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).

A massless string passes over a frictionless pulley and carries mass m

1.	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).
Answer» Solution :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))`