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Course by: Sunil Kumar Singh. E-mail the author

# Pulleys(application-I)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

## Hints for solving problems

In some questions, we are required to find relation for the accelerations of different elements in the arrangement. We, therefore, need a framework or a plan to establish this relation. This relation is basically obtained by differentiating the relation of positions of the movable elements in the arrangement.

1. Draw reference, which is fixed. This reference can be the level of fixed pulley or the ground.
2. Identify all movable elements like blocks and pulleys (excluding static ones).
3. Assign variables for the positions of movable elements from the chosen reference.
4. Write down constraint relations for the positions of the elements. Usually, total length of the string is the “constraint” that we need to make use for writing relation for the positions.
5. Differentiate the relation for positions once to get relation of velocities and twice to get relation of accelerations.

## Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the pulleys. The questions are categorized in terms of the characterizing features of the subject matter :

• Static or fixed pulley
• Moving pulley
• Combination or Multiple pulley system

## Static or fixed pulley

Problem 1 : Pulley and strings are “mass-less” and there is no friction involved in the arrangement. Find acceleration of the block of mass 3 kg and tensions “ T 1 T 1 ” and “ T 2 T 2 ” as shown in the figure.

Solution 1 : We observe here that two strings are taught. Therefore, acceleration of the all constituents of the system is same. This situation can be used to treat the system as composed of three blocks only (we neglect string as it has no mass). The whole system is pulled down by a net force as given by :

F = 2 g + 3 g g = 4 g F = 2 g + 3 g g = 4 g

Total mass of the system is :

m = 1 + 2 + 3 = 6 k g m = 1 + 2 + 3 = 6 k g

Applying law of motion, the acceleration of the system and hence acceleration of the block of mass 3 kg is :

a = F m = 4 g 6 = 40 6 = 20 3 m / s 2 a = F m = 4 g 6 = 40 6 = 20 3 m / s 2

In order to find tension “ T 1 T 1 ”, we consider FBD of mass of 1 kg. Here,

F y = T 1 - 1 X g = m a y = 1 X 20 3 F y = T 1 - 1 X g = m a y = 1 X 20 3

T 1 = 20 3 + 10 = 50 3 N T 1 = 20 3 + 10 = 50 3 N

In order to find tension “ T 2 T 2 ”, we consider FBD of mass of 3 kg. Here,

F y = 3 X g T 2 = m a y = 3 X 20 3 = 20 F y = 3 X g T 2 = m a y = 3 X 20 3 = 20

T 2 = 3 X 10 20 = 10 N T 2 = 3 X 10 20 = 10 N

Problem 2 : Pulley and string are “mass-less” and there is no friction involved in the arrangement. The blocks are released from rest. After 1 second from the start, the block of mass 4 kg is stopped momentarily. Find the time after which string becomes tight again.

Solution 1 : We need to understand the implication of stopping the block. As soon as block of 4 kg (say block “B”) is stopped, its velocity becomes zero. Tension in the string disappears. For the other block of 2 kg (say block “A”) also, the tension disappears.

The block “A”, however, has certain velocity with which it is moving up. It will keep moving up against the force of gravity. On the other hand, block “B” will resume its motion of a free fall under gravity.

The string will become taught again, when the displacements of two blocks equal after the moment the block “B” was stopped momentarily. Let “t” be that time, then :

v A t 1 2 g t 2 = 0 + 1 2 g t 2 v A t 1 2 g t 2 = 0 + 1 2 g t 2

v A 1 2 g t = 1 2 g t v A 1 2 g t = 1 2 g t

t = v A g t = v A g

Clearly, we need to know the velocity of the block “A”, when block “B” was stopped. It is given in the question that the system of blocks has been in motion for 1 s. Now, the acceleration of the system (hence that of constituent blocks) is :

a = 4 g 2 g 6 = 20 6 = 10 3 m / s 2 a = 4 g 2 g 6 = 20 6 = 10 3 m / s 2

Applying equation of motion for block “A”, we have :

v A = u A + a t = 0 + 10 3 X 1 = 10 3 m / s v A = u A + a t = 0 + 10 3 X 1 = 10 3 m / s

Now putting values in the equation for time, we have :

t = v A g = 10 3 X 10 = 1 3 s t = v A g = 10 3 X 10 = 1 3 s

## Moving pulley

Problem 3 : Pulleys and string are “mass-less” and there is no friction involved in the arrangement. The two blocks have equal mass “m”. Find acceleration of the blocks and tension in the string.

Solution 1 : In order to determine accelerations, we shall write constraint equation to find the relation for the accelerations of two blocks. Going by the hints given in the module, we see that pulley on the table is fixed and can be used as reference. However, one of the blocks is hanging and is not in the level of other movable block on the table.

