Two masses 8 kg and 12 kg are connected 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-

Question

Toppr Admin · Accepted Answer

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, 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:
$T - m_{1} g = m a$ … (i)

For mass m2:
The equation of motion can be written as:
$m_{2} g - T = m 2 a$ … (ii)

Adding equations (i) and (ii), we get:
$(m 2 - m 1) g = (m 1 + m 2) a$

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

$= \frac{(12 - 8)}{(12 + 8)} \times 10 = \frac{4 \times 10}{20} = 2 m s - 2$

Therefore, the acceleration of the masses is 2 m/s2.
Substituting the value of a in equation (ii), we get:
$m_{2} g - T = m_{2} \frac{(m_{2} - m_{1}) g}{(m_{1} + m_{2})}$

$T = (m 2 - \frac{(m_{2}^{2} - m 1 m 2)}{(m 1 + m 2)} g$

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

$T = 2 \times 12 \times 8 \times 10 / (12 + 8)$
$T = 96 N$
Therefore, the tension in the string is 96 N.

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.

Two masses $8$ $k g$ and $12$ $k g$ 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.