Q.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.
Answer:-
The given system of two masses and a pulley can be represented as shown in the following figure:
Larger mass, m2 = 12 kg
Tension in the string = T
Mass m2, owing to its weight, moves downward with acceleration a,and mass m1 moves upward.
Applying Newton’s second law of motion to the system of each mass:
For mass m1:
The equation of motion can be written as:
T – m1g = ma … (i)
For mass m2:
The equation of motion can be written as:
m2g – T = m2a … (ii)
Adding equations (i) and (ii), we get:
(m2 – m1)g = (m1 + m2)a
∴ a = ( (m2 – m1) / (m1 + m2) )g ….(iii)
= (12 – 8) / (12 + 8) × 10 = 4 × 10 / 20 = 2 ms-2
Therefore, the acceleration of the masses is 2 m/s2.
Substituting the value of a in equation (ii), we get:
m2g – T = m2(m2 – m1)g / (m1 + m2)
T = (m2 – (m22 – m1m2) / (m1 + m2) )g
= 2m1m2g / (m1 + m2)
= 2 × 12 × 8 × 10 / (12 + 8)
= 96 N
Therefore, the tension in the string is 96 N.