# Pulley system on a frictionless cart

Let's say you have a pulley set up as below on a cart, with a massless pulley and string. The mass hanging off the side is attached via a rail, and all surfaces & pulley are frictionless except between the tires and the ground (to allow for rolling, of course).

Furthermore, the mass of the weight hanging off the side is greater than that resting on top of the cart.

When the system is released from rest, will the cart begin to move or not? I would think not, because there is no force that could cause the cart to move -- since all surfaces are frictionless, it is as if the pulley and the cart are two separate entities.

However, what about the tension in the string and thus in the pulley connected to the cart? Does that not exert a lateral force capable of accelerating the cart?

Additionally, how does conservation of momentum play into this.

Since the system is initially at rest, the sum of the momentum vectors of each of the objects must add to zero at any point in time, right? So if the block on top of the cart is accelerating due to the tension in the rope due the force of gravity acting on the hanging block, does that mean the cart must move in the opposite direction so momentum is preserved?

Finally, where do non-inertial reference frames play a part in this? Since the cart is (potentially) accelerating taking the reference frame of the cart will lead to the introduction of "fictitious" forces. Is there any (possibly simpler) way to determine what will happen to the cart from this (non-inertial) reference frame?

• For this question, it doesn't matter that the hanging weight is larger than the resting weight. The result (excepting its magnitude) will be the same. – Daniel Griscom Oct 3 '15 at 11:45
• BTW: this is an example of the way to ask a very basic question on Physics. The focus is on physics and not on a particular incarnation of the problem. – dmckee Oct 3 '15 at 17:48
• Also, the final answer will depend a bit on whether the hanging mass runs on a rail down the side or is free to swing. – dmckee Oct 3 '15 at 17:49
• @dmckee In what way does it alter the outcome? – 1110101001 Oct 3 '15 at 20:24
• Well, it doesn't change the answer to "Does it move?", but it does change the answer to "How far has it gone before the falling block hits the ground? And how fast is it moving at that time?" because the falling block is no longer constrained to have the same horizontal velocity as the cart. – dmckee Oct 3 '15 at 20:28

## 2 Answers

Yes, the cart will move, due to the force applied by the string to the pulley.

To solve, calculate the string tension while the weights are moving, and then note that the pulley has to provide an opposing force in order to change the string's direction. The reaction to that force acts upon the cart, accelerating it.

Momentum is conserved, because the resting weight is accelerated to the right, while the cart is accelerated to the left.

Calculating the actual numbers will be entertaining, as you must include the cart's acceleration while calculating the string tension. I'm guessing that including a new, accelerating reference frame won't be helpful, as you won't know the magnitude of the acceleration until the problem has been solved.

• The reaction to that force acts upon the cart, accelerating it. — To clarify, the force being considered is the tension of the string pulling the block on top of the cart and the reaction force as per Newton's 3rd law is the block pulling upon the string, right? $$\\$$ However, if the pulley is frictionless why would the block exerting an opposite on the string affect the pulley? – 1110101001 Oct 3 '15 at 20:13
• @1110101001 There is tension on the string. Without the pulley, the string would be pulled towards a line between the two weights. To keep the string along its path, the pulley pushes it towards the upper-right. The reaction to that force is a force on the pulley towards the lower left. – Daniel Griscom Oct 3 '15 at 21:36
• Ah, that makes sense. As for actually calculating the tension, is it correct that it satisfies the system of equations? $$T=m_{1}\left( a-a_{M} \right)$$ $$T-m_{2}\text{g}=-m_{2}a$$ However, shouldn't this be equal to $$a_{M}\left( M+m_{2} \right)$$ $$\\$$ ($m_1$ and $m_2$ are the small blocks, $M$ is the large block, $a$ is the magnitudes of acceleration of the small blocks, and $a_m$ is the acceleration of the large block.) – 1110101001 Oct 4 '15 at 4:31
• It's best to ask follow-on questions like this as a complete new question. The goal of the site is to end up with clear questions having best answers; mixing further Q+A into the comments makes things more difficult to follow. – Daniel Griscom Oct 4 '15 at 11:35

From looking at the chart I would have to agree with everyone who said yes. But not roll but instead it would flip. Do to there being no friction stopping the cart from doing so and the hanging wait being heavier then the resting weight. If you add friction and weight of the cart you may come up with a variable that would fit the theory that the cart wouldn't move.

• To clarify, without static friction acting on the tires, the torque generated by gravity acting on the weight+pulley would cause it to flip, right? However, how would the addition of friction between the tires and ground change this scenario? There would need to be a torque in the opposite direction to counter it, but what generates this torque? – 1110101001 Oct 3 '15 at 19:27
• If you add friction to the scenario, as the larger weight falls the smaller weight would hit the pulley system sending the cart to rise on the other end and fall which would make it roll in the opposite direction of the weights. – timothy brachna Oct 3 '15 at 20:03
• Assume the cart has a uniform mass distribution and that the drawing is roughly to scale. Then even if the top mass is very light the cart could mass as little as 1/3 the hanging mass and still not flip. You could rig a scenario where it flips, but assuming that it will is a bit of a stretch. – dmckee Oct 3 '15 at 20:31
• well. if you think about it the scenario above it fully it would be that they would all be in constant movement, touching with no effect from anything they would just keep moving. Whether it would flip or not could not be thought of. adding a little friction would change the scenario all together. – timothy brachna Oct 3 '15 at 20:46