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

Here as is clear from net force on the rod
` f = (F_(1) - F_(2)) `
Acceleration of rod , along F_(1)
` =a = (f)/(M)= (F_(1) - F_(2))/(M) `
Mass of part AB of the rod
` ((M)/L)y `
Let T be the tension in the rod at B
Writing the equation of motion of part AB of the
rod : ` F_(1) - T = ma = (M)/ (L) y .((F_(1)-F_(2))/(M))`
`F_(1) - T = (F_(1) - F_(2) ) (y)/(L)`
or `T = F_(1) - (F_(1) - F_(2)) (y)/(L)`
`T = F_(1) (1- y y // L ) + F_(2) (y//L)`
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