Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A lift is tied with four iron wires and its mass is 1500Kg.If maximum acceleration of lift is $2m/s^2$ and maximum safe stress is $2\times 10^8N/m^2$ find the minimum diameter of the wire $g=10m/s^2$

$\begin{array}{1 1}(A)\;.01m\\(B)\;.005m\\(C)\;.5m\\(D)\;\text{None of above}\end{array} $

Can you answer this question?

1 Answer

0 votes
If the maximum acceleration of lift is $2m/s^2$ and then maximum force is $m(g+a)$
$\Rightarrow 1500\times 12$
$\Rightarrow 18000N$
Force on 1wire=$\large\frac{18000}{4}$N
$\Rightarrow 4500N$
Maximum stress =$\large\frac{4500}{\pi r^2}$
This should be less than given safe stress
$\large\frac{4500}{\pi r^2}$$ < 2\times 10^8$
$716.2\times 10^{-8} < r^2$
$26.76\times 10^{-4} < r$
$.00535 < D$
Hence (B) is the correct answer.
answered May 8, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App