Question

A mild steel wire of length 1.0 m and cross-sectional area 0.50 x ${10}^{-2}$cm² is stretched, well within Its elastic limit, horizontally between two pillars. A mass of 100g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.

Help __!!
Don't spam ___!!

​

Answer 1

Given :

Length of steel wire,l = 1m

cross sectional area,A = $\sf\:0.5\times{10}^{-2}$cm²

mass of 100 g is suspended from the mid of the wire .

To find :

The depression at the mid-point.

Solution :

Let x be the depression at the mid point .

and let length of wire be 2l.

From the Figure (attachment 1)

DB= x

AC = 2l

AD=DC=l= 0.5 m

AB=BC=$\sf\:\sqrt{l{}^{2}+x{}^{2}}$

m= 100g= 0.1kg

_________________________

intital length = AC=2l

Final length = AB+BC

=$\sf\:\sqrt{l{}^{2}+x{}^{2}}$

⇒Incease in legth= ∆l

= $\sf\:AB+BC-2l$

$\sf = 2 \sqrt{l {}^{2} + x { }^{2} } - 2l$

$= 2l \sqrt{1 + \sf \dfrac{x {}^{2} }{l {}^{2} } } - 2l$

Now use Binomial expansion

$= 2l(1 + \sf\dfrac{x {}^{2} }{2l {}^{2} } ) - 2l$

$\sf = \dfrac{x {}^{2} }{l}$

$\sf strain = \dfrac{change \: in \: length}{original \: length}$

$\sf \implies strain = \frac{ \triangle l}{2l}$

$\sf = \dfrac{ \frac{x {}^{2} }{l} }{2l} = \dfrac{x {}^{2} }{2l {}^{2} } ...(1)$

Now ,draw a free body diagram of suspended mass ,m (attachment 2)

If t is the tension in the wire ,then

$\sf2t \cos \theta = mg$

[from the free body diagram]

$\sf \implies t = \dfrac{mg}{2 \cos \theta}$

From the Figure;

$\sf \cos \theta = \dfrac{x}{ \sqrt{l {}^{2} + x {}^{2} } }$

Now use Binomial expansion in the Denominator

$\sf \sqrt{l {}^{2} + x {}^{2} } = l \sqrt{1 + \frac{x {}^{2} }{l {}^{2} } }$

$\sf = l(1 + \frac{x {}^{2} }{2l {}^{2} } )$

since, x<<l,$\sf so \:1 > > \frac{x {}^{2} }{2l {}^{2} }$

$\sf \implies1 + \frac{x ^{2} }{2l {}^{2} } \approx1$

Thus,

$\sf \cos \theta = \frac{x}{l}$

Therefore,

$\sf t = \dfrac{mgl}{2x}$

We know that

$\sf stess = \dfrac{norml \: force}{area}$

$\sf = \dfrac{mgl}{2Ax}$

$\sf youngs \: modulus = \dfrac{normal \: stress}{strain}$

We know that for steel Young's modulus

= $\sf\:2\times10{}^{11}$

$\sf \implies Y = \dfrac{mgl}{2ax} \times \dfrac{2l {}^{2} }{x {}^{2} }$

$\sf\implies x = l( \frac{mg}{YA}) {}^{ \frac{1}{3} }$

Now put the given values

$\sf{x=0.5\left[\dfrac{0.1\times9.8}{2\times10^{11}\times0.5\times10^{-6}}\right]^{\frac{1}{3}}}\\\\\\\sf{x=0.5\left[\dfrac{0.1\times9.8}{2\times\dfrac{5}{10}\times10^{11}\times10^{-6}}\right]^{\frac{1}{3}}}\\\\\\\sf{x=0.5\left[0.1\times10^{-11}\times10^6\times9.8\right]^{\frac{1}{3}}}\\\\\\\sf{x=0.5\left(9.8\times10^{-6}\right)^{\frac{1}{3}}}\\\\\\\sf{x=0.5\times(9.8)^{\frac{1}{3}}\times10^{-2}}\\\\\\\sf{x=0.5\times2.14\times10^{-2}}\\\\\\\sf{x=1.07\times10^{-2}}\\\\\\\sf{x=1.07\ cm}$

Therefore;

$\purple{\boxed{\large{\bold{Depression\:at\:the\:mid\:point=1.074cm}}}}$

__________________________

Binomial Expansion:

