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## Collection of Solved Problems in Physics

A pulley system, task number: 508.

A bucket with mass m 2 and a block with mass m 1 are hung on a pulley system. Find the magnitude of the acceleration with which the bucket and the block are moving and the magnitude of the tension force T by which the rope is stressed. Ignore the masses of the pulley system and the rope. The bucket moves up and the block moves down.

## Hint 1 – the forces in the pulley system and the force equations

Figure out which forces affect the bucket and the block. Draw a picture. Write down the force equations for the bucket and the block.

## Solution to Hint 1 – the forces in the pulley system and the force equations

Forces that affect the bucket m 2 :

\(\vec{F}_{G_2}\)…weight

\(\vec{T}\)…tension force by which the rope affects the bucket

Forces that affect the block m 1 :

\(\vec{F}_{G_1}\)…weight

\(\vec{T}\prime\)…tension force by which the rope affects the block

\(\vec{T^{'}}\prime\prime\)…tension force by which the rope affects the block

The vector force equation for the bucket:

\[\vec{F}_{G2}+\vec{T}=m_2\vec{a}_2\tag{1}\]

The vector force equation for the block:

\[\vec{F}_{G1}+\vec{T}\prime+\vec{T}\prime\prime=m_1\vec{a_1}\tag{2}\]

## Hint 2 – the scalar force equations, tension forces

Choose a suitable coordinate system and rewrite the force equations to scalar form. Decide what relations will hold for the magnitude of the tension forces T , T' and T'' .

## Solution to Hint 2 – the scalar motion equation, tension forces

We choose the y- axis the way it is marked on the picture. We rewrite equations (1) and (2) to scalar form:

\[T-F_{G2}= m_2a_2\tag{3}\]

\[F_{G1}-T\prime-T\prime\prime = m_1a_1\tag{4}\]

Because we leave the mass of the pulley system and the rope out of account they have no moment of inertia and don’t affect the tension forces. The following holds for the magnitude of the tension forces:

\[|\vec{T}|=|\vec{T}\prime|=|\vec{T}\prime\prime|\]

We rewrite equations (3) and (4):

\[T-F_{G2}= m_2a_2\tag{5}\]

\[F_{G1}-2T= m_1a_1\tag{6}\]

## Hint 3 – acceleration

Decide what the magnitude of the acceleration a 1 will be in relation to a 2 . Imagine that the bucket goes up a distance s what distance does the block go down?

## Solution to Hint 3 – acceleration

Relation between the magnitudes of acceleration a 1 and a 2 :

If the bucket goes up a distance s , the block goes down a distance \(\frac{s}{2}\).

\[s=\frac{1}{2}a_2t^2\]

\[\frac{s}{2}=\frac{1}{2}a_1t^2\]

We divide the first equation by the second equation, yielding:

\[2=\frac{a_2}{a_1}\]

\[a_2=2a_1\tag{7}\]

Equation (5) and equation (6) can be written as:

\[T-m_2g= m_22a_1\tag{8}\]

\[m_1g-2T= m_1a_1\tag{9}\]

## Hint 4 – the magnitude of the acceleration, the force T

We have two equations (8) and (9) in two variables T and a 1 . Solve these equations and from equation (7) determine the magnitude of the acceleration a 2 .

## Solution to Hint 4 – the magnitude of the acceleration, the force T

We have two equations (8) and (9) in two variables T and a 1 . From them we can determine the magnitude of the acceleration a 1 and the force T . We multiply equation (8) by 2 and add both equations up:

\[m_1g-2m_2g= m_1a_1+4m_2a_1\]

\[(m_1-2m_2)g= (m_1+4m_2)a_1\]

\[a_1=\frac{(m_1-2m_2)g}{m_1+4m_2}\tag{10}\]

According to equation (7):

\[a_2=2a_1=2\frac{(m_1-2m_2)g}{m_1+4m_2}=\frac{(2m_1-4m_2)g}{m_1+4m_2}\tag{11}\]

The magnitude of the tension force T is given for example by equation (8):

\[T-m_{2g}= m_22a_1\]

\[T=m_22a_1+m_{2g}=m_2\frac{2(m_1-2m_2)g}{m_1+4m_2}+m_2g\]

\[T=\frac{2(m_2m_1-2m_2^2)g}{m_1+4m_2}+m_2g =\frac{(2m_2m_1-4m_2^2+m_2(m_1+4m_2))g}{m_1+4m_2}\]

\[T=\frac{(2m_2m_1-4m_2^2+m_2m_1+4m_2^2)g}{m_1+4m_2}= \frac{3m_2m_1g}{m_1+4m_2}\]

\[T=\frac{3m_2m_1g}{m_1+4m_2}\tag{12}\]

## COMPLETE SOLUTION

Draw all the forces which affect the bucket and the block in the picture. Write the force equations for them.

\(\vec{T}\prime\prime\)…tension force by which the rope affects the block

We choose the y -axis the way it is marked on the picture.

We rewrite equations (1) and (2) to scalar form:

## COMPLETE ANSWER

The magnitude of the acceleration with which the block moves is \(a_1=\frac{(m_1-2m_2)g}{m_1+4m_2}\).

The magnitude of the acceleration with which the bucket moves is \(a_2=2a_1=\frac{(2m_1-4m_2)g}{m_1+4m_2}\).

The magnitude of the tension force T which affects the rope is \(T=\frac{3m_2m_1g}{m_1+4m_2}\).

