Pulley system on a frictionless cart

Question

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?

Answer 1

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).

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.