Q19  A mild steel wire of length 1.0 m and cross-sectional area 0.50 \times 10 ^{-2} 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 midpoint.

Answers (1)
S Sayak

Let the ends of the steel wire be called A and B.

length of the wire is 2l = 1 m.

The cross-sectional area of the wire is A=0.50\times 10^{-2}cm^{2}

Let the depression at the midpoint due to the suspended 100 g be y.

Change in the length of the wire is \Delta l

\\\Delta l=AD+DB-AB\\ \Delta l=\sqrt{l^{2}+y^{2}}+\sqrt{l^{2}+y^{2}}-2l\\ \Delta l=2\sqrt{l^{2}+yx^{2}}-2l\\ \Delta l=2l\left (\left ( 1+\frac{y^{2}}{l^{2}} \right )^{\frac{1}{2}}-1 \right )\\ \Delta l=2l(1+\frac{y^{2}}{2l^{2}}-1)\\ \Delta l=\frac{y^{2}}{l}

The strain is \frac{\Delta l}{2l}=\frac{y^{2}}{2l^{2}}

The vertical components of the tension in the arms balance the weight of the suspended mass, we have

\\2Fcos\theta =mg\\ 2F\frac{y}{\sqrt{l^{2}+y^{2}}}=mg\\ F=\frac{mg\sqrt{l^{2}+y^{2}}}{2y}\\ F=\frac{mgl}{2y}

The stress in the wire will be 

\frac{F}{A}=\frac{mgl}{2Ay}

The Young's Modulus of steel is Y=2\times 10^{11}Nm^{-2}

\\Y=\frac{Stress}{Strain}\\ Y=\frac{\frac{mgl}{2Ay}}{\frac{y^{2}}{2l^{2}}}\\ Y=\frac{mgl^{3}}{Ay^{3}}\\

\\y=\left ( \frac{mgl^{3}}{AY} \right )^{\frac{1}{3}}\\ y=\left ( \frac{100\times 10^{-3}\times 9.8\times (0.5)^{3}}{0.5\times 10^{-2}\times 10^{-4}\times 2\times 10^{11}} \right )\\ y=1.074\times 10^{-2}m\\ y=1.074cm

The depression at the mid-point of the steel wire will be 1.074 cm.

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