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Pulleys

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

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Summary: 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.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

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 different combination with other elements – almost always with strings and blocks.

It is relatively difficult 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 find 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 fulfill 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 fixed pulley
  • Moving pulley
  • Combination or Multiple pulley system

Static pulley

The pulley is fixed 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 figure. Neglecting friction between surfaces, find acceleration of the blocks and tension in the string (consider g = 10 m / s 2 g = 10 m / s 2 ).

Figure 1
Static pulley system
Static pulley system (fap1.gif)

Solution : The blocks are connected by a taut string. Hence, their accelerations are same. Let us assume directions of accelerations as shown in the figure. Also, let the magnitude of accelerations be “a”.

Figure 2
Static pulley system
Static pulley system (fap2.gif)

Free body diagram of body of mass 10 kg 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.

Figure 3
Free body diagram
Free body diagram (fap3.gif)

∑ F y = T - 10 g = 10 a ⇒ T - 10 x 10 = 10 a ⇒ T = 100 + 10 a ∑ F y = T - 10 g = 10 a ⇒ T - 10 x 10 = 10 a ⇒ T = 100 + 10 a

Free body diagram of body of mass 20 kg Free body diagram of body of mass 20 kg

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

Figure 4
Free body diagram
Free body diagram (fap4.gif)

∑ F y = 20 g - T = 20 a ⇒ 20 x 10 - T = 20 a ⇒ T = 200 - 20 a ∑ F y = 20 g - T = 20 a ⇒ 20 x 10 - T = 20 a ⇒ T = 200 - 20 a

From two equations, we have :

⇒ 200 - 20 a = 100 + 10 a ⇒ a = 100 30 = 3.33 m / s 2 ⇒ 200 - 20 a = 100 + 10 a ⇒ a = 100 30 = 3.33 m / s 2

Putting value of "a" in either of above two equations, we determine tension as :

⇒ T = 100 + 10 X 3.3 = 133 N ⇒ T = 100 + 10 X 3.3 = 133 N

There 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 " m 1 m 1 " and " m 2 m 2 " connected by a string as in the previous example. Let us also consider that m 2 > m 1 m 2 > m 1 so that block of mass " m 2 m 2 " is pulled down and block of mass " m 1 m 1 " is pulled up. Let "a" be the acceleration of the two block system.

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

F = m 2 g - m 1 g = m 2 - m 1 g F = m 2 g - m 1 g = m 2 - m 1 g

The total mass of the system is :

m = m 1 + m 2 m = m 1 + m 2

Applying law of motion, the acceleration of the system is :

a = F m = m 2 - m 1 g m 1 + m 2 a = F m = m 2 - m 1 g m 1 + m 2 (1)

Clearly, this method to find 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 efficacy of this technique, using the data of previous example. Here,

m 1 = 10 k g ; m 2 = 20 k g m 1 = 10 k g ; m 2 = 20 k g

The acceleration of the block is :

a = m 2 - m 1 g m 1 + m 2 = 20 - 10 X 10 10 + 20 a = m 2 - m 1 g m 1 + m 2 = 20 - 10 X 10 10 + 20

a = 10 3 = 3.33 m / s 2 a = 10 3 = 3.33 m / s 2

Moving pulley

Moving pulley differs 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 fixed at one end and the Pulley is pulled by a force in horizontal direction as shown in the figure.

Figure 5
Moving pulley system
Moving pulley system (fa20.gif)

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 figure. To analyze the situation, we use the fact that the length of string remains same in two situations. Now,

Figure 6
Moving pulley system
Moving pulley system (fa21.gif)

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

L = AB + BC = AB + BD + ED L = AB + BC = AB + BD + ED

⇒ BC = BD + ED ⇒ BC = BD + ED

⇒ ED = BC - BD ⇒ ED = BC - BD

Now, displacement of the block is :

⇒ CE = CD - ED = CD - BC + BD = BD + BD = 2 BD = 2 x ⇒ CE = CD - ED = CD - BC + BD = BD + BD = 2 BD = 2 x

Note that for every displacement “x” of pulley, the displacement of block is 2x. We can appreciate this fact pictorially as shown in the figure below :

Figure 7
Moving pulley system
Moving pulley system (fa19.gif)

Further, as acceleration is second derivative of displacement with respect to time, the relation between acceleration of the block ( a B a B ) and pulley ( a P a P ) is :

⇒ a B = 2 a P ⇒ a B = 2 a P (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 fixed, while other end is moving with acceleration. There is, in fact, no anomaly. Simply, the acceleration of the pulley is also reflected in the motion of the loose end of the string as they are in contact and that the motion of the string is affected 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 figure. If a force “F” acts in horizontal direction, find the accelerations of the pulley and block.

