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Consider the small weight is attached to a big diameter wheel and the big weight is attached to a smaller diameter cilinder. I've read that if you ignore friction and all the forces are balanced perfectly, you could consider the possibility of this machine lowering the lighter weight through the distance of one wheel circumference at uniform speed while the heavier weight is raised by the smaller distance of one circumference of the cylinder. In this situation, what would be the forces that balance out the differences in the gravitational forces that act on each mass?


1 Answer 1


Let's assume the weights are $m_1g$ and $m_2g$, the tensions in the chains are $T_1$ and $T_2$ and the radii of the pulleys are $r_1$ and $r_2$.

If the weights are moving up/down at a constant speed then they are in equilibrium, so we must have

$T_1 = m_1g \\ T_2=m_2g$

The rotational moments acting on the pulleys are $T_1r_1$ in one direction and $T_2r_2$ in the opposite direction. If the pulleys are rotating at a constant speed then these moments must net to zero, so

$T_1r_1 = T_2r_2 \\ \Rightarrow m_1r_1 = m_2r_2$

This is the condition for the weights to move at constant speed (in the absence of friction). There must, of course, also be a vertical force to support the axle of the hoist. If the hoist itself has weight $W$ then the vertical force required to support it is $T_1+T_2+W$, which is $(m_1+m_2)g + W$


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