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Two opposite forces F1 = 120 N and F2 = 80 N act on an elastic plank of modulus of elasticity Y=2×1011N/m2 and length l = 1 m placed over a smooth horizontal surface. If the cross-sectional area of the plank is S = 0.5 m2, the change in length of the plank is _______ nm.

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Detailed Solution

Consider an element of thickness dx.Change in the length of the element is dl=TSdxY and T=F1−F1−F2xl∫0Δldl=∫0lF1−F1−F2xldxSYΔl=F1+F2l2SY=200×12×0.5×2×1011=1×10-9m∴ x=1

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Two opposite forces F1 = 120 N and F2 = 80 N act on an elastic plank of modulus of elasticity Y=2×1011N/m2 and length l = 1 m placed over a smooth horizontal surface. If the cross-sectional area of the plank is S = 0.5 m2, the change in length of the plank is _______ nm.