solved problems with pulleys

OpenStax-CNX module: m14060 1

Pulleys

Sunil Kumar Singh

This work is produced by OpenStax-CNX and licensed under the

Creative Commons Attribution License 2.0

Abstract

The basic consideration in making useful mechanical arrangements are two folds (i) improve the

convenience of applying force and (ii) reduce the magnitude of force required.

A pulley is a part of a convenient arrangement that makes it possible to transfer force with change of

direction. Unless otherwise stated, a pulley is considered to have negligible mass and friction. This is a

relative approximation with respect to mass and friction involved with other elements. Pulleys are used in

dierent combination with other elements almost always with strings and blocks.

It is relatively dicult to fetch a bucket of water from a well with a string as compared to a pulley and

string system. The basic consideration in making useful mechanical arrangements are two folds (i) improve

the convenience of applying force and (ii) reduce the magnitude of force. The example of fetching of a bucket

of water with pulley and string achieves the goal of improving convenience as we nd it easier applying

force in the level of arms horizontally rather than applying force vertically.

Had it been possible to reduce force for doing a mechanical activity, then that would have been wonder

and of course against the well founded tenets of physical laws. What is meant here by reducing force is that

we can fulll a task (which comprises of force and motion) by reducing force at the expense of extending

motion.

The important characterizing aspects of pulley are discussed in the sections named :

Static or xed pulley Moving pulley Combination or Multiple pulley system

1 Static pulley

The pulley is xed to a frame. In this situation, we are only concerned with the accelerations of the bodies

connected to the string that passes over the pulley. Since string is a single piece inextensible element, the

accelerations of the bodies attached to it are same.

We are at liberty to choose the direction of acceleration of the blocks attached to the pulley. A wrong

choice will be revealed by the sign of acceleration that we get after solving equations. However, it is a straight

forward choice here as it is obvious that the bigger mass will pull the blocks - string system down.

Example 1

Problem : Two blocks of masses 10 kg and 20 kg are connected by a string that passes over a

pulley as shown in the gure. Neglecting friction between surfaces, nd acceleration of the blocks

and tension in the string (consider g = 10m / s2 ).Version 1.20: Dec 4, 2008 10:17 am -0600

http://creativecommons.org/licenses/by/2.0/

http://cnx.org/content/m14060/1.20/


Static pulley system

Figure 1

Solution : The blocks are connected by a taut string. Hence, their accelerations are same. Let

us assume directions of accelerations as shown in the gure. Also, let the magnitude of accelerations

be a.



Static pulley system

Figure 2

Free body diagram of body of mass 10 kg

The external forces are (i) weight of block, 10g, and (ii) tension, T, in the string.



Free body diagram

Figure 3

Fy = T 10g = 10a

T 10 x 10 = 10a T = 100 + 10aFree body diagram of body of mass 20 kg

The external forces are (i) weight of block, 20g, and (ii) tension, T, in the string.



Free body diagram

Figure 4

Fy = 20g T = 20a

20 x 10 T = 20a T = 200 20aFrom two equations, we have :

200 20a = 100 + 10a a = 10030 = 3.33m / s2Putting value of "a" in either of above two equations, we determine tension as :

T = 100 + 10 X 3.3 = 133NThere is a very useful technique to simplify the solution involving "mass-less" string and pulley. As string

has no mass, the motion of the block-string system can be considered to be the motion of a system comprising

of two blocks, which are pulled down by a net force in the direction of acceleration.

Let us consider two blocks of mass " m1 " and " m2 " connected by a string as in the previous example.Let us also consider that m2 > m1 so that block of mass " m2 " is pulled down and block of mass " m1 " ispulled up. Let "a" be the acceleration of the two block system.

Now the force pulling the system in the direction of acceleration is :

F = m2g m1g = (m2 m1) g



The total mass of the system is :

m = m1 +m2Applying law of motion, the acceleration of the system is :

a =F

m=

(m2 m1) g(m1 +m2)(1)

Clearly, this method to nd acceleration is valid when the block - string system can be combined i.e.

accelerations of the constituents of the system are same. We can check the ecacy of this technique, using

the data of previous example. Here,

m1 = 10 kg; m2 = 20 kg

The acceleration of the block is :

a =(m2 m1) g(m1 +m2)

=(20 10)X10(10 + 20)

a =103

= 3.33 m/s2

2 Moving pulley

Moving pulley diers to static pulley in one important respect. The displacements of pulley and block,

which is attached to the string passing over it, may not be same. As such, we need to verify this aspect

while applying force law. The point is brought out with clarity in the illustration explained here. Here, we

consider a block attached to a string, which passes over a mass-less pulley. The string is xed at one end

and the Pulley is pulled by a force in horizontal direction as shown in the gure.

