By means of a rope whose mass is negligible, two blocks are suspended over a pul

Question

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.3 kg and m2 = 45.0 kg. The pulley can be treated as a uniform, solid, cylindrical disk. The downward acceleration of the 45.0 kg block is observed to be exactly one-half the acceleration due to gravity. Noting that the tension in the rope is not the same on each side of the pulley, find the mass of the pulley.

Explanation / Answer

Given that m1 = 12.3 kg m2 = 45.0 kg a = 4.9 m/s2 M = 2.0 kg The equation motion for blocks and pulley are t = t2 - t1 = I a = r ( T2-T1) = 0.5 M r2 a ==> T2 - T1 = 0.5 M r a T2 - T1 = 0.5 M a ...........1 For m1 block T1 - m1 g = m1 a .......2 For m2 block m2 g - T2 = m2 a .......3 ==> T1 - T2 -m1g + m2g = m1a+m2a ==> - 0.5 M a -m1g + m2g = m2a+m1a ==> M = - 2 [m2a+m1a + m1g - m2g] / a = - 2 [m2 + m1] - 2[m1-m2] (g/a) = -2[57.3] - 2[-32.7](2) = 16.2 kg

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