Question

A lift is tied with thick iron wires and its mass is $1000kg$ The minimum diameter of wire, if the maximum acceleration of lift is $1.2m/s^2$ and the maximum safe stress is $1.4\times {10}^{8}N/m^2$, is ($g=9.8m/s^2$)

• A. $0.00141m$
• B. $0.00282m$
• C. $0.005m$
• D. $0.01m$

Answer 1

The correct option is D $0.01m$

When the lift is accelerated upwards with acceleration $a$, then tension in the rope is
$T=m(g+a)$
$T=1000(9.8+1.2)=11000N$
Now, stress$=\frac{F}{A}=\frac{T}{\pi r^2}$
or $r^2=\frac{T}{\pi \times stress}=\frac{11000\times 7}{22\times 1.4\times {10}^8}$
or $r^2=\frac{1}{4\times {10}^4}$
or $r=\frac{1}{200}$
So, $D=2r=2\times \frac{1}{200}$
$D=\frac{1}{100}=0.01m$