Figure 8
Moving pulley system
Moving pulley system (fap5.gif)

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
Moving pulley system
Moving pulley system (fap6.gif)

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

Free body diagram of pulley 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.

Figure 10
Free body diagram
Free body diagram (fap7.gif)

∑ F x = F - 2 T = m p a p ∑ F x = F - 2 T = m p a p

As mass of the pulley is zero,

⇒ F = 2 T ⇒ T = F 2 ⇒ F = 2 T ⇒ T = F 2

This 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 Free body diagram of block

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

Figure 11
Free body diagram
Free body diagram (fap8.gif)

∑ F x = T = m a x = 2 m a ∑ F x = T = m a x = 2 m a

Combining two equations, we have :

⇒ a = F 4 m ⇒ a = F 4 m

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

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 figure.

Figure 12
Multiple pulley system
Multiple pulley system (fap9.gif)

Let us consider that accelerations of the blocks are as shown in the figure. 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
Multiple pulley system
Multiple pulley system (fap10.gif)

We observe that for given masses, there are five unknowns a 1 , a 2 , a 3 , T 1 and T 2 a 1 , a 2 , a 3 , T 1 and T 2 . Whereas the free body diagram corresponding to three blocks provides only three independent relations. Thus, all five 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 five equations for five 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 different 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 first establish the relation among the positions of moving blocks and pulley with respect to some fixed reference. Then we can obtain relation among accelerations by taking second differentiation of position with respect to time. Important to understand here is that these positions are measured with respect to a fixed reference. As such, their differentiation will yield accelerations of the blocks with respect to fixed reference i.e. ground reference. Let the positions be determined from a horizontal datum drawn through the static pulley as shown in the figure.

Figure 14
Multiple pulley system
Multiple pulley system (fap11.gif)

Now, the lengths of the two strings are constant. Let they be L 1 and L 2 L 1 and L 2 .

L 1 = x 0 + x 1 L 2 = ( x 2 - x 0 ) + ( x 3 - x 0 ) = x 2 + x 3 - 2 x 0 L 1 = x 0 + x 1 L 2 = ( x 2 - x 0 ) + ( x 3 - x 0 ) = x 2 + x 3 - 2 x 0

Eliminating x 0 x 0 , we have :

⇒ L 2 = x 2 + x 3 - 2 L 1 + 2 x 1 ⇒ x 2 + x 3 + 2 x 1 = 2 L 1 + L 2 = constant ⇒ L 2 = x 2 + x 3 - 2 L 1 + 2 x 1 ⇒ x 2 + x 3 + 2 x 1 = 2 L 1 + L 2 = constant

This is the needed relation for the positions. We know that acceleration is second derivative of position with respect to time. Hence,

⇒ a 2 + a 3 + 2 a 1 = 0 ⇒ a 2 + a 3 + 2 a 1 = 0 (3)

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

2: Tensions in the strings

We can, now, find the relation between tensions in two strings by considering the free body diagram of pulley “B” as shown in the figure.

Figure 15
Free body diagram
Free body diagram (fap14.gif)

Here,

T 1 = 2 T 2 T 1 = 2 T 2 (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 figure.

Figure 16
Free body diagram
Free body diagram (fap15.gif)

m 1 g - T 1 = m 1 a 1 m 2 g - T 2 = m 2 a 2 m 3 g - T 2 = m 3 a 3 m 1 g - T 1 = m 1 a 1 m 2 g - T 2 = m 2 a 2 m 3 g - T 2 = m 3 a 3 (5)

Thus, we have altogether 5 equations for 5 unknowns.

There is one important aspect of the motions of blocks of mass " m 2 m 2 " and " m 3 m 3 " with respect to moving pulley "B". The motion of blocks take place with respect to an accelerating pulley. Thus, interpretation of the acceleration must be specific 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 first change value with respect to the ground.

Example 3

Problem : In the arrangement shown in the figure, 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.

