A mild steel wire of cross-sectional area 0.60 x 10 -2 cm2 and length 2 m is stretched ( not beyond its elastic limit ) horizontally between two columns. If a 100g mass is hung at the midpoint of the wire, find the depression at the midpoint.

Asked by Pragya Singh | 1 year ago |  142

1 Answer

Solution :-

Let YZ be the mild steel wire of length 2l = 2m and cross sectional area A = 0.60 x 10 -2 cm. Let  the mass of m = 100 g = 0.1 kg be hung from the midpoint O, as shown in the figure. And let x be the depression at the midpoint  i.e OD

From the figure;

ZO =YO = l = 1 m ;

M = 0.1 KG

ZD = YD =  (l2 + x2)1/2

Increase in length, ∆l = YD + DZ – ZY

= 2YD – YZ    ( As DZ = YD)

\( 2(l^2 + x^2)\dfrac{1}{2}\)– 2l

∆l = 2l(\(\dfrac{x^2}{2I^2}\)) = \( \dfrac{x^2}{I}\)

Therefore, longitudinal strain =\( \dfrac{∆l}{2l}\)

\( \dfrac{x^2}{2I^2}\). . . . . . . . ( i )

If T is the tension in the wires, then in equilibrium 2Tcosθ = 2mg

Or,          

T = \( \dfrac{mg}{2cos θ}\)

\( \dfrac{ mg (l^2 + x^2)^\dfrac{1}{2}}{2x}\) 

\( \dfrac{mgl}{2x}\)

Therefore, Stress = \( \dfrac{T}{A}\)

\( \dfrac{mgl}{2Ax}\)   . . . . . . . . . . ( ii )

Y = \( \dfrac{stress}{strain} = \dfrac{mgl}{2Ax} \times \dfrac{2l^{2}}{x^{2}}\)

\( \dfrac{mgl^{3}}{2Ax^{3}}\)

x = \( l[\dfrac{mg}{YA}]^{\dfrac{1}{3}} = 1[\dfrac{0.1 \times 10}{20 \times 10^{11} \times 0.6 \times 10^{-6}}]^{\dfrac{1}{3}}\)

= 9.41 x 10-3 m

Answered by Pragya Singh | 1 year ago

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