A block of mass m2 = 100 kg is to be hoisted up using the pulley system shown be
ID: 1783827 • Letter: A
Question
A block of mass m2 = 100 kg is to be hoisted up using the pulley system shown below. The pulley consists of two cylinders of different radii, R1 = 0.75 m and R2 = 0.25 m, that are stuck together and which have a total moment of inertia of 10 kg·m2 about their central axes. The pulley itself is secured to the back wall in such a way that it is free to rotate about its center without friction. (a) Assuming that the ropes do not slip, with what force must someone pull on the rope wrapped around the R1 cylinder to just lift the block off the floor? (b) If this person pulls with a force of 500 N, what would be the linear acceleration of the block? (Careful: how are linear/tangential and rotational quantities related for spinning objects?)
Explanation / Answer
In equilibrium
net torque = 0
F*R1 - M2*g*R2 = 0
F*R1 = M2*g*R2
F = M2*g*(R2/R1)
F = 100*9.8*(0.25/0.75)
F = 326.7 N
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if F = 500 N
net torque = F*R1 - M2*g*R2
from newtons law
net torque = I*alpha
alpha = angular acceleration = a/R2
F*R1 - M2*g*R2 = I*a/R2
500*0.75 - (100*9.8*0.25) = 10*a/0.25
linear acceleration a = 3.25 m/s^2 <<<<-------ANSWER
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