First slide

Stress and strain

Question

A lift is tied with thick iron and its mass is 314 kg. What should be the minimum diameter of wire if the maximum acceleration of lift is 1.2 $m / \sec^{2}$ and the maximum safe stress of the wire is 1 x $10^{7} N / m^{2}$ ?

Moderate

A

2 cm
B

1 cm
C

1.5 cm
D

none of these

Solution

The tension, T in the rope of the lift when it goes upward is given by
$T = m (g + a) = 314 \times 11 N$
Let r be the radius of the wire, then maximum stress will be $\frac{T}{{πr}^{2}}$
Hence, $\frac{T}{{πr}^{2}} = 1.1 \times 10^{7}$
or $r^{2} = \frac{T}{π \times 1.4 \times 10^{8}} = \frac{314 \times 11}{3.14 \times (1.1 \times 10^{7})} = \frac{1}{10^{4}}$
Now $r = \frac{1}{10^{2}} m = 1 cm$
$∴$ Diameter of the wire = 2r = 2cm.

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