Figure 17: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq10.gif)

Solution : Let the acceleration of block “A” be “a” in the downward direction. Let the tensions in the string be “ T 1 T 1 ” and “ T 2 T 2 ”. See figure below showing forces acting on the blocks and moving pulley. The force analysis of the block “A” yields :

Figure 18: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq11.gif)

⇒ 5 g − T 1 = 5 a ⇒ 5 g − T 1 = 5 a ⇒ 50 − T 1 = 5 a ⇒ 50 − T 1 = 5 a

From constraint of string length, we see that :

Figure 19: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq12.gif)

x 1 − x 0 + x 2 − x 0 + x 2 = L x 1 − x 0 + x 2 − x 0 + x 2 = L ⇒ x 1 + 2 x 2 = L + 2 x 0 = Constant ⇒ x 1 + 2 x 2 = L + 2 x 0 = Constant

Differentiating twice with respect to time, we get relation between accelerations of two blocks as :

⇒ a 1 = - 2 a 2 ⇒ a 1 = - 2 a 2 ⇒ a 2 = - a 1 2 = - a 2 ⇒ a 2 = - a 1 2 = - a 2

Force analysis of the moving mass-less pulley yields :

⇒ T 2 = 2 T 1 ⇒ T 2 = 2 T 1

Force analysis of the block B results (refer to the force diagram shown in the beginning) :

⇒ 2 T 1 − 10 X 10 = 10 X a 2 = 5 a ⇒ 2 T 1 − 10 X 10 = 10 X a 2 = 5 a ⇒ 2 T 1 − 100 = 5 a ⇒ 2 T 1 − 100 = 5 a

Simultaneous equations of forces on blocks A and B are :

⇒ 50 − T 1 = 5 a ⇒ 50 − T 1 = 5 a ⇒ 2 T 1 − 100 = 5 a ⇒ 2 T 1 − 100 = 5 a

Solving for “a”, we have :

⇒ a = 0 ⇒ a = 0

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

Exercises

Exercise 1

In the arrangement shown, the acceleration of block “A” is 1 m / s 2 m / s 2 . Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Then,

Figure 20: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq6.gif)

(a) The acceleration of block “B” is 1 m / s 2 m / s 2 .

(b) The acceleration of block “B” is 2 m / s 2 m / s 2 .

(c) The acceleration of block “B” is 0.5 m / s 2 m / s 2 .

(d) The acceleration of block “B” is 0.75 m / s 2 m / s 2 .

Solution

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 / s 2 m / s 2 , the acceleration of “B” is 2 m / s 2 m / s 2 .

Hence, option (b) is correct.

Exercise 2

Given : mass of block “A” = 0.5 kg and acceleration of pulley = 1 m / s 2 m / s 2 . Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Then,

Figure 21: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq3.gif)

(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.

Solution

Let the acceleration of block “A” is “a”. We know that the acceleration of block “A” is two times that of pulley. Hence,

a = 1 X 2 = 2 m / s 2 a = 1 X 2 = 2 m / s 2

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

Figure 22: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq4.gif)

T = m a = 0.5 X 2 = 1 N T = m a = 0.5 X 2 = 1 N

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

⇒ F = 2 T = 2 X 1 = 2 N ⇒ F = 2 T = 2 X 1 = 2 N

Hence, options (b) and (c) are correct.

Exercise 3

In the arrangement shown in the figure, accelerations of blocks A,B and C are a A a A , a B a B and a C a C respectively. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Then

Figure 23: Blocks and pulleys arrangement
Blocks and pulleys arrangement
Blocks and pulleys arrangement (npq13.gif)

(a) 2 a A + 2 a B + a C = 0 (a) 2 a A + 2 a B + a C = 0

(b) a A + a B + 2 a C = 0 (b) a A + a B + 2 a C = 0

(c) a A + a B + a C = 0 (c) a A + a B + a C = 0

(d) 2 a A + a B + a C = 0 (d) 2 a A + a B + a C = 0

Solution

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

Figure 24: Blocks and pulleys arrangement
Blocks and pulleys arrangement
Blocks and pulleys arrangement (npq14.gif)

x A + x A − x 0 + x B − x 0 + x B − x 0 + x C − x 0 = L x A + x A − x 0 + x B − x 0 + x B − x 0 + x C − x 0 = L ⇒ 2 x A + 2 x B + x C = L + 3 x 0 = constant ⇒ 2 x A + 2 x B + x C = L + 3 x 0 = constant

Differentiating two times, we have :

⇒ 2 a A + 2 a B + a C = 0 ⇒ 2 a A + 2 a B + a C = 0

Hence, option (a) is correct.

Exercise 4

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,

Figure 25: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq5.gif)

(a) The force “F” is 40 N.

