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?

enter image description here

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?

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    $\begingroup$ 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. $\endgroup$ Oct 3, 2015 at 11:45
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    $\begingroup$ 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. $\endgroup$ Oct 3, 2015 at 17:48
  • $\begingroup$ 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. $\endgroup$ Oct 3, 2015 at 17:49
  • $\begingroup$ @dmckee In what way does it alter the outcome? $\endgroup$
    – 1110101001
    Oct 3, 2015 at 20:24
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    $\begingroup$ 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. $\endgroup$ Oct 3, 2015 at 20:28

5 Answers 5


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.

Edit: as noted in a comment by dmmckee, the answer will depend on whether the hanging weight is constrained to stay in contact with the cart, or is free to swing away from it (which it would do if allowed).

  • $\begingroup$ 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? $\endgroup$
    – 1110101001
    Oct 3, 2015 at 20:13
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    $\begingroup$ @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. $\endgroup$ Oct 3, 2015 at 21:36
  • $\begingroup$ 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.) $\endgroup$
    – 1110101001
    Oct 4, 2015 at 4:31
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    $\begingroup$ 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. $\endgroup$ Oct 4, 2015 at 11:35
  • $\begingroup$ @Buraian That's the smallest edit I've ever seen! And, thanks for adding the bounty (whether or not my answer ends up getting it). $\endgroup$ Jan 2, 2021 at 17:26

Let the mass of cart be $M$, mass of hanging weight be $m_1$ and mass on top of cart be $m_2$. Choosing and inertial coordinate system, let the x coordinate of $M$ , $m_1$ and $m_2$ be $X$ , $x_1$ and $x_2$ respectively. Let $T$ be the tension in the string and $a$ be the acceleration of the masses ( horizontal for $m_2$ and vertical for $m_1$).

Equations of motion give:

$$ m_1g -T=m_1a $$ $$ T=m_2a \tag{0}$$ Hence, $a=m_1g/(m_1+m_2) \tag{1}$

let $X_{cm}$ be the X centre of mass coordinate of the system. $$X_{cm}=\frac{m_1x_1+m_2x_2+MX}{m_1+m_2+M}$$

differentiating twice; $$\ddot{X}_{cm}=\frac{m_1\ddot{x}_1+m_2\ddot{x}_2+M\ddot{X}}{m_1+m_2+M} \tag{2}$$ Since $\ddot{X}_{cm} =0 $ , $\ddot{x}_1=\ddot{X}$,($m_1$ does not swing ) and $\ddot{x}_2=a$ , $$m_1\ddot{X} + m_2a+ M\ddot{X} = 0 \tag{3}$$ We can use eq(1) and eq(3) to find $X, x_1, x_2$ as a function of time.

Method 2:

Since @Buraian wants equations with the method @Daniel Griscom suggested, here they are: Consider the part of the string that is in contact with the pulley. It experiences a force $T$ downwards and $T$ towards the left. Say the pulley applies a force of $N_1$ on the string ( towards upper right). By Newton's third law, the string applies $N_1$ on mass $M$ (towards bottom left).

Since $m_1$ doesn't swing, the rail (part of mass $M$) applies a force $N_2$ (towards left) on $m_1$ and $m_1$ applies a force of $N_2$ on $M$ towards right. Let $M$ (and $m_1$)accelerate with $a^{'}$, horizontally.

Since the net force on a mass less string is always $0$, $$N_1cos(45^0)=T \tag{4}$$ from eq(1),eq(0) and eq(4) we can get the value of $N_1$.

Equation of motion for $M$: $$N_1cos(45^0)-N_2 = Ma^{'} \tag{5}$$ Equation of motion for $m_1$ (X direction ): $$N_2=m_1a^{'}\tag{6}$$

Needless to say, by eq(5) and eq(6) we can obtain all the quantities and predict the motions of blocks. we get $a^{'}$ which is the same as $\ddot{X}=m_1m_2g/((m_1+m_2)(m_1+M))$ from first method.


Okay the pulley would definitely move, I will set up the laws to be used to show that it will.

First consider the three particle system, the cart and the two blocks. It is clear that the net momentum must be conserved along the x direction (in fact it should be zero). And since the net momentum is zero, we could also state it as the center of mass doesn't move along the x direction.

Next since the two blocks are connected by the string remember that they are constrained to move by the same amount with respect to the block.

Now say the B goes down by x, which means A must move to the right by x (with respect to the cart C)

Now that brings an imbalance to the $x_{com}$, so the Cart itself must move to the right by some amount. Qualitatively this is enough to say that the cart will move, and it will move to the left.

  • $\begingroup$ Describe the force which causes it to move @SK Dash , I understood the existence of this phenomena already $\endgroup$ Jan 3, 2021 at 7:41

Use an inertial reference frame. $T$ is tension in string, $M$ is mass of hanging weight, $m$ is mass of weight on top of cart, $M_c$ is mass of cart . $Mg - T = Ma$ and $T = ma$ where a is the acceleration of $m$ to the right and $M$ down in the inertial frame (relative to an observer on the ground). So $a = Mg/(M + m)$. Considering the cart and the two masses as a system, the only external forces are gravity and the constraint from the surface on the cart; both these forces are in the vertical direction. Since there is no net external force in the horizontal direction, the momentum in the horizontal direction is constant. $m$ moves with acceleration $a$ in the horizontal direction (taken as positive to the right). For a simple solution, assume $M$ is constrained to not "swing". To conserve momentum in the horizontal direction, $M_c$ and $M$ move with velocity $v$ such that $m\int_{0}^{t}a \enspace dt + (M_c + M)v = 0$, so $M_c$ and $M$ have velocity in the horizontal direction of ${-(mMg)t \over (M + m)(M_c+m)}$; this is the velocity of the cart (to the left) as $M$ is falling.

I see this this is basically the same answer @Sai Srikar Valiveru has provided.

In a non-inertial reference frame with acceleration $a = Mg/(M + m)$ to the right, $m$ is stationary and experiences a fictitious force $-ma$ (to the left) and in this frame $-ma + T = 0$. (The tension is the same in both the inertial and non-inertial reference frames as it must be.) For the system consisting of $m$, $M$, and $M_c$, in this non-inertial frame momentum in the horizontal direction is not conserved because of the external fictitious force $-ma$ present in this frame. So the problem is more difficult to solve in this frame than in the inertial frame previously used.

  • $\begingroup$ I don't get this , if the mass on top of the big block moves with some acceleration, how does that the big block get same velocity? There is no friction and all $\endgroup$ Jan 8, 2021 at 9:43
  • $\begingroup$ But I think I get the idea now, in a real case, if the surface of big block was frictionless then the block on top of big block's horizontal motion woulc cause a swing , is that correct?) $\endgroup$ Jan 8, 2021 at 9:44
  • $\begingroup$ Not quite. The normal reaction $N_1$ and $N_2$ accelerate $M$. Since my previous answer wasn't satisfactory, I have updated it. $\endgroup$ Jan 10, 2021 at 14:11

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.

  • $\begingroup$ 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? $\endgroup$
    – 1110101001
    Oct 3, 2015 at 19:27
  • $\begingroup$ 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. $\endgroup$ Oct 3, 2015 at 20:03
  • $\begingroup$ 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. $\endgroup$ Oct 3, 2015 at 20:31
  • $\begingroup$ 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. $\endgroup$ Oct 3, 2015 at 20:46

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