To facilitate writing of constraint relation, we imagine here that hanging block “D” is raised on the same level as block “A”. Now, we assign variables to denote positions of two movable elements as shown in the figure.

x A + x D = L x A + x D = L

Differentiating twice,

a A + a D = 0 a A + a D = 0

a A = a D a A = a D

This means that the accelerations of two blocks are same. Since string is single piece, tension in the string is same. Considering free body diagrams, we have :

For A :

T = m a T = m a

For D :

m g T = m a m g T = m a

Combining, we have :

m g m a = m a m g m a = m a

a = g 2 a = g 2

T = m a = m g 2 T = m a = m g 2

## Combination or Multiple pulley system

Problem 4 : Pulleys and string are “mass-less” and there is no friction involved in the arrangement. If at a given time velocities of block “1” and “2” are 2 m/s upward. Find the velocity of block “3” at that instant.

Solution 1 : We see here that velocities of block “1” and “2” are given and we have to find the velocity of remaining block “3”. It is apparent that if we have the relation of velocities of movable blocks, then we can get the result. We draw the reference as a horizontal line through the center of the static pulley and denote variables for all the movable elements of the system. Let " L 1 L 1 " and " L 1 L 1 " be the lengths of two strings, then :

x 1 + x 4 = L 1 x 1 + x 4 = L 1

x 2 x 4 + x 3 x 4 = L 2 x 2 x 4 + x 3 x 4 = L 2

Differentiating once, we get the relations for velocities of the movable elements,

v 1 + v 4 = 0 v 1 + v 4 = 0

v 2 v 4 + v 3 v 4 = 0 v 2 v 4 + v 3 v 4 = 0

We need to eliminate " v 4 v 4 " to get the required relation of velocities of the blocks. Rearranging second equation, we have :

v 2 + v 3 2 v 4 = 0 v 2 + v 3 2 v 4 = 0

Substituting for " v 4 v 4 " from the first equation,

v 2 + v 3 2 v 1 = 0 v 2 + v 3 2 v 1 = 0

v 2 + v 3 + 2 v 1 = 0 v 2 + v 3 + 2 v 1 = 0

Now, considering upward direction as positive, v 2 = v 3 = 2 m / s v 2 = v 3 = 2 m / s . Putting these values in the relation for velocities,

2 + v 3 + 2 X 2 = 0 2 + v 3 + 2 X 2 = 0

v 3 = 6 m / s v 3 = 6 m / s

Negative sign means that velocity of the block “3” is vertically downward.

Problem 5 : Pulleys and string are “mass-less” and there is no friction involved in the arrangement. Find the relation for the accelerations between the hanging plank (marked “1”) and the block (marked “2”).

Solution 1 : In order to obtain the relation for the accelerations of given elements, we first need to develop constraint relations for the three string having fixed lengths. For this, we choose a horizontal reference through the center of topmost fixed pulley as shown in the figure.

With reference to positions as shown in the figure, the constraint relations are :

x 1 + x 4 = L 1 x 1 + x 4 = L 1

x 1 x 4 + x 3 x 4 = L 2 x 1 x 4 + x 3 x 4 = L 2

x 1 x 3 + x 2 x 3 = L 3 x 1 x 3 + x 2 x 3 = L 3

Differentiating above relations twice with respect to time, we have three equations :

a 1 + a 4 = 0 a 1 + a 4 = 0

a 1 a 4 + a 3 a 4 = 0 a 1 a 4 + a 3 a 4 = 0

a 1 a 3 + a 2 a 3 = 0 a 1 a 3 + a 2 a 3 = 0

Rearranging second and third equation,

a 1 + a 2 2 a 3 = 0 a 1 + a 2 2 a 3 = 0

a 1 + a 3 2 a 4 = 0 a 1 + a 3 2 a 4 = 0

Substituting for " a 3 a 3 " from second equation.

a 1 + a 2 2 2 a 4 a 1 = 0 a 1 + a 2 2 2 a 4 a 1 = 0

Substituting for " a 4 a 4 " from first equation,

a 1 + a 2 2 2 a 1 a 1 = 0 a 1 + a 2 2 2 a 1 a 1 = 0

a 1 + a 2 + 6 a 1 = 0 a 1 + a 2 + 6 a 1 = 0

a 2 + 7 a 1 = 0 a 2 + 7 a 1 = 0

a 2 = 7 a 1 a 2 = 7 a 1

Thus, we need to pull the block a lot to raise the plank a bit. This is a mechanical arrangement that allows to pull a heavy plank with smaller force.

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