Two forces `F_(1)` and `F_(2) ( F_2 gt F_(1))` are applied at the free ends of uniform rod kept on a horizontal frictionless surface. Find tension in rod at a distance x from end ‘A’,

Answer

`T=F_(2)-((F_(2)-F_(1)))/(L).x`

Explanation

`a=(F_(2)-F_(1))/(m)`

`T-F_(1)=m_(2)a`

`rArrT-F_(1)=m/L(L-x)(F_(2)-F_(1))/(m)" "(m_(2)=m/L(L-x))`
`rArrT=F_(1)+(1-x/L)(F_(2)-F_(1))`
`=F+F_(2)-F_(1)-x/L(F_(2)-F_(1))=F_(2)-x/L(F_(2)-F_(1))`