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_(3)g-T)` downward, therefore , we have
`m_(3g-T_(2)= m_(3)a` ....(1)
Resultant force on mass `m_(2)` is
`(m_(2) g +T_(2) - T_(1)) ` downward , therefore , we have
` m_(2) g + T_(2)-T_(1)=m_(2) a ` ....(2)
Resultant 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_(3)g+m_(2)g-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))`
`:. "Required Tension , " T_(2)=(2m_(1)m_(3)g)/(m_(1)+m_(2)+m_(3))`