# Pulley problem - unable to understand the concept

Given:

• The pulley is moving towards the right.
• All blocks have different masses. (The pulley and the strings are massless.)

What I don't understand:

• Is the tension the same for both A and B?
• If the tension is the same, then both blocks should have different accelerations, but this is not true?
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Is the tension the same for both A and B?

Yes. A massless string needs to have constant tension, otherwise it would feel a net force and have infinite acceleration.

If the tension is the same, then both blocks should have different accelerations

Yes.

but this is not true?

Why do you think the boxes should have the same acceleration? I can't help find the flaw in your reasoning if you don't say why you think that.

Incidentally, the details of the situation are rather unclear. It is not obvious which way gravity points or what the precise nature of the box is. However, I can tell you that the detail that the pulley is moving to the right is immaterial, due to the principle of relativity.

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A quick look at this problem reveals a few simple relations:

The pulley is experiencing forces both to the left and the right.

To the right it experiences tension from the string due to the block C pulling downwards due to gravity (I guess block C is suspended?) Let this tension be T. The value of tension can be obtained by $T = m_Cg$

To the left it experiences twice the tension from the string to the left due to the string dragging masses A and B along the table. Let this tension be $T'$. However the pulley experiences twice the tension as it is pulled by twice by the same string (eg: the string is looped around the pulley)

As the sum of forces on the pulley is

$F = T + 2T'$

...and since the pulley is massless $m = 0$.

Therefore, the sum of forces on the pulley $=0$ ($m = 0$ hence $F = ma = 0$).

...and from this we can say that $T = -2T'$ from our previous relation! (The negative sign simply implies that the tensions are acting in opposite directions)

Yes, the tension in the same string will be the same.

Now, since the only force acting on A is $T'$, acceleration of A can be calculated as:

$T' = m_Aa_A$

Therefore:

$a_A = T' / m_A$

Similarly with B:

$T' = m_Ba_B$

Therefore:

$a_B = T' / m_B$

As the acceleration of A and B vary due to their different masses, we can now say that they will move with different accelerations!

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shortstheory, another problem i am facing is that a massless pulley can never accelerate as net force is always zero on it.like here t and 2t' act on it but pulley accelerates towards right and t=2t'(magnitude). explain, thank you. –  Chris Sep 8 '13 at 4:04
It's somewhat counter-intuitive. The net force on the pulley is the sum of $T$ and $-2T'$ (which act in opposite directions, hence the -ve sign in front of $2T'$. Although the net force on the pulley is 0, you must remember that the mass of the pulley is 0. So even if $a$ is non-zero, the product of $m$ and $a$ will be 0. So, you can accelerate a massless object without applying any force. Of course, this doesn't apply in real life situations because massless objects do not exist. You can only see this mathematically. –  shortstheory Sep 8 '13 at 13:18
thanks for reply.i got it. can you suggest some good lectures or websites on all constraints in mechanics from the very fundamentals to high level.thanks. –  Chris Sep 8 '13 at 17:24
Welcome mate :) With regards to your questions about websites, I think khanacademy.org is a good place to start to build up your fundamentals for high school level problems. But if you really want a wide range of levels, I highly suggest you watch Walter Lewin's lectures on classical mechanics at MIT's OpenCourseWare site here: ocw.mit.edu/courses/physics/… –  shortstheory Sep 8 '13 at 18:49
Also, since you've understood the answer, mark this as this as the accepted answer by clicking the checkbox on side. –  shortstheory Sep 8 '13 at 19:03

The masses will have different accelerations because the pulley will rotate.

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