Three masses $$m_1 = 4$$ kg, $$m_2 = 4$$ kg and $$m_3 = 6$$ kg are suspended from a fixed smooth frictionless pully as shown in the figure below. The value of $$T_1/T_2$$ is _______. (take $$g = 10$$ m/s$$^2$$)

Solution :

Let acceleration of masses $$m_2$$ and $$m_3$$ downward be $$a$$.

Then mass $$m_1$$ moves upward with same acceleration $$a$$.

Given :

$$m_1 = 4\text{ kg}$$

$$m_2 = 4\text{ kg}$$

$$m_3 = 6\text{ kg}$$

For mass $$m_1$$ :

$$T_1 - m_1g = m_1a$$

$$T_1 - 40 = 4a$$

$$T_1 = 40 + 4a \quad ...(1)$$

For mass $$m_2$$ :

$$m_2g + T_2 - T_1 = m_2a$$

$$40 + T_2 - T_1 = 4a \quad ...(2)$$

For mass $$m_3$$ :

$$m_3g - T_2 = m_3a$$

$$60 - T_2 = 6a$$

$$T_2 = 60 - 6a \quad ...(3)$$

Substituting (1) and (3) into (2) :

$$40 + (60-6a) - (40+4a) = 4a$$

$$60 - 10a = 4a$$

$$14a = 60$$

$$a = \frac{30}{7}\text{ m s}^{-2}$$

Now,

$$T_1 = 40 + 4\left(\frac{30}{7}\right)$$

$$= \frac{400}{7}\text{ N}$$

and,

$$T_2 = 60 - 6\left(\frac{30}{7}\right)$$

$$= \frac{240}{7}\text{ N}$$

Therefore,

$$\frac{T_1}{T_2} = \frac{400/7}{240/7}$$

$$= \frac{5}{3}$$

Final Answer :

$$\frac{5}{3}$$