$\bf (1+x)^{n}= 1+ nx + \frac{n(n-1)}{2!}x^{2} +\frac{n(n-1)(n-2)}{3!} x^{3}+..$

Answer 2

The fig. given below shows the steel wire of length $\sf{L=1\ m}$ which becomes $\sf{L'}$ after stretching. Let the depression at the midpoint be $\sf{d.}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(-25,0){\line(1,0){50}}\put(0,0){\line(0,-1){5}}\put(0,-5){\line(5,1){25}}\put(0,-5){\line(-5,1){25}}\multiput(-12.5,3)(25,0){2}{\scriptsize\text{ \sf{\dfrac{L}{2}} }}\multiput(-12.5,-8)(25,0){2}{\scriptsize\text{ \sf{\dfrac{L'}{2}} }}\put(1,-3){\scriptsize\textsf{d}}\put(-3,-3.5){ \theta }\end{picture}$

Thus we have,

$\longrightarrow\sf{L'=2\sqrt{\left(\dfrac{L}{2}\right)^2+d^2}}$

$\longrightarrow\sf{L'=2\sqrt{\dfrac{L^2}{4}+d^2}}$

$\longrightarrow\sf{L'=2\sqrt{\dfrac{L^2+4d^2}{4}}}$

$\longrightarrow\sf{L'=\sqrt{L^2+4d^2}}$

$\longrightarrow\sf{L'=\left[L^2+4d^2\right]^{\frac{1}{2}}}$

Putting $\sf{L=1,}$

$\longrightarrow\sf{L'=\left[1+4d^2\right]^{\frac{1}{2}}\quad\quad\dots(1)}$

We consider a small depression $\sf{d,}$ so that $\sf{d^2}$ can be neglected.

We know that, for $\sf{|x|\ \textless\textless\ 1,\ \left[1+x\right]^\frac{1}{2}\approx1+\dfrac{x}{2}.}$ Hence (1) becomes,

$\longrightarrow\sf{L'=1+2d^2\right\quad\quad\dots(2)}$

Let $\sf{\theta}$ be the angle made by the depression with the half of new length of the wire, as shown in the fig. Then,

$\longrightarrow\sf{\cos\theta=\dfrac{d}{\left(\dfrac{L'}{2}\right)}}$

$\longrightarrow\sf{\cos\theta=\dfrac{2d}{L'}}$

From (2),

$\longrightarrow\sf{\cos\theta=\dfrac{2d}{1+2d^2}}$

Well, $\sf{d^2}$ could be neglected, right?!

$\longrightarrow\sf{\cos\theta=2d\quad\quad\dots(3)}$

Let the extension in the length of the steel wire be $\sf{\Delta L,}$ which is,

$\longrightarrow\sf{\Delta L=L'-L}$

From (2),

$\longrightarrow\sf{\Delta L=1+2d^2-1\quad\quad[\,L=1\ m\,]}$

$\longrightarrow\sf{\Delta L=2d^2}$

The fig. given shows the steel wire of cross sectional area $\sf{A=0.5\times10^{-2}\ cm^2=0.5\times10^{-6}\ m^2,}$ with $\sf{100\ g=0.1\ kg}$ mass suspended from its center.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(-25,0)(4,0){13}{\line(1,0){2}}\put(0,0){\line(0,-1){5}}\put(0,-5){\line(5,1){25}}\put(0,-5){\line(-5,1){25}}\put(0,-5){\vector(0,-1){10}}\put(-1,-19){ \sf{F} }\put(25,0){\vector(4,1){15}}\put(-25,0){\vector(-4,1){15}}\multiput(-25,0)(50,0){2}{\vector(0,1){5}}\multiput(-43,0)(84,0){2}{ \sf{A} }\multiput(-25.6,6)(42,0){2}{ \sf{A\cos\theta} }\multiput(-3,-3.5)(4,0){2}{ \theta }\multiput(-28,2)(54.5,0){2}{ \theta }\end{picture}$

The stretching force is nothing but the weight of the mass, i.e.,

$\longrightarrow\sf{F=1\ N\quad[\,g=10\ m\,s^{-2}\,]}$

As in the fig., we should not consider vertical component of $\sf{A}$ because the vectors $\sf{A\ \&\ F}$ are not acting on same line thus vertical component of

We take the net area acting along the vertical, thus,

$\longrightarrow\sf{A'=2A\cos\theta}$

From (3),

$\longrightarrow\sf{A'=2d\times10^{-6}}$

We know, Young's Modulus of Steel is,

$\longrightarrow\sf{Y=2\times10^{11}\ Pa}$

$\longrightarrow\sf{\dfrac{FL}{A'\,\Delta L}=2\times10^{11}\ Pa}$

Putting values for each,

$\longrightarrow\sf{\dfrac{1}{2d\times10^{-6}\times2d^2}=2\times10^{11}\ Pa}$

$\longrightarrow\sf{\dfrac{1}{d^3}=0.8\times10^{6}\ m^{-3}}$

$\longrightarrow\sf{d^3=1.25\times10^{-6}\ m^3}$

$\longrightarrow\sf{\underline{\underline{d=1.077\ cm}}}$

Hence the depression is nearly $\mathbf{1.077\ cm.}$