  Calculate the tension and acceleration in a string during the motion of two bodies attached to the ends of a string that passes over a frictionless pulley such that one body moves vertically and the other moves on a smooth horizontal surface:

Difficulty: Medium

The motion of two bodies attached to the ends of a string that passes over a frictionless pulley such that one body moves vertically and the other moves on a smooth horizontal surface:

Consider two bodies A and B of masses m1 and m2 respectively attached to the ends of an inextensible string.

Let body A move downwards with an acceleration a. Since the string is inextensible, therefore, body B also moves over the horizontal surface with the same acceleration a. As the pulley is frictionless, hence tension T will be the same throughout the string. Since body A moves downwards, therefore, its weight m1g is greater than the tension T in the string.

:      Net force acting on body $A =m _{1}g -T$

According to Newton’s second law of motion: $m _{1}g -T =m_{1}a$ ......(i)

The forces acting on body B are:

1. Weight m2g of body B acting downward.
2. Reaction R of the horizontal surface acting on body B in the upwards direction.
3. Tension T in the string pulls body B horizontally over the smooth surface.

As body B has no vertical motion, hence resultant vertical forces (m2g and R) must be zero. Thus, the net force acting on body B is T.

According to Newton's second law of motion;

T = m2a ......(2)

Adding Eq. (i) and (ii), we get acceleration a as:

$m _{1}g -m_{2}a =m _{2}a$

$m _{1}g = m_{1}a +m _{2}a$

$m _{1}g = a \left(m_{1}a +m _{2}\right)$

$a = \frac{m_{1}}{m_{1}+m_{2}}g$ .......(iii)

Putting the value of a in equation (ii) to get tension T as

$T = \frac{m_{1}m_{2}}{m_{1}+m_{2}}g$