Question

A 2 kg and a 4 kg hang freely at opposite ends of a light inextensible string which passes over a small , light pulley fixed onto a rigid support. Calculate the acceleration of the system.

Answer 1

Answer on Question #40810, Physics, Mechanics

A $2\mathrm{kg}$ and a $4\mathrm{kg}$ hang freely at opposite ends of a light inextensible string which passes over a small, light pulley fixed onto a rigid support. Calculate the acceleration of the system.

Solution:

Given:

$m_{1} = 4\mathrm{kg},$

$m_{2} = 2\mathrm{kg},$

$W_{1} = m_{1}g$

$W_{1} = m_{2}g$

The equations of motion are:

$m_{1}a = m_{1}g - T$

$m_{2}a = T - m_{2}g$

The adding of two equations gives:

$m_{1}a + m_{2}a = m_{1}g - T + T - m_{2}g$

$m_{1}a + m_{2}a = m_{1}g - m_{2}g = g(m_{1} - m_{2})$

Thus, the acceleration is

$a = \frac{g(m_1 - m_2)}{m_1 + m_2}$

$a = \frac{9.81\cdot(4 - 2)}{4 + 2} = 3.27\mathrm{m / s^2}$

Answer. $a = 3.27 \, \text{m/s}^2$ .