A pulley is nothing but a wheel which is kept on an axle with a support of a string or wire. It is used to lift the heavy materials and moving the large material to from one place to another etc. Using this set up we can find out the acceleration of the suspending object and the tension of the string using equations. The equations and some sample problems are given in the following section.

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The sample pulley problems to find out the acceleration and tension are given below:

### Solved Examples

**Question 1: **The objects of masses 3 Kg and 5 Kg are suspended to two ends of a string which passes through a pulley. Initially this system is at rest. Calculate the acceleration and the tension?

** Solution: **

we know that, T=mg+ma

In the case of 3 Kg mass

T=3g+3a

T-3g=3a...........(1)

In the case of 5 Kg mass

T=5g-5a

5g-T=5a.........(2)

Adding these two equations we get,

2g=8a

a=$\frac{2g}{8}$

a=$\frac{2\times 9.81}{8}$= 2.45 m/s^{2}

Substitute this value in (1)

T=3g+3a=3$\times$ 9.81+3$\times$ 2.45

T=29.43+7.35=36.78 N

**Question 2: **An object of weight 5 Kg is placed on an inclined plane at 30º to the horizontal. This object is connected by a string which is passing through a pulley kept at the top of the inclined plane. An object of 3 Kg is attached on the other end of the string. Determine the acceleration and tension of the string?

** Solution: **

The equation is given by,

T=mgsinθ+ma

T-5gsin30=5a

T-2.5g=5a..............(1)

and

3g-T=3a.................(2)

Adding (1) and (2)

0.5g=8a

a=$\frac{0.5g}{8}$

a=$\frac{0.5\times 9.81}{8}$=0.6125 m/s^{2}

Substitute this value in (1)

T-2.5g=5a⇒ T= 2.5×9.81=3×0.6125

T=27.6 N

The formula to find out the tension is given by,

T = mg + ma

Where**T** is the tension

**m** is the mass of the object

** g** is the acceleration due to gravity

**a** is the acceleration of the object

T = mg + ma

Where

we know that, T=mg+ma

In the case of 3 Kg mass

T=3g+3a

T-3g=3a...........(1)

In the case of 5 Kg mass

T=5g-5a

5g-T=5a.........(2)

Adding these two equations we get,

2g=8a

a=$\frac{2g}{8}$

a=$\frac{2\times 9.81}{8}$= 2.45 m/s

Substitute this value in (1)

T=3g+3a=3$\times$ 9.81+3$\times$ 2.45

T=29.43+7.35=36.78 N

The equation is given by,

T=mgsinθ+ma

T-5gsin30=5a

T-2.5g=5a..............(1)

and

3g-T=3a.................(2)

Adding (1) and (2)

0.5g=8a

a=$\frac{0.5g}{8}$

a=$\frac{0.5\times 9.81}{8}$=0.6125 m/s

Substitute this value in (1)

T-2.5g=5a⇒ T= 2.5×9.81=3×0.6125

T=27.6 N