In the arrangement shown in show that tension in the string between masses m_(2) and m_(3) is T = (2 m_(1) m_(3) g)/(m_(1) + m_(2) + m_(3))

In the arrangement shown in show that tension in the string between ma

1.	In the arrangement shown in show that tension in the string between masses m_(2) and m_(3) is T = (2 m_(1) m_(3) g)/(m_(1) + m_(2) + m_(3))
Answer» SOLUTION :Let `T_(1)` be the tension in the STRING between `m_(1)` and `m_(2)` and `T` be the tension in the string between `m_(2)` and `m_(3)` If a is acceleration of the SYSTEM in the direction shown then `m_(1) a =T_(1) - m_(1) g` ….(i) `(m_(2) + m_(3)) a = (m_(2) + m_(3)) g - T_(1)` ....(ii) and `m_(3) a = m_(3) g - T` ...(iii) Adding (i) and(ii) `(m_(1)+ m_(2) +m_(3) a = (m_(2) + m_(3) - m_(1)) g` `a = (m_(2) + m_(3) - m_(1)) g// (m_(1) + m_(2) + m_(3))` PUT in (iii) `(m_(3) (m_(2) + m_(3) -m_(1)) g)/(m_(1) + m_(2) + m_(3)) =m_(3) g - T` `T = m_(3) g-(m_(3) (m_(2) + m_(3) -m_(1)) g)/(m_(1) + m_(2) + m_(3))` `T = 2 m_(1) m_(3) g//(m_(1) + m_(2) + m_(3))` .