Question

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 1

The given system of two masses and a pulley can be represented as shown in the following figure:

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:
$m_2\,g \,– \,T$= $m_2a$ ………….… (ii)
Adding equations (i) and (ii), we get:
(m_2\,-m_1)$$g = (m_1+m_2)$$a

$\therefore a$ = $\bigg(\frac{m_2-m_1}{m1+m2}\bigg)g$ .................(iii)

= $\bigg(\frac{12-8}{12+8}\bigg) \times\ 10$

= $\frac{4}{20} \times 10$ = $2 \; m/s^2$
Therefore, the acceleration of the masses is $2 \;m/s^2.$
Substituting the value of $a$ in equation (ii), we get:
$m^2g - T$ = $m_2\bigg(\frac{m_2-m_1}{m_1+m_2}\bigg)g$

$T$ = $\bigg(\frac{m_2 - m_2^2-m_1m_2}{m_1+m_2}\bigg)g$

= $\bigg(\frac{2m_1m_2}{m_1+m_2}\bigg)g$

= $\bigg(\frac{2 \times 12 \times 8}{12+8}\bigg) \times 10$

= $\bigg(\frac{2 \times 12 \times 8}{20}\bigg) \times 10$
= $96 \,\text N$

Therefore, the tension in the string is $96 \,\text N$.