Question

A steel wire with cross-section $3 \,cm^2$ has elastic limit 2.4 $\times$ 10⁸ N $m^{-2}$. The maximum upward acceleration that can be given to a $1200 \,kg$ elevator supported by this cable wire if the stress is not to exceed one-third of the elastic limit is (Take $g \,= \,10\, ms^{-2})$

• A. 12 $m\,s^{-2}$
• B. 10 $m\,s^{-2}$
• C. 8 $m\,s^{-2}$
• D. 7 $m\,s^{-2}$

Answer 1

Answer: (B)

Maximum tension an elevator can tolerate is
$T=\frac{1}{3}stress\times area of cross-section$
$\quad=\frac{1}{3}\times\left(2.4\times10^{8}\right)\times\left(3\times10^{-4}\right)=2.4\times10^{4} N$
If a is the maximum upward acceleration of elevator then T = m(g + a)
or 2.4 $\times$ 10⁴ = 1200(10 + a)
On solving, a = 10 m s^-2.