## Browse Course Material

Course info, instructors.

- Prof. Deepto Chakrabarty
- Dr. Peter Dourmashkin
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## Departments

As taught in.

- Classical Mechanics

## Learning Resource Types

12.1 pulley problems - part i, set up the equations.

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In the figure, pulley A is fixed to the ceiling. An ideal rope is attached at one end to block 1 of mass \(m_1\), it passes around a second pulley, labelled B, and its other end is fixed to the ceiling. A second ideal rope attaches pulley B to a second block of mass \(m_2\). As a result, pulley B and block 2 move together.

The goal of the problem is to calculate the accelerations of blocks 1 and 2.

The solution of this problem is divided into four parts:

- Part I : Set up the system of equations.
- Part II: Constraint condition - find the relationship between the accelerations.
- Part III: Constraint condition using a virtual displacement argument.
- Part IV: Solving the system of equations.

The video below will help you to set up the equations needed to answer the question.

- Download video
- Download transcript

- JEE Articles

## Pulley Problems

Here, we will look at some problems based on the pulley and learn how to solve them. Also, we will understand a few equations involved and the method of applying them in solving different problems.

## Pulley Problems and Constraint Equation Videos

## Concept Based on Rotation of Pulley

## Constraint Relation for Strings

## Finding Tension in the Rope

## Pulley Systems DPP

## Atwood Machine

## How to Solve Pulley Problems?

For solving any pulley problem, the first step is to understand the given conditions and write down the constraint equations accordingly.

CASE – 1

Let M 1 & M 2 be the mass attached to pulley A. Now, consider that mass M 1 is moving down with acceleration a 1 and mass M 2 is moving up with acceleration a 2.

Now, from the Free Body Diagram (FBD):

The equation of force from FBD:

M 1 .g – T = M 1 .a 1 . . . . . . . . . . (1)

T – M 2 .g = M 2 .a 2 . . . . . . . . . . . (2)

From the above force equation, we have three unknown values, but there are only two equations (Equation (1) & Equation (2)), so we need a third equation relating the two unknowns. The relation is known as the constraint equation because the motion of M 1 and M 2 is interconnected.

The following assumptions must be considered before writing the equation:

1. The string is taut and inextensible at each and every point in time.

2. The string is massless, and hence the tension is uniform throughout.

3. Pulley is massless.

The string is inextensible; hence the total change in length of the string should be zero. Suppose mass M 1 goes down by x 1 distance and mass M 2 moves up by x 2 distance. Then, change in length = x 1 – x 2 (Taking elongation as positive and compression as negative).

On equating it to zero, we get, x 1 – x 2 = 0 (x = displacement)

On differentiating equation twice, we get,

a 1 – a 2 = 0 . . . . . . . . . (3) (a = acceleration)

Using Equations (1), (2) and (3), we can calculate the values of T, a 1 and a 2 . This is an example of a Fixed Pulley System.

CASE – 2

Consider the following pulley system:

First, we have to relate the tension between string 1 and string 2; consider the FBD of pulley B:

2T 1 – T 2 = M B .a B

2T 1 = T 2 . . . . . . . . . (4) (Since the pulley is massless)

Now, consider the accelerations, as shown in the below figure:

For String 2:

Movable pulley B and block M 2 will have the same acceleration, otherwise, the string will stretch or compress.

a B = a 2 . . . . . . . . . . . . (5)

Increase in length to the left of fixed pulley A = L 1

Decrease in length to the right of fixed pulley A = L 2

Decrease in length to the right of movable pulley B = L 3

And, L 1 + L 2 + L 3 = Constant.

Therefore, the total change in length = L 1 – 2L 2 (Since L 2 = L 3 )

On differentiating the above equation twice, we get,

⇒ a 1 = 2a 2 . . . . . . . . (6)

Similarly, if we write the force equation on both blocks:

M 1 .g – T = M 1 .a 1 . . . . . . . . . . (7)

2T – M 2 .g = M 2 .a 2 . . . . . . . . . (8)

Using Equations (6), (7) and (8), we get,

\(\begin{array}{l}T = \frac {3 M_1 M_2 g}{4M_1~ +~ M_2} \end{array} \)

\(\begin{array}{l}a_1 = \frac {(4M_1 ~-~ 2 M_2)g}{4M_1 ~+~M_2}\end{array} \)

\(\begin{array}{l}a_2 = \frac {(2 M_1~-~M_2)g}{4 M_1~+~M_2} \end{array} \)

Instead of using the above method, we can also use a shortcut to find the constraint relationship for saving time in JEE.

If we consider all the blocks + pulleys + string as one system, then the tension becomes an internal force, and we know that the work done by the internal force is zero.

\(\begin{array}{l}\Rightarrow \sum T.x = 0\end{array} \)

Work done by tension on block M 1 = -T.x 1

Work done by tension on block M 2 = 2T.x 2

Work by tension on pulley A = T.0 = 0

Work done by tension on pulley B = 2T.x 2 – 2T.x 2 = 0

Hence, we get 2T.x 2 – T.x 1 = 0

⇒ x 1 = 2.x 2

Differentiating twice with respect to time, we get,

⇒ a 1 = 2.a 2

The above technique can be used in almost all possible pulley systems, consisting of fixed pulleys or movable pulleys.

Stay tuned to BYJU’S to learn more about the laws of motion.

## More Pulley Problem Videos

Pulley systems.

## Problem-based on Pulley-mass System (No-slip condition)

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