Confusion on mechanics problem involving cart, blocks, and pulley

[jbriggs444]

I am convinced that the solution to this problem has to be a mistake for the reason that the accelerations of the top block $m_2$ and the hanging block $m_1$ are simply inconsistent with each other. To reiterate, the solution in item (b) says that the acceleration of the top block is $a$ whereas the solution in item (d) says that the acceleration of the hanging block is $a-A$. There is simply no conceivable way for the vertical acceleration of the hanging block to be $a-A$ if it was stated that the hanging block can only move in the vertical direction.

What reference frame are you adopting when you assert that the horizontal acceleration of $m_2$ is equal to the vertical acceleration of $m_2$?

If you are using the accelerating frame anchored to $M$ then you would be correct. What happens if you shift to the inertial rest frame of the floor? Is the horizontal acceleration of $m_2$ unaffected by the change in reference frame?

 

[niko_niko]

What reference frame are you adopting when you assert that the horizontal acceleration of $m_2$ is equal to the vertical acceleration of $m_2$?

If you are using the accelerating frame anchored to $M$ then you would be correct. What happens if you shift to the inertial rest frame of the floor? Is the horizontal acceleration of $m_2$ unaffected by the change in reference frame?

I shall refer to $m_2$ as the top block for convenience. As for the horizontal acceleration of the top block relative to the ground frame, it should be $a-A$. However, as I have said, the solution says it is simply $a$. As for $m_1$ which I shall refer to as the hanging block, the solution states that it's acceleration is $a-A$, however this can't be the case because the hanging block moves only vertically so the acceleration A of the cart must not be relevant for the case of the hanging block. Please see the solution for (b) and (d), I do not want to post it for the third time.

 

[jbriggs444]

I shall refer to $m_2$ as the top block for convenience. As for the horizontal acceleration of the top block relative to the ground frame, it should be $a-A$. However, as I have said, the solution says it is simply $a$. As for $m_1$ which I shall refer to as the hanging block, the solution states that it's acceleration is $a-A$, however this can't be the case because the hanging block moves only vertically so the acceleration A of the cart must not be relevant for the case of the hanging block. Please see the solution for (b) and (d), I do not want to post it for the third time.

Let me go back and review the proffered solution with which I think you disagree.

So contrary to your assertion $a$ is taken as the vertical acceleration of $m_1$ on the right. Not $m_2$ up top.

$a-A$ is then the horizontal acceleration of $m_2$ up on top. Just as you agree it should be. The pulley is accelerating leftward while block $m_2$ is accelerating rightward.

The rate at which the length of horizontal cord is shortening is given by an acceleration rate of $(a-A) - A = a$. The rate at which the length of vertical cord is lengthening is given by an acceleration rate of $a$. This consistency check passes!

Note that mass $m_1$ will be remaining horizontally in place while mass $M$ moves away to the left. From the point of view of the accelerating rest frame of $M$, mass $m_1$ will be swinging horizontally away.

Are we in agreement so far?

If we examine the equations, we have: $$m_2(a-A) = T$$The acceleration term is relative to the ground frame. We have no inertial pseudo forces to account for. The only horizontal force on $m_2$ is T. So that equation is fine.

We also have: $$m_1a = m_1g - T$$That is, the net downward force on $m_1$ is the downward force from gravity plus the upward force from tension. That seems straightforward enough.

What is your objection?

 

[niko_niko]

Let me go back and review the proffered solution with which I think you disagree.

View attachment 329177

So contrary to your assertion $a$ is taken as the vertical acceleration of $m_1$ on the right. Not $m_2$ up top.

$a-A$ is then the horizontal acceleration of $m_2$ up on top. Just as you agree it should be.

Note that mass $m_1$ will be remaining horizontally in place while mass $M$ moves away to the left. From the point of view of the accelerating rest frame of $M$, mass $m_1$ will be swinging horizontally away.

Are we in agreement so far?

If we examine the equations, we have: $$m_2(a-A) = T$$The acceleration term is relative to the ground frame. We have no inertial pseudo forces to account for. The only horizontal force on $m_2$ is T. So that equation is fine.

We also have: $$m_1a = m_1g - T$$That is the net downward force on $m_1$ is the downward force from gravity plus the upward force from tension. That seems straightforward enough.

What is your objection?

To everything you have stated, I agree. However, my objection does not come from the portion of the solution you have pointed out. Rather, here:

I believe the continued solution here contradicts with the established accelerations in the previous portion of the solution.

 

[jbriggs444]

To everything you have stated, I agree. However, my objection does not come from the portion of the solution you have pointed out. Rather, here:
View attachment 329180
View attachment 329179
I believe the continued solution here contradicts with the established accelerations in the previous portion of the solution.

What I see here is a formula for (b) that evaluates to $a$ which you have agreed is the acceleration of $m_1$. However, (b) was supposed to be about the acceleration of $m_2$.

I suspect that it is a correct answer for (d). Though I have not verified the formula.

What I also see here is a formula for (d) that evaluates to $a-A$ which you have agreed is the acceleration of $m_2$. However, (d) was supposed to be about the acceleration of $m_1$.

I suspect that this is a correct answer for (b). Though I have not verified the formula.

 

[niko_niko]

What I see here is a formula for (b) that evaluates to $a$ which you have agreed is the acceleration of $m_1$. However, (b) was supposed to be about the acceleration of $m_2$.

I suspect that it is a correct answer for (d). Though I have not verified the formula.

What I also see here is a formula for (d) that evaluates to $a-A$ which you have agreed is the acceleration of $m_2$. However, (d) was supposed to be about the acceleration of $m_1$.

I suspect that this is a correct answer for (b). Though I have not verified the formula.

That is exactly the crux of my issue. I was able to obtain the same equations in my solution but I was utterly confused why the author seemingly did a switcheroo.

 

[jbriggs444]

That is exactly the crux of my issue. I was able to obtain the same equations in my solution but I was utterly confused why the author seemingly did a switcheroo.

So your concern all along was not with the consistency of the accelerations. Or the correctness of the solutions. It was with labelling.

 

[haruspex]

What I see here is a formula for (b) that evaluates to $a$ which you have agreed is the acceleration of $m_1$. However, (b) was supposed to be about the acceleration of $m_2$.

I suspect that it is a correct answer for (d). Though I have not verified the formula.

What I also see here is a formula for (d) that evaluates to $a-A$ which you have agreed is the acceleration of $m_2$. However, (d) was supposed to be about the acceleration of $m_1$.

I suspect that this is a correct answer for (b). Though I have not verified the formula.

Did you see post #27?
I have verified the formula for $a$.