Question

An elevator cable is to have a maximum stress of 7x10^7 N/m^2 to allow for appropriate safety factors. Its maximum upward acceleration is 1.5 m/s^2. If the cable has to support the total weight of 2000 kg of a loaded elevator, the area of cross-section of the cable should be

Answer 1

Step 1: Given that:

Maximum stress = $7\times {10}^{7}N{m}^{-2}$

Maximum upward acceleration(a) = $1.5m{s}^{-2}$

The weight that can be supported by the cable = $2000kgwt.$

Step 2: Concept and formula used:

Stress($\sigma$) is given as the ratio of the normal force acting on the unit area.

That is; $\sigma =\frac{F}{A}$

In case of an elevator;

When the elevator is in rest,

The weight of a body in the elevator = $mg$

When the lift moves in an upward direction with an acceleration a is given as;

The net weight of a body in the elevator = $(mg+ma)=m(g+a)$

Step 3: Calculation of the area of the cross-section of the cable:

Let the area of the cross-section of the cable is A, then

Using relation,

$\sigma =\frac{m(g+a)}{A}$ and putting the given values, we get;

$7\times {10}^{7}N{m}^{-2}=\frac{2000kg\times (9.8m{s}^{-2}+1.5m{s}^{-2})}{A}$

$A=\frac{2000\times 11.3}{7\times {10}^7}$

$A=\frac{22600}{7}\times {10}^{-7}$

$A=3228.57\times {10}^{-7}$

$A=3.23\times {10}^{-4}{m}^2$

Thus,

The area of the cross-section of the cable is $3.23\times {10}^{-4}{m}^2$ .