(b) The acceleration of block A is 4 m / s 2 m / s 2 .

(c) The acceleration of pulley is 1 m / s 2 m / s 2

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

Solution

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

T = k x = 200 X 0.1 = 20 N T = k x = 200 X 0.1 = 20 N

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

Figure 26: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq4.gif)

⇒ T = m a = 10 X a ⇒ T = m a = 10 X a

⇒ a = T 10 = 20 10 = 2 m / s 2 ⇒ a = T 10 = 20 10 = 2 m / s 2

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

⇒ F = 2 T = 2 X 20 = 40 N ⇒ F = 2 T = 2 X 20 = 40 N

Now, acceleration of pulley is half of the acceleration of block. Hence, acceleration of pulley is 1 m / s 2 m / s 2 .

Hence, options (a) and (c) are correct.

Exercise 5

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,

Figure 27: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq7.gif)

(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.

Solution

Let the tensions in the strings connected to blocks “A” and “B” be “ T 1 T 1 ” and “ T 2 T 2 ” respectively. We see here 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 :

Figure 28: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq8.gif)

⇒ T 1 = m 1 X a = 1 X a = a ⇒ T 1 = m 1 X a = 1 X a = a

Consideration of mass-less pulley,

⇒ T 2 = 2 T 1 = 2 X a = 2 a ⇒ T 2 = 2 T 1 = 2 X a = 2 a

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

⇒ m 2 g - T 2 = m 2 X a 2 ⇒ m 2 g - T 2 = m 2 X a 2 ⇒ 2 X 10 - 2 a = 2 X a 2 = a ⇒ 2 X 10 - 2 a = 2 X a 2 = a ⇒ a = 20 3 m / s 2 ⇒ a = 20 3 m / s 2

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

⇒ T 1 = a = 20 3 N ⇒ T 1 = a = 20 3 N ⇒ T 2 = 2 a = 40 3 N ⇒ T 2 = 2 a = 40 3 N

Hence, options (b) is correct.

Exercise 6

Two blocks of mass “m” and “nm” are hanging over a pulley as shown in the figure. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. If n>1 and acceleration of the blocks is g/3, then value of “n” is :

Figure 29: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq9.gif)

a 3 2 b 5 3 c 3 d 2 a 3 2 b 5 3 c 3 d 2

Solution

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

F = n m g − m g = n − 1 m g F = n m g − m g = n − 1 m g

Total mass of the system, M, is :

M = n m + m = n + 1 m M = n m + m = n + 1 m

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

⇒ a = n - 1 m g n + 1 m = n - 1 g n + 1 ⇒ a = n - 1 m g n + 1 m = n - 1 g n + 1

According to question,

⇒ a = n - 1 g n + 1 = g 3 ⇒ a = n - 1 g n + 1 = g 3 ⇒ 3 n - 3 = n + 1 ⇒ 3 n - 3 = n + 1 ⇒ 2 n = 4 ⇒ 2 n = 4 ⇒ n = 2 ⇒ n = 2

Hence, option (d) is correct.

Exercise 7

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,

Figure 30: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq1.gif)

(a) The acceleration of block “B” is 1 m / s 2 m / s 2 .

(b) The acceleration of block “A” is 1 m / s 2 m / s 2 .

(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.

Solution

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

Figure 31: Pulley block arrangement
Pulley block arrangement
Pulley block arrangement (npq2.gif)

F − T 1 = M a F − T 1 = M a ⇒ 2.5 − T 1 = 1 X a = a ⇒ 2.5 − T 1 = 1 X a = a

Considering free body diagram of mass-less pulley,

⇒ T 1 = 2 T 2 ⇒ T 1 = 2 T 2

The 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 :

⇒ T 2 = m X 2 a = 0.5 X 2 a = a ⇒ T 2 = m X 2 a = 0.5 X 2 a = a

Thus,

⇒ T 1 = 2 X a = 2 a ⇒ T 1 = 2 X a = 2 a

Putting this value in the force equation of block “A”,

2.5 − 2 a = 0.5 a 2.5 − 2 a = 0.5 a ⇒ a = 1 m / s 2 ⇒ a = 1 m / s 2

Clearly,

⇒ T 1 = 2 a = 2 X 1 = 2 N ⇒ T 1 = 2 a = 2 X 1 = 2 N ⇒ T 2 = T 1 / 2 = 1 N ⇒ T 2 = T 1 / 2 = 1 N

Hence, option (a) is correct.

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