First slide

Elastic behaviour of solids

Question

Two opposite forces F₁ = 120 N and F₂ = 80 N act on an elastic plank of modulus of elasticity $Y = 2 \times 10^{11} N / m^{2}$ 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.

Moderate

Solution

Consider an element of thickness dx.
Change in the length of the element is $dl = \frac{T}{S} \frac{dx}{Y}$ and $T = F_{1} - (F_{1} - F_{2}) \frac{x}{l}$
$\int_{0}^{Δl} dl = \int_{0}^{l} \frac{F_{1} - \frac{(F_{1} - F_{2}) x}{l} dx}{SY}$
$Δl = \frac{(F_{1} + F_{2}) l}{2 SY} = \frac{200 \times 1}{2 \times 0 .5 \times 2 \times 10^{11}} = 1 \times 10^{- 9} m$
$∴ x = 1$

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