Finding acceleration of the pulley block system

[Kashmir]

What is the initial acceleration of mass 5M .The pulleys are ideal and the string inextensible.

My attempt-

2Mg-T=2Ma (for 2M)

T=Ma (for M)

Solving we get T=2Mg/3

T-N=5MA (for 5M)

N=2MA (for 2M)

Solving we get A=2g/21
but the given ans. is 2g/23

 

[scottdave]

Draw some diagrams. Which direction will mass 5 be moving? How much mass is it moving in that direction?

 

[Chestermiller]

The 23 must be a typo.

 

[Kashmir]

The 23 must be a typo.

It seems 23 is correct.
Found one solution here https://www.toppr.com/ask/question/...ial-acceleration-of-the-wedge-of-mass-5m-the/

 

[Juanda]

I got 2g/21 as well.
Typos are not too common but it's already 3 of us getting that number.

 

[jbriggs444]

What is the initial acceleration of mass 5M .The pulleys are ideal and the string inextensible.
[...]
Solving we get A=2g/21
but the given ans. is 2g/23

You do not say, but it seems likely that all of surfaces are frictionless.

Let me go through your work line by line... With the result that I agree with your answer.

Oh shoot, but that answer is indeed wrong. If we are adopting the inertial frame then the pulleys are moving. The horizontal acceleration of the top block will not match the vertical acceleration of the right side block.

The posted answer addresses this by temporarily adopting the accelerating frame and including the resulting inertial pseudo-force on the top block in its calculation for $T$.

Which means that the remainder of this post is erroneous.

2Mg-T=2Ma (for 2M)

This is solid. It is based on a force balance in the vertical direction for the $2M$ block on the right. There is, we assume, no vertical friction from this horizontal contact.

T=Ma (for M)

Yes. This time it is a force balance in the horizontal direction for the $M$ block on the top. We do not care about the vertical direction. We assume that there is no horizontal friction.

Solving we get T=2Mg/3

Yes. The two tensions are equal because the pulley is ideal. The two accelerations are equal because the string is inextensible. You solve the simultaneous equations and determine the tension, $T$.

One can obtain the same result from a physical intuition rather than brute force algebra. You have total mass $3M$ accelerating under net force $2M/g$ so the acceleation will be $\frac{2g}{3}$. The tension required to produce this acceleration is then $\frac{2Mg}{3}$

T-N=5MA (for 5M)

Now you seem to be doing a horizontal force balance on the $5M$ block. Those horizontal forces will be: $2T$ to the right from the left hand pulley, $T$ to the left from the right hand pulley and $N$ to the left from the contact force with the $2M$ block. So yes, this appears to be correct.

N=2MA (for 2M)

This is the force balance in the horizontal direction for the right hand $2M$ block. All seems to be in order.

Solving we get A=2g/21

I will solve this differently. With a conservation of momentum argument. (Or a third law argument -- the two are equivalent).

We have the top $M$ block accelerating leftward under a net force of $\frac{2Mg}{3}$. Its rate of change of momentum over time is identical to this.

If momentum is conserved for the entire system and if the bottom $5M$ and $2M$ blocks accelerate together to the right then their rate of change of momentum over time must be equal and opposite: $\frac{2Mg}{3}$.

Given the bottom block's combined mass of $7M$, we can achieve this rate of change of momentum with an acceleration of:$$A = \frac {\frac{2Mg}{3}} {7M} = \frac{2g}{21}$$

 

[Kashmir]

I think 2g/23 is correct.

The problem i think is that we are assuming both accelerations for M and 2M to be same,which isn't true in this particular case.
2Mg-T=2Ma (for 2M)

T=Ma (for M)

The above two equations use the same value of a, which isn't correct.

 

[Juanda]

I think 2g/23 is correct.

The problem i think is that we are assuming both accelerations for M and 2M to be same,which isn't true in this particular case.
2Mg-T=2Ma (for 2M)

T=Ma (for M)

The above two equations use the same value of a, which isn't correct.

But that's true because they're connected by a perfect string that's always in tension.
If you move the M block an x distance to the right, the 2M block will move the same distance upwards.
If their positions are directly related like that then their velocities and accelerations are equally linked. You'd obtain that conclusion through derivation.

 

[jbriggs444]

But that's true because they're connected by a perfect string that's always in tension.

That would be true if the pulleys were stationary. They are not.

 

[Juanda]

That would be true if the pulleys were stationary. They are not.

Since pulleys are massless, it is indifferent whether they are stationary or not.

It could be assumed they're "made of ice" and the ropes are sliding on them without any friction obtaining the same result in the process.

