a .12 kg meterstick is supported at its 32.4cm mark by astring attached to the c

Question

a .12 kg meterstick is supported at its 32.4cm mark by astring attached to the ceiling. a .63 kg mass hangs vertically fromthe 4.99 cm mark. a mass is attached somewhere on the meterstick tokeep it horizontal and in both rotational and translationalequilibrium. the force applied by the string attaching the meterstick to the ceiling is 23.8N. find the value of the unknown mass.the acceleration of gravity is 9.81m/s2. answer in unitsof kg.

Explanation / Answer

Given that M = 0.12kg m1 = 0.63 kg m2 = ? T = 23.8N The vertical component of forces        (M+ m1 + m2 )g = T ==> m2 = (T/g) - M - m1 = 2.428 - 0.12-0.63 =1.678kg Therefore unknown mass m2 = 1.678 kg

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