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, $m_1=10$ kg

Mass of body B, $m2=20$ kg

Total mass of the system, $m=m1+m2=30$ kg

Using Newtons second law of motion, the acceleration (a) produced in the system can be calculated as:

$F=ma$

$a=\frac{F}{m}=\frac{600}{30}=20m/s^2$

When force F is applied on body A:

The equation of motion can be written as: $F-T=m_{1}a$

$T=F-\left(m_1\right)a$

$=600-10\times 20=400$ N (i)

When force F is applied on body B:

The equation of motion can be written as:

$F-T=m_{2}a$

$T=F-\left(m_2\right)a$

$T=600-20\times 20=200$ N (ii)

which is different from value of T in case (i). Hence our answer depends on which mass end, the force is applied.