# Are the accelerations the same at either end of a moveable pully? [closed]

Given a moveable pulley with a fixed pulley on either side,

Is the acceleration of the left weight (m1) the same as that of the right weight (m2)?

Intuitively, I would imagine it to be, since if m1 drops by 10 metres, then m3 would rise by 5 metres, and thus m2 would drop by 10 metres. But mathematically the accelerations don't appear to be the same... Perhaps there is a mistake in my reasoning.

Assume the system is released from rest, an inextensible and massless string is used for the pulleys, and there is no friction forces acting on the system.

• Re, "I would imagine [that] if m1 drops by 10 metres, then m3 would rise by 5 metres, and thus m2 would drop by 10 metres." If m1 drops by 10 meters, then that's 10 meters of rope going over the left-hand pulley. Same goes for m2 and the right-hand pulley. If both of those masses drop by 10m, then you've removed 20m of rope from between the two upper pulleys.... Jun 10, 2021 at 14:00
• Also note, You seem to assume that there are two independent lengths of rope that both are rigidly attached to m3, but that's not what the drawing looks like to me. To my eyes, m3 looks like it is a third pulley, and so far, I see two answers below that both assume that m3 is a third pulley. If this is a homework problem, then you might want to ask for clarification on that point. Jun 10, 2021 at 14:02
• Just curious, but did my hint answer your question "Is the acceleration of the left weight (m1) the same as that of the right weight (m2)?"? Jun 11, 2021 at 23:01
• @BobD I think so because since they tension force is the same the net force must be different, is that correct? Jun 12, 2021 at 9:38
• Since the tension is the same the accelerations can only be the same if $m_{1}=m_2$. Do you see that? Jun 12, 2021 at 14:53

Hint: The tension in the string is the same throughout.

Hope this helps.

While it is certainly possible that $$x_1 = - \frac{1}{2} x_3$$ and that $$x_2 = - \frac{1}{2} x_3$$, as your intuition suggests, it is not necessarily the case. (We define $$x_1$$, $$x_2$$, and $$x_3$$ to be the displacements from the original positions of the masses.) If both of the equations were automatically true, then rope would never move relative to $$m_3$$, since the same amount of rope would be going over the side pulleys at all times. But hopefully it seems intuitive that you could come up with some highly unbalanced set of masses (with $$m_1 \gg m_2$$, for example) where the rope should slide over $$m_3$$.

Instead, you should look to find one equation that relates $$x_1$$, $$x_2$$, and $$x_3$$ all together. This equation will be a mathematical statement that the overall length of the rope does not change. Once you have this equation, you can differentiate it to get a parallel statement concerning the accelerations of each mass.

• @BobD: I guess I assumed that "frictionless" referred to the bearings of the pulley, and that there was the usual assumption that the rope moved in tandem with the pulleys (so static friction there, which does no work.) If the rope-pulley interface was truly frictionless, then you're right that the pulleys would never rotate. I would submit, however, that a situation where the pulleys move with the ropes is more realistic than one in which they slide frictionlessly over the pulley. Jun 10, 2021 at 15:19
• Do you agree the tension is the same throughout the string? Jun 10, 2021 at 15:37
• @BobD: Assuming that the pulleys have negligible mass, then yes. Jun 10, 2021 at 15:39
• OK, then if the tension is the same throughout the string, the net torque on the movable pulley will be zero. On the other hand, I suppose that if the pulley is massless its moment of inertia will be zero in which case it would require zero torque to rotate it. So I guess we can conclude the pulley can rotate provided it is massless. The trouble is the OP has assigned it a mass of m3 Jun 10, 2021 at 15:47
• No problem. That was initially in my head too! Jun 10, 2021 at 15:57