Deflection?

Solution

$First\quad note\quad that\quad AC=CB=l/2\quad where\quad l\quad is\quad the\quad total\quad length\\ of\quad the\quad wire.\quad By\quad observing\quad the\quad FBD\quad on\quad the\quad load\quad and\quad \\ writing\quad equation\quad for\quad equilibrium\quad we\quad get:\\ 2Fsin(\theta )=P\quad \quad \quad (i)\\ Now\quad AD=lsec(\theta )/2\quad \\ \Rightarrow Elongation=\Delta l=AD-AC=l/2(sec(\theta )-1)\quad (ii)\\ Now\quad writing\quad the\quad equation\quad \\ \frac { Stress }{ Strain } =Modulus\quad of\quad elasticity\\ we\quad get\quad \frac { F/A }{ \Delta l/(l/2) } =E\quad hence\quad F=\frac { 2\Delta lEA }{ l } .\\ Putting\quad the\quad value\quad of\quad \\ \Delta l\quad we\quad get\quad F=EA(sec(\theta )-1)=EA(\frac { 1-cos(\theta ) }{ cos(\theta ) } )=\frac { 2EA{ sin }^{ 2 }(\theta /2) }{ cos(\theta ) } \quad (iii)\\ Here\quad comes\quad the\quad trick\quad since\quad \theta \approx 0\quad we\quad can\quad write\\ sin(\theta )\approx \theta \quad and\quad cos(\theta )\approx 1\quad and\quad tan(\theta )\approx \theta \quad so\\ equation\quad (iii)\quad becomes\\ \\ F=EA{ \theta }^{ 2 }/2\quad putting\quad value\quad of\quad F\quad in\quad eq.\quad (1)\quad we\quad get\\ \\ P=(2\theta )(EA{ \theta }^{ 2 }/2)=EA{ \theta }^{ 3 }\quad \quad \\ Solving\quad \theta =\sqrt [ 3 ]{ \frac { P }{ EA } } \quad \quad (iv)\\ Now\quad our\quad vertical\quad elongation\quad \delta =ltan(\theta )/2\approx l\theta /2\quad \\ Using\quad equation\quad (iv)\quad we\quad get\quad \delta =\frac { l }{ 2 } \sqrt [ 3 ]{ \frac { P }{ EA } } \quad \\ So\quad m=2,n=3\quad hence\quad m+n=5\\ \\ \\ \\ \\ \\$

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For equilibrium,
$2T\sin\theta = P$

Using given condition, $\displaystyle \sin\theta \approx \theta \approx \tan\theta$

$2T\tan\theta = P$

$2T\frac{x}{\frac{l}{2}} = P$

$\frac{4Tx}{l} = P$

Now, let us find the extension in the wire.

$\Delta l = 2\sqrt{\left(\frac{l}{2}\right)^2 + x^2} - l$

Using Binomial expansion and simplifying the result,

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

Using the definition of Young's Modulus,

$E = \frac{\text{Stress}}{\text{Strain}} = \frac{\frac{T}{A}}{\frac{\Delta l}{l}}$

$E = \frac{Pl^3}{8Ax^3}$

Thus,

$x = \frac{l}{2} \left(\frac{P}{AE}\right)^{\frac{1}{3}}$

Hence,

$\boxed{m+n = 5}$