# 9.3 Working with moving pulleys  (Page 2/3)

 Page 2 / 3

## Example

Problem : In the arrangement shown the accelerations of blocks “1”, “2” and “3” are 2 $m/{s}^{2}$ , 8 $m/{s}^{2}$ and 2 $m/{s}^{2}$ respectively in upward direction. The masses of the pulleys and string are negligible. Friction is absent everywhere. Find the acceleration of the block “4”.

Solution : The accelerations of three blocks are given with respect to ground. For analysis in moving frames of pulleys, we assign directions of relative accelerations as shown in the figure below. The pulley “B” and “C” are hanging from fixed pulley, “A”. If pulley “B” moves up, then pulley “C” moves down. The blocks “1” and “2”, in turn, are hanging from moving pulley “B”. If relative acceleration of “1” is up, then relative acceleration of “2” is down as they are connected by one string. Similarly, the blocks “3” and “4” are hanging from moving pulley “C”. If relative acceleration of “3” is up, then relative acceleration of “4” is down as they are connected by one string.

Let us now consider upward direction as positive. Since relative acceleration of block "1" is equal and opposite to relative acceleration of block "2", we have :

${a}_{1B}=-{a}_{2B}$

Expanding in terms of absolute accelerations, we have :

$⇒{a}_{1}-{a}_{B}=-\left({a}_{2}-{a}_{B}\right)={a}_{B}-{a}_{2}$

Putting given values,

$⇒2-{a}_{B}={a}_{B}-8$

$⇒2{a}_{B}=10$

$⇒{a}_{B}=5\phantom{\rule{1em}{0ex}}m/{s}^{2}$

Now, pulleys “B” and “C” are connected through a string passing over a static pulley “A”. The accelerations of the moving pulleys “B” and “C” are, therefore, equal in magnitude but opposite in direction. From the analysis above, the pulley “B” has upward acceleration of 5 $m/{s}^{2}$ . It, then, follows that pulley “C” has acceleration of 5 $m/{s}^{2}$ in downward direction,

${a}_{C}=-{a}_{B}=-5\phantom{\rule{1em}{0ex}}m/{s}^{2}$

Again following the earlier reasoning that relative accelerations are equal and opposite, we have :

${a}_{3C}=-{a}_{4C}$

$⇒{a}_{3}-{a}_{C}=-\left({a}_{4}-{a}_{C}\right)={a}_{C}-{a}_{4}$

Putting values,

$⇒2-\left(-5\right)=-5-{a}_{4}$

$⇒{a}_{4}=-5-7=-12\phantom{\rule{1em}{0ex}}m/{s}^{2}$

The negative sign shows that the block “4” moves down i.e. opposite to the positive reference direction, which is upward.

## Limitation of analysis in moving frame of reference

Analysis of accelerations of blocks attached to moving block is based on the fact that blocks are attached to a taut inflexible string. These accelerations of blocks are accelerations with respect to moving pulley, which is itself accelerating. We can not ,however, use these accelerations directly to analyze forces on the blocks as moving pulley system is an accelerated frame of reference. We need to determine acceleration with respect to ground i.e. inertial frame of reference in order to apply Newton's second law of motion .

We, therefore, need to first determine acceleration with respect to ground using relation for relative motion . Considering the moving pulley "B" and attached blocks "1" and "2", we have :

$T-{m}_{1}g={m}_{1}{a}_{1}={m}_{1}\left({a}_{1B}+{a}_{B}\right)$

${m}_{2}g-T={m}_{2}{a}_{2}=-{m}_{2}{a}_{1}=-{m}_{2}\left({a}_{1B}+{a}_{B}\right)$

There is yet another approach for analyzing forces in accelerating frame of reference. We apply the concept of pseudo force. We shall study about pseudo force subsequently in a separate module. Here, we shall limit our discussion to only the working steps as applied in the analysis of motion of moving pulley.

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