Question

A 50kg mass is suspended from one end of a wire of length 4m and diameter of 3mm whose other end is fixed. What will be the elongation of the wire. Take the Young's modulus for the material of the wire to be 7 x 10 Nm ​

Answer 1

The elongation of the wire is $3.96 \times 10^{-3} \ m$

The given parameters;

• mass of the object = 50 kg

• length of the wire, l = 4 m

• diameter of the wire, d = 3 mm = 0.003 m

• Young's modulus = $7 \times 10^{10 \ Nm$

The Young's modulus is given as;

$E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae}$

The area of the wire is calculated as;

$A = \frac{\pi d^2}{4} = \frac{\pi (0.003)^2}{4} = 7.069\times 10^{-6} \ m^2$

The elongation of the wire is calculated as;

$E = \frac{FL}{Ae} \\\\e = \frac{FL}{AE} = \frac{(50\times 9.8) \times 4}{(7.069\times 10^{-6}) \times 7\times 10^{10}} \\\\e = 3.96 \times 10^{-3} \ m$

Thus, the elongation of the wire is $3.96 \times 10^{-3} \ m$

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Answer 2

We have that  the elongation of the wire is

$e=3.7*10^5m$

From the question we are told that

A 50kg mass

Length 4m

Diameter of 3mm

Young's modulus for the material of the wire to be $7 x 10 Nm$

Generally the equation for the elongation is mathematically given as

$e=\frac{\rho}{lE}$

Where

$\rho=\frac{F}{s}$

Therefore

$e=\frac{4x50x0.8}{\pi*(3*10^{-3})^2x7x10^10}$

$e=3.7*10^5m$

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