Question

A steel wire of length

1 m

and diameter

3 mm

is stretched horizontally between supports attached at its ends. What load hung from the mid-point of the wire would be required to produce a depression of

1 cm

?

(Y = 2 \times 10^{11} N / m^{2}, g = 9.8 m / s^{2})

.

Accepted Answer

Given,

The length of the wire, $L = 1 m$

Diameter of the wire, $d = 3 mm$ , radius $r = 1.5 mm$

Depressed length, $x = 1 cm$

Young's modulus, $Y = 2 \times 10^{11} N / m^{2}$

Acceleration due to gravity, $g = 9.8 m / s^{2}$

Let's consider, $2 L =$ Total length of the wire

$x =$ Depression

$T =$ Tension in the wire

$W =$ Load

Now, the length of wire, $L' = 2 O X'$

$L' = 2 {(L^{2} + x^{2})}^{\frac{1}{2}} L' = 2 L {(1 + \frac{x^{2}}{L^{2}})}^{\frac{1}{2}} I f, x^{2} < < < < < L^{2} L' = 2 L (1 + \frac{1}{2} \frac{x^{2}}{L^{2}}) L' - 2 L = \frac{x^{2}}{L^{2}} △ L = \frac{x^{2}}{L^{2}}$

Strain $= \frac{△ L}{2 L} = \frac{x^{2}}{2 L^{2}}$

Load is

$W = 2 T \sin θ, If \sin θ is very small, θ = \frac{x}{L} W = 2 T \frac{x}{L} T = \frac{W L}{2 x}$

Stress $= \frac{T}{{πr}^{2}} = \frac{W L}{2 {πr}^{2} x}$

We know that, $Y = \frac{Stress}{Strain}$

$Y = \frac{\frac{W L}{2 {πr}^{2} x}}{\frac{x^{2}}{2 L^{2}}} = \frac{m g L^{3}}{{πr}^{2} x^{3}}$

$m = \frac{Y {πr}^{2} x^{3}}{g L^{3}} m = \frac{2 \times 10^{11} \times 3.14 \times {(1.5 \times 10^{- 3})}^{2} \times {(1 \times 10^{- 2})}^{3}}{9.8 \times {(1)}^{3}} m = 1.15 kg$

Exercise 1

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