's sum torques around the midpoint... the string generates a torque of 23.9N*0.112m clockwise (the string is 11.2 cm from the midpoing) the 0.622 mass generates a torque of 0.622kg*9.81m/s/s*0.453m counterclockwise the mass of the stick generates no torque around its center, so we have 2.68 -2.76=1.70kg*9.81m/s/s*d where d is the distance from the midpoint d=-0.0052 or 0.5cm to the right of the midpoint, or 50.5 cm

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