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 is clear from (APC) 3 Net force on the rod ` f = F_(1) - F_(2) `
Acceleration of rod
` a = (f)/(M) = ((F_(1) - F_(2)))/(M) , along F_(1) `
Mass of part AB of the rod ` = ((M)/(L))y `
If T is tension in the rod at B then equation of motion of portion AB of the rod
` F_(1) - T = ma = ((m)/(L) y) ((F_(1) - F_(2))/(M)) `
` = (F_(1) - F_(2)) (y)/(L)`
`T = F_(1) (1- (y)/(L)) + F_(2) ((y)/(L))`
` T = F_(1) (1 - (y)/(L))+ F_(2) (y/(L)) `
This is the desired result