What is the tension in a rod of length length L and mass M at a distance y from`F_(1)` when the rod is acted on by two unequal force `F_(1)` and `F_(2) ( lt F_(1))` as shown in.

Explanation

As net force on the rod `=(F_(1)-F_(2))` and its mass is M. So, acceleration of the rod will be
`a=(F_(1)-F_(2))//M` …(i)
Now considering the motion of part AB of the rod [which has mass `(M//L)y` and acceleration a given by Eqn. (i)] assuming that tension at B is T,
`F_(1)-T=(M//L)yxxa [from F=ma]`
Substituting a from Eqn (i)
`F_(1)-T=(M/Ly)((F_(1)-F_(2))/M)`
or `T=F_(1)(1-y/L)+F_(2)(y/L)`
To calculate tension at B we can also consider the motion of the other part of rod, i.e., BC. However, then equation of motion will be,
`T-F_(2)=(M//L)(L-y)xxa`
`"i.e " T=F_(2)+(M(L-y))/Lxx((F_(1)-F_(2)))/M`
which of solving gives again,
`T=F_(1)(1-y/L)+F_(2)(y/L)`