A massless string passes over a frictionless pulley and carries mass `m_1` hanging at one end and mass `m_2` connected by another massless string to mass `m_3` at other end as shown in figure. Calculate the tension in string joining masses `m_2 and m_3`.

Explanation

Let `T_1` be the tension in the string joining `m_1 and m_2`, while `T_2` the tension is string joining `m_2 and m_3`.
Let a be acceleration of masses. The resultant force on mass `m_3` is `(m_3g-T)` downward, therefore, we have
`m_3g - T_2 = m_3a` ......... (1)
Resultant force on mass ` m_2 ` is
` (m_2 g + T_2-T_1)` downwards , therefore , we have
`m_2 g + T_2-T_1 = m_2 a ` ...... (2)
Reusultant force on mass ` m_1 ` is ` (T_1 - m_1 g )` upward , therefore , we have
` T_1 -m_1 g = m_1 a ` ........ (3)
Adding (1) , (2) and (3) , we get , `m_3g + m_2g - m_1 g = (m_3 + m_2 + m_1) a `
`:. ` acceleration ` a=((m_2 + m_3-m_1))/( m_1 + m_2 + m_3) g `
substituting this value of a in (1) , we get
`m_3 g - T_2 = m_3 . ((m_2 + m_3 - m_1)g)/(m_1 + m_2 + m_3)`
This gives `T_2 = m_3 g - (m_3 (m_2 + m_3 - m_1)g)/( m_1 + m_2 + m_3)`
`:. ` Requried Tension , `T_2 =(2m_1 m_3 g)/( m_1 + m_2 + m_3)`.