Moving pulley system

Figure 5

In order to understand the relation of displacements, we visualize that pulley has moved a distance x

to its right. The new positions of pulley and block are as shown in the gure. To analyze the situation, we

use the fact that the length of string remains same in two situations. Now,




Figure 6

Length of the string, L, in two situations are given as :

L = AB + BC = AB + BD + ED

BC = BD + ED

ED = BC BDNow, displacement of the block is :

CE = CD ED = CD BC + BD = BD + BD = 2BD = 2xNote that for every displacement x of pulley, the displacement of block is 2x. We can appreciate this

fact pictorially as shown in the gure below :




Figure 7

Further, as acceleration is second derivative of displacement with respect to time, the relation between

acceleration of the block ( aB ) and pulley ( aP ) is :

aB = 2aP (2)

This is an important result that needs some explanation. It had always been emphasized that the

acceleration of a taut string is always same through out its body. Each point of a string is expected to have

same velocity and acceleration! What happened here ? One end is xed, while other end is moving with

acceleration. There is, in fact, no anomaly. Simply, the acceleration of the pulley is also reected in the

motion of the loose end of the string as they are in contact and that the motion of the string is aected by

the motion of the pulley.

But the point is made. The accelerations of two ends of a string need not be same, when in contact with

a moving body with a free end.

Example 2

Problem : A block of mass, m is connected to a string, which passes over a smooth pulley as

shown in the gure. If a force F acts in horizontal direction, nd the accelerations of the pulley

and block.




Figure 8

Solution : Let us consider that the acceleration of pulley is a in the direction of applied

force. Now as analyzed before, acceleration of the block is 2a and is in the same direction as that

of pulley.


Figure 9

As motion is conned to x-direction, we draw free body diagram considering forces in x-direction

only.

Free body diagram of pulley

The external forces are (i) force, F, (ii) tension, T, in the string and (ii) tension, T, in the string.



Free body diagram

Figure 10

Fx = F 2T = mpapAs mass of the pulley is zero,

F = 2T T = F2This is an unexpected result. The pulley is actually accelerated, but the forces on it form a

balanced force system. This apparent contradiction of force law is due to our approximation that

pulley is mass-less. Free body diagram of block

The external force is (i) tension, T, in the string.

Free body diagram

Figure 11

Fx = T = max = 2ma

Combining two equations, we have :

a = F4m



Thus, acceleration of the pulley is F/4m and that of block is F/2m.

3 Combination or Multiple pulley system

Multiple pulleys may involve combination of both static and moving pulleys. This may involve combining

characterizing aspects of two systems.

Let us consider one such system as shown in the gure.

Multiple pulley system

Figure 12

Let us consider that accelerations of the blocks are as shown in the gure. It is important to note that we

have the freedom to designate direction of acceleration without referring to any other consideration. Here,

we consider all accelerations in downward direction.




Figure 13

We observe that for given masses, there are ve unknowns a1 , a2 , a3 , T1 and T2 . Whereas thefree body diagram corresponding to three blocks provides only three independent relations. Thus, all ve

unknowns can not be evaluated using three equations.

However, we can add two additional equations; one that relates three accelerations and the one that

relates tensions in the two strings. Ultimately, we get ve equations for ve unknown quantities.

1: Accelerations

The pulley A is static. The accelerations of block 1 and pulley B are, therefore, same. The pulley

B, however, is moving. Therefore, the accelerations of blocks 2 and 3 may not be same as discussed for the

case of static pulley. The accelerations of blocks 2 and 3 with respect to moving pulley are dierent than

their accelerations with respect to ground reference.

We need to know the relation among accelerations of block 1, 2 and 3 with respect ground. In order

to obtain this relation, we rst establish the relation among the positions of moving blocks and pulley

with respect to some xed reference. Then we can obtain relation among accelerations by taking second

dierentiation of position with respect to time. Important to understand here is that these positions are

measured with respect to a xed reference. As such, their dierentiation will yield accelerations of the blocks

with respect to xed reference i.e. ground reference. Let the positions be determined from a horizontal datum

drawn through the static pulley as shown in the gure.




Figure 14

Now, the lengths of the two strings are constant. Let they be L1 and L2 .

