A mild steel wire of length 1.0 m and ...

Mechanical Properties of Solids

Question : 19 of 21

Marks: +1, -0

A mild steel wire of length 1.0 m and cross-sectional area

0.50 \times 10^{-2} \mathrm{cm}^{2}

is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.

Solution:

Refer figure, let x be the depression at the mid point i.e.

CD = x.

In figure,

AC = CB = l = 0.5 \text{ m};

m=100 \text{ g}=0.100 \text{ kg},

A D = B D = (l^{2}+x^{2})^{1/2}

Increase in length,

\Delta l = AD + DB - AB = 2AD - AB

=2(l^{2}+x^{2})^{1/2}-2l

=2l\left(1+\frac{x^{2}}{l^{2}}\right)^{1/2}-2l

=2l\left[1+\frac{x^{2}}{2l^{2}}\right]-2l

=\frac{x^{2}}{l}

\therefore

Strain

=\frac{\Delta l}{2l}=\frac{x^{2}}{2l^{2}}

T

is the tension in the wire, then

2T\cos\theta = mg

T=\frac{mg}{2\cos\theta}

Here,

\cos\theta = \frac{x}{\left(l^{2}+x^{2}\right)^{1/2}}

=\frac{x}{l\left(1+\frac{x^{2}}{l^{2}}\right)^{1/2}}

=\frac{x}{l\left(1+\frac{1}{2}\frac{x^{2}}{l^{2}}\right)}

As,

x \ll l,

1 \gg \frac{1}{2}\frac{x^{2}}{l^{2}}

and

1+\frac{1}{2}\frac{x^{2}}{l^{2}} \approx 1 \therefore \cos\theta = \frac{x}{l}

Hence,

T=\frac{mg}{2\left(\frac{x}{l}\right)}=\frac{mgl}{2x},

Stress

=\frac{T}{A}=\frac{mgl}{2Ax}

Y=\frac{\text{stress}}{\text{strain}}

=\frac{mgl}{2Ax} \times \frac{2l^{2}}{x^{2}}

=\frac{mgl^{3}}{Ax^{3}}

\therefore x = l \left[ \frac{mg}{YA} \right]^{1/3}

=0.5 \left[ \frac{0.1 \times 10}{20 \times 10^{11} \times 0.5 \times 10^{-6}} \right]^{1/3}

=1.074 \times 10^{-2}

=1.074 \text{ cm}

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