#### Question

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

#### Solution 1

The given system of two masses and a pulley can be represented as shown in the following figure:

Smaller mass, *m*_{1} = 8 kg

Larger mass, *m*_{2} = 12 kg

Tension in the string = *T*

Mass *m*_{2}, owing to its weight, moves downward with acceleration *a*,and mass *m*_{1}moves upward.

Applying Newton’s second law of motion to the system of each mass:

__For mass ____m___{1}__:__

The equation of motion can be written as:

*T* – *m*_{1}g = *ma* … (*i*)

__For mass ____m___{2}__:__

The equation of motion can be written as:

*m*_{2}g* –* *T* = *m*_{2}*a* … (*ii*)

Adding equations (*i*) and (*ii*), we get:

(m_2-m_1)g = (m_1+ m_2)a

`:.a = ((m_2-m_1)/(m_1+m_2))g ...(iii)`

`=((12-8)/(12+8)) xx 10= 4/20 xx 10 = 2 "m/s"^2`

Therefore, the acceleration of the masses is 2 m/s^{2}.

Substituting the value of *a* in equation (*ii*), we get:

`m_2g - T = m_2 ( (m_2-m_1)/(m_1+m_2))g`

`=((2m_1m_2)/(m_1+m_2))g`

`=(2xx12xx8)/(12+8)xx10`

`=(2xx12xx8)/20 xx 10 = 96 N`

Therefore, the tension in the string is 96 N.

#### Solution 2

For block `m_2 -> m_2g - T = m_2a` ...(i)

and for block m_1 -> T - m_1g = m_1a ...(ii)

Adding i and ii we obtain

`(m_2 - m_1)g = (m_2+m_1)a`

or `a = ((m_2-m_1)/(m_2+m_1))g`

`= (12 - 8)/(12 + 8) xx 10`

`= (4xx10)/20 = 2 ms^(-2)`

Substituting value of a in equation ii we obtain

`T = m_1(g+a)`

`= 8 xx (10 + 2)`

= 8 x12 = 96 N