L1 = x0 + x1

L2 = ( x2 x0 ) + ( x3 x0 ) = x2 + x3 2x0Eliminating x0 , we have :

L2 = x2 + x3 2L1 + 2x1 x2 + x3 + 2x1 = 2L1 + L2 = constantThis is the needed relation for the positions. We know that acceleration is second derivative of position

with respect to time. Hence,

a2 + a3 + 2a1 = 0 (3)

This gives the relation of accelerations involved in the pulley system.

2: Tensions in the strings

We can, now, nd the relation between tensions in two strings by considering the free body diagram of

pulley B as shown in the gure.



Free body diagram

Figure 15

Here,

T1 = 2T2 (4)

note: This result appears to be simple and on expected line. But it is not so. Note that pulley

"B" itself is accelerated. The result, on the other hand, is exactly same as for a balanced force

system. In fact this equality of forces in opposite direction is possible, because we have considered

that pulley has negligible mass. This aspect has been demonstrated in the force analysis of the

example given earlier (you may go through the example again if you have missed the point).

3: Free body diagrams of the blocks

The free body diagrams of the blocks are as shown in the gure.



Free body diagram

Figure 16

m1g T1 = m1a1m2g T2 = m2a2m3g T2 = m3a3(5)

Thus, we have altogether 5 equations for 5 unknowns.

There is one important aspect of the motions of blocks of mass " m2 " and " m3 " with respect to movingpulley "B". The motion of blocks take place with respect to an accelerating pulley. Thus, interpretation

of the acceleration must be specic about the reference (ground or moving pulley). We should ensure that

all measurements are in the same frame. In the methods, described above we have considered accelerations

with respect to ground. Thus, if acceleration is given with respect to the moving pulley in an analysis, then

we must rst change value with respect to the ground.

Example 3

Problem : In the arrangement shown in the gure, mass of A = 5 kg and mass of B = 10 kg.

Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Find

the accelerations of blocks.



Pulley block arrangement

Figure 17: Pulley block arrangement

Solution : Let the acceleration of block A be a in the downward direction. Let the tensions

in the string be T1 and T2. See gure below showing forces acting on the blocks and movingpulley. The force analysis of the block A yields :





5g T1 = 5a

50 T1 = 5aFrom constraint of string length, we see that :





x1 x0 + x2 x0 + x2 = L

x1 + 2x2 = L+ 2x0 = ConstantDierentiating twice with respect to time, we get relation between accelerations of two blocks

as :

a1 = 2a2

a2 = a12 = a

2Force analysis of the moving mass-less pulley yields :

T2 = 2T1Force analysis of the block B results (refer to the force diagram shown in the beginning) :

2T1 10X10 = 10Xa2 = 5a

2T1 100 = 5aSimultaneous equations of forces on blocks A and B are :



50 T1 = 5a

2T1 100 = 5aSolving for a, we have :

a = 0Thus, accelerations of two blocks are zero. Note, however, that tensions in the strings are not

zero.

4 Exercises

Exercise 1 (Solution on p. 25.)

In the arrangement shown, the acceleration of block A is 1 m/s2 . Consider string and pulley tobe mass-less and friction absent everywhere in the arrangement. Then,



(a) The acceleration of block B is 1 m/s2 .(b) The acceleration of block B is 2 m/s2 .(c) The acceleration of block B is 0.5 m/s2 .(d) The acceleration of block B is 0.75 m/s2 .


Given : mass of block A = 0.5 kg and acceleration of pulley = 1 m/s2 . Consider string andpulley to be mass-less and friction absent everywhere in the arrangement. Then,





(a) The force F is 1 N.

(b) The force F is 2 N.

(c) The tension in the string is 1 N.

(d) The tension in the string is 0.5 N.


In the arrangement shown in the gure, accelerations of blocks A,B and C are aA , aB and aCrespectively. Consider string and pulley to be mass-less and friction absent everywhere in the

arrangement. Then



Blocks and pulleys arrangement

Figure 22: Blocks and pulleys arrangement

(a) 2aA + 2aB + aC = 0

(b) aA + aB + 2aC = 0

(c) aA + aB + aC = 0

(d) 2aA + aB + aC = 0


Given : mass of block A = 10 kg, spring constant = 200 N/m, extension in the string = 0.1

m. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement.

Then,





(a) The force F is 40 N.

(b) The acceleration of block A is 4 m/s2 .(c) The acceleration of pulley is 1 m/s2

(d)The tension in the string is 10 N.


In the arrangement shown, the mass of A and B are 1 kg and 2 kg respectively. Consider string

and pulley to be mass-less and friction absent everywhere in the arrangement. Then,



(a) The tension in the string connected to A is 1 N.