 

[Chestermiller]

As reckoned by an observer traveling to the right with the same acceleration "a" as that of the block relative to the ground (non-inertial frame of reference), the situation is the same as if there were a body force to the left on each of the masses, with a horizontal pseudo-gravitational acceleration equal to a. So, in this frame of reference, the force balances on the masses would be $$2Mg-T=2Ma_s$$and $$T+aM=Ma_s$$where $a_s$ is the downward acceleration of the 2M mass. If we eliminate T between these two equations, we obtain:$$a_s=\frac{2g+a}{3}$$So, $$T=M(a_s-a)=M\frac{2(g-a)}{3}=7Ma$$What does this give you for a?

 

[jbriggs444]

Since pulleys are massless, it is indifferent whether they are stationary or not.

It could be assumed they're "made of ice" and the ropes are sliding on them without any friction obtaining the same result in the process.

If you adopt the accelerating frame of reference where the big $5M$ block is at rest then you need an extra pseudo-force to explain the motion of the top $2M$ block.

If you adopt the inertial frame of reference then the pulleys are accelerating. The acceleration of the $1M$ top block will not match the acceleration of the $2M$ right hand block because the pulleys are accelerating.

Look at it this way:

In the accelerating frame, it is abundantly clear that the horizontal acceleration of the top ($1M$) block matches the vertical acceleration of the right hand ($2M$) block.

If we shift to the inertial frame, then we will have changed the acceleration of the top block without having changed the acceleration of the right hand block. So it is clear that they can no longer match.

 

[Chestermiller]

If you adopt the accelerating frame of reference where the big $5M$ block is at rest then you need an extra pseudo-force to explain the motion of the top $2M$ block.

If you adopt the inertial frame of reference then the pulleys are accelerating. The acceleration of the $1M$ top block will not match the acceleration of the $2M$ right hand block because the pulleys are accelerating.

Look at it this way:

In the accelerating frame, it is abundantly clear that the horizontal acceleration of the top ($1M$) block matches the vertical acceleration of the right hand ($2M$) block.

If we shift to the inertial frame, then we will have changed the acceleration of the top block without having changed the acceleration of the right hand block. So it is clear that they can no longer match.

See my post # 11.

 

[Juanda]

If you adopt the accelerating frame of reference where the big $5M$ block is at rest then you need an extra pseudo-force to explain the motion of the top $2M$ block.

If you adopt the inertial frame of reference then the pulleys are accelerating. The acceleration of the $1M$ top block will not match the acceleration of the $2M$ right hand block because the pulleys are accelerating.

Look at it this way:

In the accelerating frame, it is abundantly clear that the horizontal acceleration of the top ($1M$) block matches the vertical acceleration of the right hand ($2M$) block.

If we shift to the inertial frame, then we will have changed the acceleration of the top block without having changed the acceleration of the right hand block. So it is clear that they can no longer match.

I misunderstood what you meant. I thought you were talking about the rotation of the pulleys, not their linear movement.

However, I still cannot agree with you. Even if the pulley is linearly accelerating because it is part of the body 5M, the x movement of M (top block) will be the same as the y movement of 2M (vertical block) because the length of the string is fixed. Then, if the pulleys are massless or frictionless (the point is they don't contribute through rotation) the tension of the string will be the same on both sides.

As an example of what I mean, see the following:

The length of the string is independent of the acceleration of the bodies so the relation must stay true whether any of the bodies are accelerating or not.

If still in doubt, let's see the problem with forces being involved and the pulley being able to accelerate. External forces are shown in orange (weight and force on the pulley) and internal forces are in blue (tension in the strings).

TO SUM UP, in the problem from OP and the one I'm showing the tension is the same on both sides of the pulley as long as the pulley has no rotational inertia or it is not rotating at all (rope sliding without friction over it). Also, the movement of one block will be the same as the other in the axial direction of the rope.

If the pulley had rotational inertia it would be necessary to add how its angle relates to the position of the blocks. All the relevant independent equations are shown in the following picture:

See how I didn't use accelerated frames at any point. To be honest, I think there is no point on doing so. They add no insight and just make everything more confusing in my opinion.

 

[jbriggs444]

However, I still cannot agree with you. Even if the pulley is linearly accelerating because it is part of the body 5M, the x movement of M (top block) will be the same as the y movement of 2M (vertical block) because the length of the string is fixed. Then, if the pulleys are massless or frictionless (the point is they don't contribute through rotation) the tension of the string will be the same on both sides.

We agree that the tension throughout the string is the same. That much is not in question.

As an example of what I mean, see the following:
View attachment 327865

The length of the string is independent of the acceleration of the bodies so the relation must stay true whether any of the bodies are accelerating or not.

The length of the string is fixed independent of the acceleration of the bodies. We agree on this as well.

