Question

A load of 10 kg is suspended by a metal wire 3 m long and having a cross-sectional area 4 mm2. Find (a) the stress (b) the strain and (c) the elongation. Young modulus of the metal is 2.0 × 1011 N m−2.

Answer 1

(a). The stress is $2.5\times10^{7}\ N/m^2$

(b). The strain is $1.25\times10^{-5}$

(c). The elongation is $3.75\times10^{-5}\ m$

Explanation:

Given that,

Mass of load = 10 kg

Length = 3 m

Area =4 mm²

Young modulus $Y=2.0\times10^{11}\ N/m^2$

(a). We need to calculate the stress

Using formula of stress

$stress = \dfrac{F}{A}$

$stress =\dfrac{mg}{A}$

Put the value into the formula

$stress = \dfrac{10\times9.8}{4\times10^{-6}}$

$stress =24500000\ N/m^2$

$stress=2.5\times10^{7}\ N/m^2$

(b). We need to calculate the strain

Using formula of young modulus

$Y=\dfrac{FL}{A\times \Delta L}$

$\dfrac{\Delta L}{L}=\dfrac{F}{AY}$

$strain =\dfrac{\Delta L}{L}$

$strain=\dfrac{F}{AY}$

Put the value into the formula

$strain=\dfrac{10}{4\times10^{-6}\times2.0\times10^{11}}$

$strain=0.0000125$

$strain=1.25\times10^{-5}$

(c). We need to calculate the elongation

Using formula of stress

$elongation =\Delta L$

$stress=\dfrac{\Delta L}{L}$

$\Delta L=stress\times L$

Put the value into the formula

$\Delta L=1.25\times10^{-5}\times3$

$\Delta L=0.0000375\ m$

$\Delta L=3.75\times10^{-5}\ m$

Hence, (a). The stress is $2.5\times10^{7}\ N/m^2$

(b). The strain is $1.25\times10^{-5}$

(c). The elongation is $3.75\times10^{-5}\ m$

Learn more :

Topic : Young modulus

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