Question

a canteliver of rectangular cross section has a length of 50cm it is breath is 3cm and thickness 0.6cm.a weight of 1kg is attached at the free end.the depression produced is 4.2cm . calculate Young's modulus of the material of the bar.given g=9.8 m/sec^2​

Answer 1

The Young's modulus of the material of the bar is $\boxed{1.8\times 10^{10}}$ N/m²

Explanation:

Given:

A cantilever has length 50 cm, breadth 3 cm and thickness 0.6 cm

A weight of 1 kg is attached at the free end

Depression is 4.2 cm

To calculate:

Young's modulus of elasticity

Solution:

We know that for a load P and length L of the cantilever, the depression at any distance x is given by

$\delta_x=\frac{Px^2(3L-x)}{6EI}$

Thus, depression at distance L (at the end of the cantilever)

$\delta=\frac{PL^2(3L-L)}{EI}$

$\implies \delta=\frac{PL^3}{3EI}$

Here, E is the Young's modulus of elasticity and I is the moment of inertia

For the given cantilever

$L=50$ cm

$b=3$ cm

$d=0.6$ cm

$P=1\times 9.8$ N

$\delta=4.2$ cm

The moment of inertia of the cross section will be

$I=\frac{bd^3}{12}$

$\implies I=\frac{3\times 0.6^3}{12}$

$\implies I=0.054$ cm^4

$\implies I=0.054\times 10^{-8}$ m^4

Thus,

$4.2\times 10^{-2}=\frac{9.8\times 0.5^3}{3\times E\times 0.054\times 10^{-8}}$

$\implies E=\frac{9.8\times 0.125}{3\times 4.2\times 0.054\times 10^{-10}}$

$\implies E=1.8\times 10^{10}$ N/m²

Hope this answer is helpful.

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