Question

A rod of length L kept on a smooth horizontal surfaceis pulled along its length by a force F. The area ofcross-section is A and Young's modulus is Y. Theextension in the rod is​

Answer 1

The elongation in the rod is δ = 1 / 2 FL/YA

Explanation:

Let us consider a small element of length dx at a distance x from the free end of the rod. The magnitude of force at this section is F' = Fx/L.

Therefore, the stress at this section is

σ=F'/A= F/A . x/ L

Elongation dδ produced in this differential element is

dδ= F / YAL . xdx

Thus, total elongation is

δ = F / YAL∫1-0 xdx

= F / YAL [x^2/2]1-0

δ = 1 / 2 FL/YA

Thus the elongation in the rod is δ = 1 / 2 FL/YA

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