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In the given question, I'm required to figure out the velocity of block $C$, given that both blocks $A$ and $B$ move towards left with velocities $1$ $m/s$ each.

Its easy to figure out that in horizontal direction, as block $C$ is in contact with block $B$, it would move with $1$ $m/s$.

Doubt: How can I possibly find the velocity of block $C$ in the vertical direction. I know I have to relate the motion of blocks with the pulleys, but how can I possibly do this? Also, if I'm assuming the tension as $T$ in the longest string, what would the tensions in the other smaller strings? How can these tensions be related to one another?

I've literally tried to answer this question for 2 weeks but I'm failing time and again. Please let me know the general approach to solve problems of similar type, also if you happen to solve it just by simple observation , please let me know the logic . This really matters to me. Thanks for giving this your precious time

EDIT :Can I use the method in the answer in this post to solve this problem ? If yes , then how and if no then why not ? I tried but I couldn't solve it by this . Finding the acceleration of Block attached using tricky string setup

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  • $\begingroup$ Is the bottom left pulley moving relative to block B? Is the bottom left pulley moving relative to the bottom right pulley? Is the length of string between the block B and the bottom right pulley changing? $\endgroup$ – Farcher Oct 29 '17 at 14:44
  • $\begingroup$ Well , nothing of that sort is specified in the question , though like other questions , it can be assumed that all velocities are in the frame of reference of earth(ground).So , the length of the bottom string should not change . $\endgroup$ – Tanuj Oct 29 '17 at 14:47
  • $\begingroup$ Put another way., if you are sitting on block B do you see block A moving? $\endgroup$ – Farcher Oct 29 '17 at 14:54
  • $\begingroup$ No , I don't think so , No. $\endgroup$ – Tanuj Oct 29 '17 at 14:56
  • $\begingroup$ So you have answered the question as sitting on block B you see nothing moving. $\endgroup$ – Farcher Oct 29 '17 at 15:02
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Given that all three blocks $A, B, C$ are moving left with the same velocity, and that the string attached to $C$ is fixed at the other end to $B$, and that all of the pulleys are fixed to blocks $A$ and $B$, the only way block $C$ could have a vertical velocity would be if blocks $A$ and $B$ had differing velocities (changing separation).

We know that $A$ and $B$ are moving with the same velocity, which means that the string cannot love around any of the pulleys. Therefore, $C$ has no vertical component of its velocity.

Elaboration

In the general case, the only way such a block can move in a pulley system like this is if the string moves around the pulley. If the string is fixed at one end, and given that the string cannot be stretched, the only way it can move around the pulleys is if the pulleys themselves move.

Since the blocks are moving with the same velocity, we can consider them at rest relative to one another. Given that the pulleys are fixed to the blocks, they can also be considered at rest relative to one another. By definition, because they are at rest, they are not moving, and thus, the string is not moving over them.

In The Case of Differing Velocities

This problem is made simpler by the fact that all of the pulleys cause the string to turn at right angles.

In the case of, for example, block $B$ moving left at $4\text{ms}^{-1}$, the velocity of the leftmost pulley relative to the two rightmost pulleys is $3\text{ms}^{-1}$ to the left. The means that the length that the string has to travel to pass through the pulley system is decreasing by $6\text{ms}^{-1}$.

Since the string remains the same length, but the path through which it must travel does not, this allows block $C$ to have a vertical component of its velocity of $6\text{ms}^{-1}$ downwards.

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  • $\begingroup$ This is the correct answer dude ! Congrats. Can you possibly put a more elaborate and clearer explanation to it ? I'm a beginner and would really appreciate all of your help ! Thanks man ! :) $\endgroup$ – Tanuj Oct 29 '17 at 14:53
  • $\begingroup$ Here I would like to ask why the following method cannot be used to solve the following problem ? (Method in the answer )physics.stackexchange.com/questions/43875/… $\endgroup$ – Tanuj Oct 29 '17 at 14:57
  • $\begingroup$ This explains it pretty clear , but in the similar problem , if block A has a velocity 1 m/s and block B has a velocity 4 m/s , then how would you figure out the velocity of block C in this case ? $\endgroup$ – Tanuj Oct 29 '17 at 15:06
  • $\begingroup$ Just one word , WOW ! Hey is there anyway I can "ping" you for help with my future physics doubts here ? $\endgroup$ – Tanuj Oct 29 '17 at 15:28
  • $\begingroup$ What did you mean by this - "This problem is made simpler by the fact that all of the pulleys cause the string to turn at right angles." ? $\endgroup$ – Tanuj Oct 29 '17 at 15:29

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