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A homogeneous rod of length L is acted u...

A homogeneous rod of length L is acted upon by two forces `F_(1) and F_(2)` applied to its ends and directed opposite to each other. With what force F will the rod be stretched at the cross section at a distance l from the end where `F_(1)` is applied?

A

`(F_(1) + (F_(1) - F_(2)) (x)/(l)) = T`

B

`(F_(1) - (F_(1) + F_(2)) (x)/(l)) = T`

C

`(F_(1) + (F_(1) + F_(2)) (x)/(l)) = T`

D

`(F_(1) - (F_(1) - F_(2)) (x)/(l)) = T`

Text Solution

Verified by Experts

The correct Answer is:
D

Tension at a distance `x` is given by
`(F_(1) - (F_(1) - F_(2)) (x)/(l))`
Shearing stress `= (T cos theta )/(a// sin theta) = (T sin^(2) theta)/(A_(0))`
where `T = (F_(1) - (F_(1) - F_(2))(x)/(l))`
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