Question

A steel cable whose cross sectional area is 2.5 cm² supports a 1,000 kg elevator. The elastic limit of the cable is 3.0 X 10⁸ Pa. What is the maximum upward acceleration that be given the elevator if the tension in the cable is to be no more than 20 % of the elastic limit?

Answer 1

From Newton's second law:

$ma_{max}=T_{max}-mg,$

where m - the mass of the elevator;

$a_{max}$ - the maximum upward acceleration of the elevator;

$T_{max}$ - the maximum tension force of the cable.

Since the tension in the cable is to be no more than 20 % of the elastic limit:

$T_{max}=0.2\sigma_{lim}A,$

where $\sigma_{lim}$ - the elastic limit of steel;

A - the cross sectional area of the cable.

So,

$a_{max}=\frac{0.2\sigma_{lim}A}{m}-g=\frac{0.2\cdot 3\cdot10^8\cdot 2.5\cdot 10^{-4}}{1000}-9.8=5.2\space m/s^2.$

Answer: 5.2\space m/s^2.