A rod of length L and mass M is acted on by two unequal forces `F_1` and `F_2 ( lt F) `as shown in the figure.

The tension in the rod at a distance y from the end A is given by

Options

A. `F_1 (1- (y)/(L))+F_2 ((y)/(L))`
B. `F_2 (1-(y)/(L ) )+ F_1 ((Y)/(L))`
C. `(F_1 -F_2) (y)/(L)`
D. None of these

Answer

A

Explanation

Net force on the rod `= F_1 -F_2`
As mass of the rod is M , so acceleration of the rod is
` a=((F_1 -F_2))/(M)`
Now considering the motion of the rod [ whose mass is equal to `(m//L)`y] then the equation of motion is
` F_1 -T = (M )/(L ) y xx a`
where T is the tension in the rod at B .
`or F_1 -T =(M )/(L) y xx ((F_1-F_2)/(M))`
`or T= F_1 (1- (y)/(L ) ) +F_2 ((y)/(L ))`
second method : to calculate tension at B we can also consider the motion of the other of the rod i.e., BC . then the equation of motion will be
` T- F_2 =(M )/(L ) (l-y) xx a`
` or T =F_2 (M )/(L ) (L-y) xx ((F_1 -F_2))/(M)` ( using (i))
`or T=F_1 ( 1- (y)/(L ) )+F_2 ((y)/(L ))`