Question

Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

Answer 1

Horizontal force, F = 600 N
Mass of body A, $m_1=10kg$
Mass of body B, $m_2=20kg$
Total mass of the system, $m=m_1+m_2=30kg$
Using Newton's second law of motion, the acceleration (a) produced in the system can be calculated as:
$F=ma\therefore a=\frac{F}{m}=\frac{600}{30}=20m/s^2$
When force F is applied on body A:

The equation of motion can be written as:
$F-T=m_{1}a\therefore T=F-m_{1}a=600-10\times 20=400N\dots \dots \left(i\right)$
When force F is applied on body B:


The equation of motion can be written as:
$F-T=m_{2}aT=F-m_{2}a$
$\therefore T=600-20\times 20=200N\dots \dots \left(ii\right)$