First slide

Stress and strain

Question

The elastic limit of an elevator cable is 2 × $10^{9}$ N/ $m^{2}$ . The maximum upward acceleration that an elevator of mass 2 × $10^{3}$ kg can have when supported by a cable whose cross-sectional area is $10^{- 4}$ $m^{2}$ , provided the stress in cable would not exceed half of the elastic limit would be

Moderate

A

$10 m s^{- 2}$
B

$50 m s^{- 2}$
C

$40 m s^{- 2}$
D

Not possible to move up

Solution

From the free-body diagram of the elevator,
T - mg = ma T = m(g + a)
Stress in cable is $α = \frac{T}{A} = \frac{m (g + a)}{a}$
From the given condition, $α \leq \frac{α_{\max}}{2}$
$\frac{m (g + a)}{A} \leq \frac{σ_{\max}}{2}$ , $g + a \leq \frac{2 \times 10^{9}}{2} \times \frac{10^{- 4}}{2 \times 10^{3}} = 50$

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