However, your diagram here is a good one. You would say, I am sure, that the upward acceleration of $m_1$ here is equal to the downward acceleration of $m_2$. Right?

But if the pulley is accelerating upward then this is no longer the case.
Instead, if the upward acceleration of the pulley is $a_p$, the upward acceleration of $m_1$ is $a_1$ and the downward acceleration of $m_2$ is $a_2$ then can you agree that the rate that the string is shortening on the left hand side is $a_1 - a_p$ and the rate that the string is lengthening on the right hand side is $a_2 + a_p$?

Setting those equal, we have $a_1 = a_2 + 2a_p$. The acceleration of the pulley matters!

See how I didn't use accelerated frames at any point. To be honest, I think there is no point on doing so. They add no insight and just make everything more confusing in my opinion.

You are mistaken.

 

[Chestermiller]

Both approaches are equivalent. But, for this particular problem, solving it using a non-inertial frame of reference seemed much easier to me.

 

[Juanda]

But if the pulley is accelerating upward then this is no longer the case.

I reviewed the problem and you are right. The accelerations of the two bodies are not necessarily the same since it depends on the linear acceleration of the pulley too.

I needed to plot the diagram again to fully understand the situation and this is what I got. For clarity's sake, I defined a global inertial frame positioned lower than all the bodies.

Therefore, the example I proposed would be solved with the 6 following equations that allow us to calculate the 6 variables $T_1, T_2, \ddot{x}_1, \ddot{x}_2, \ddot{x}_3, \ddot{\theta }_3$:

NOTE: The problem could be simplified by assuming the pulley doesn't have rotational inertia or that its angular acceleration is 0 because there is no friction with the rope so it will simply slide over it.

You are mistaken.

I don't see how I could be mistaken on a personal preference. Is it mistaken to prefer pizza over burgers? What I meant by non-inertial frames not being especially useful to me is that any situation in classical physics can be described from an inertial frame instead of a non-inertial frame and it makes things clearer in my opinion. Sometimes this actually might make the math involved a little harder but clearer nonetheless. Fortunately, I no longer need to use "tricks" to solve problems quicker during tests so I can choose the inertial frame even if the process is significantly longer. Maybe one day I'll trust myself enough to solve problems using non-inertial frames but such day is yet to come.

See my post # 11.

I did and most likely it is correct. However, I didn't quite get it. I've always had troubles with non-inertial frames and that's a fault of my own.

Both approaches are equivalent. But, for this particular problem, solving it using a non-inertial frame of reference seemed much easier to me.

I believe I was initially wrong as discussed in this message. @jbriggs444 was right in calling out my mistake.
But it's true that the problem can be solved using both inertial and non-inertial frames.
Now, going back to OP's initial post, I solved the problem again following the same procedure as with the example I provided with an inertial frame, and this time I did get that the acceleration of the 5M body will be $a_{5M} =\frac{2g}{23}$.

 

[Chestermiller]

ANALYSIS USING INERTIAL FRAME OF REFERENCE

The kinematics of the motion are such that, if $a_s$ is the downward acceleration of the 2M mass, then the horizontal acceleration of the upper block of mass M relative to the left pulley is $-a_s$ (i.e., the mass M is accelerating in the negative x direction relative to the pulley). The left pulley is attached to the block 5M, which is accelerating to the right with horizontal acceleration $a_b$ in the inertial frame of the table top. So, relative to the inertial frame of the table top, the block has an acceleration of $-a_s+a_b$ in the positive x direction. So using an inertial frame of reference, the horizontal force balance on the mass M must read: $$-T=M(-a_s+a_b)$$or equivalently, $$T=M(a_s-a_b)$$

This is equivalent to the force balance I wrote in Post #11 for the mass M as reckoned from the non-inertial frame of the 5M block using a pseudo-force in the negative x-direction.

 

[erobz]

Just to show some more steps:

The global frame of reference is fixed to the ground, and the FBD's show forces relevant to solving the problem.

Horizontal Direction for Block 1M

$$ T = M( a_1 - a_5) \tag{1}$$

Edit: $a_1$ is w.r.t the block 5M.

