__Question 5.16__:
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__: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) X 10 = 4 X 10 / 20 = 2 ms

^{-2}

Therefore, the acceleration of the masses is 2 m/s

^{2}.

Substituting the value of

*a*in equation (

*ii*), we get:

m

_{2}g - T = m

_{2}(m

_{2}- m

_{1})g / (m

_{1}+ m

_{2})

T = (m

_{2}- (m

_{2}

^{2}- m

_{1}m

_{2}) / (m

_{1}+ m

_{2}) )g

= 2m

_{1}m

_{2}g / (m

_{1}+ m

_{2})

= 2 X 12 X 8 X 10 / (12 + 8)

= 96 N

Therefore, the tension in the string is 96 N.

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