(b) The tension in the string connected to A is 20/3 N.

(c) The tension in the string connected to B is 2 N.

(d) The tension in the string connected to B is 20/3 N.


Two blocks of mass m and nm are hanging over a pulley as shown in the gure. Consider

string and pulley to be mass-less and friction absent everywhere in the arrangement. If n>1 andacceleration of the blocks is g/3, then value of n is :



(a)32

(b)53

(c) 3 (d) 2


Given : mass of block A = 0.5 kg, mass of block B=1 kg and F = 2.5 N. Consider string and

pulley to be mass-less and friction absent everywhere in the arrangement. Then,





(a) The acceleration of block B is 1 m/s2 .(b) The acceleration of block A is 1 m/s2 .(c) The tension in the string connecting block B and pulley is 1 N.

(d) The tension in the string connected to block A is 2 N.



Solutions to Exercises in this Module

Solution to Exercise (p. 19)

We are required to determine acceleration of block "B" to answer this question. By inspection, we observe

that if block A moves x distance towards right, then the string length is distributed in two equal halves

about hanging pulley. Clearly, if A moves by x, then B moves by x/2. As such, acceleration of A is

twice that of B. As it is given that acceleration of A is 1 m/s2, the acceleration of B is 2 m/s2.Hence, option (b) is correct.


Let the acceleration of block A is a. We know that the acceleration of block A is two times that of

pulley. Hence,

a = 1X2 = 2 m/s2

Let the tension in the string be T. Considering forces on block A, we have :



T = ma = 0.5X2 = 1 N

Considering forces on the mass-less pulley, we have :

F = 2T = 2X1 = 2 NHence, options (b) and (c) are correct.


Here, we need to determine relation of accelerations of blocks using constraint of string length.



Blocks and pulleys arrangement

Figure 28: Blocks and pulleys arrangement

xA + xA x0 + xB x0 + xB x0 + xC x0 = L

2xA + 2xB + xC = L+ 3x0 = constantDierentiating two times, we have :

2aA + 2aB + aC = 0Hence, option (a) is correct.


Let T be the tension in the string. Here, spring force is equal to tension in the string.

T = kx = 200X0.1 = 20 N

Let acceleration of block A is a. Considering forces on block A, we have :





T = ma = 10Xa

a = T10

=2010

= 2 m/s2

Considering forces on the mass-less pulley, we have :

F = 2T = 2X20 = 40 NNow, acceleration of pulley is half of the acceleration of block. Hence, acceleration of pulley is 1 m/s2 .Hence, options (a) and (c) are correct.


Let the tensions in the strings connected to blocks A and B be T1 and T2 respectively. We seehere that only force making block A to move on the table is tension in the string connected to it. Let

acceleration of block A is a. From constraint relation, we know that the acceleration of A is twice the

acceleration of B. Thus, acceleration of block B is a/2. Applying law of motion for block A, we have :





T1 = m1Xa = 1Xa = aConsideration of mass-less pulley,

T2 = 2T1 = 2Xa = 2aApplying law of motion for block A, we have :

m2g T2 = m2Xa2

2X10 2a = 2Xa2= a

a = 203m/s2

Putting this value in the expressions of tensions, we have :

T1 = a = 203 N

T2 = 2a = 403 NHence, options (b) is correct.


We consider blocks and string as one system. The net external force on the system is

F = nmg mg = (n 1)mgTotal mass of the system, M, is :

M = nm+m = (n+ 1)m

The acceleration of the system i.e. the acceleration of either block is :



a = (n 1)mg(n+ 1)m

=(n 1) g(n+ 1)

According to question,

a = (n 1) g(n+ 1)

=g

3

3n 3 = n+ 1

2n = 4

n = 2Hence, option (d) is correct.


Let the acceleration of block B is a. Considering forces on block of mass B, we have :



F T1 =Ma

2.5 T1 = 1Xa = aConsidering free body diagram of mass-less pulley,

T1 = 2T2The pulley and the block are connected by inextensible string. As such, their accelerations are same.

Further, we know by constraint analysis that acceleration of block is twice that of pulley. Hence, acceleration

of block B is 2a. Considering forces on block A, we have :

T2 = mX2a = 0.5X2a = a



Thus,

T1 = 2Xa = 2aPutting this value in the force equation of block A,

2.5 2a = 0.5a

a = 1 m/s2

Clearly,

T1 = 2a = 2X1 = 2 N

T2 = T1/2 = 1 NHence, option (a) is correct.


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