Vertical Direction for Block 2M:

$$ \begin{aligned} 2Mg - T &= 2Ma_2 \\ & \implies T = 2M(g-a_2) \quad\quad \text{(2)} \end{aligned}$$Given a positive acceleration $a_2$ will result in a positive acceleration $a_1$, hence $a_2 = a_1$, combine (1) and (2) with the constraint:

$$ \begin{aligned} M( a_1 -a_5) &=2M(g-a_2) \\ &= 2M(g-a_1) \\ & \implies a_1 = \frac{2g+a_5}{3} \quad\quad \text{(3)} \end{aligned}$$Horizontal Direction of Block 2M:

$$ \begin{aligned} -N &= -2Ma_5 \\ & \implies N = 2Ma_5 \quad\quad \text{(4)} \end{aligned}$$Horizontal Direction of Block 5M:

$$ \begin{aligned} T -2T + N &= -5Ma_5 \\ -T + N &= -5Ma_5\quad\quad \text{(5)} \end{aligned} $$

Combine (4) and (5) (substituting for N) to yield:

$$T = 7Ma_5 \tag{6}$$

Lastly combine (1) , (3) and (6) to eliminate $T$ and $a_1$:

$$ \begin{aligned} \cancel{M}(a_1-a_5) = 7\cancel{M}a_5 \\ \frac{2g+a_5}{3} - a_5 = 7a_5 \\ & \implies a5 = \frac{2}{23}g \end{aligned}$$

I initially kept getting the wrong answer until realized that I had assumed a negative acceleration for block 5M in the horizontal force balance for block 1M. As a result, I needed to keep that assumption throughout the analysis (hence the negative signs on the RHS of (4) and (5) become important). Anyhow, thought I would share my trials and tribulations in case others were having the same difficulty there.

 

[Chestermiller]

Just to show some more steps:

View attachment 327898The global frame of reference is fixed to the ground, and the FBD's show forces relevant to solving the problem.

Horizontal Direction for Block 1M

$$ T = M( a_1 - a_5) \tag{1}$$

Vertical Direction for Block 2M:

$$ \begin{aligned} 2Mg - T &= 2Ma_2 \\ & \implies T = 2M(g-a_2) \quad\quad \text{(2)} \end{aligned}$$Given a positive acceleration $a_2$ will result in a positive acceleration $a_1$, hence $a_2 = a_1$, combine (1) and (2) with the constraint:

$$ \begin{aligned} M( a_1 -a_5) &=2M(g-a_2) \\ &= 2M(g-a_1) \\ & \implies a_1 = \frac{2g+a_5}{3} \quad\quad \text{(3)} \end{aligned}$$Horizontal Direction of Block 2M:

$$ \begin{aligned} -N &= -2Ma_5 \\ & \implies N = 2Ma_5 \quad\quad \text{(4)} \end{aligned}$$Horizontal Direction of Block 5M:

$$ \begin{aligned} T -2T + N &= -5Ma_5 \\ -T + N &= -5Ma_5\quad\quad \text{(5)} \end{aligned} $$

Combine (4) and (5) (substituting for N) to yield:

$$T = 7Ma_5 \tag{6}$$

Lastly combine (1) , (3) and (6) to eliminate $T$ and $a_1$:

$$ \begin{aligned} \cancel{M}(a_1-a_5) = 7\cancel{M}a_5 \\ \frac{2g+a_5}{3} - a_5 = 7a_5 \\ & \implies a5 = \frac{2}{23}g \end{aligned}$$

I initially kept getting the wrong answer until realized that I had assumed a negative acceleration for block 5M in the horizontal force balance for block 1M. As a result, I needed to keep that assumption throughout the analysis (hence the negative signs on the RHS of (4) and (5) become important). Anyhow, thought I would share my trials and tribulations in case others were having the same difficulty there.

It is important to mention that $a_1$ is the acceleration of block M relative to block 5M.

 

[erobz]

It is important to mention that $a_1$ is the acceleration of block M relative to block 5M.

I’m not sure it is? If the block 5M was fixed $a_5 =0$ the equations 1 and 2 reduce to $a_1 = 2/3g$. That’s w.r.t. to the inertial frame. To me the quantity $( a_1-a_5)$ is the acceleration of block 1 w.r.t. the inertial frame. Maybe I’m under thinking it.

 

[Chestermiller]

I’m not sure it is? If the block 5M was fixed $a_5 =0$ the equations 1 and 2 reduce to $a_1 = 2/3g$. That’s w.r.t. to the inertial frame. To me the quantity $( a_1-a_5)$ is the acceleration of block 1 w.r.t. the inertial frame. Maybe I’m under thinking it.

That’s correct. Isn’t that consistent with my comment?

 

[erobz]

That’s correct. Isn’t that consistent with my comment?

Yeah, I think it is. I was under thinking it. $a_1$ is not with respect to the inertial frame, it’s relative to a fixed position on block 5M as you suggest.

 

[haruspex]

I let a be the vertically down acceleration of 2M and a' be the acceleration to the right of 5M (and thus of 2M). T is the tension.
$2Ma'=2Mg-T$
M accelerates at a to the left relative to 5M:
$M(a'-a)=T$
Considering the horizontal forces on the 5M+2M+pulleys system:
$7Ma=T$ ( the two Ts from the top string cancel).
Easy from there.