Answer to Question #249402 in Mechanics | Relativity for Jdbacani

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?if the acceleration of the elevator upward is 6 m/s2, what is the tension force in the cable

Answer 1

The maximum net force can be found by Newton's second law:

$$F_\text{net}=ma_\text{max}=T-mg,\\\space\\ a_\text{max}=\frac{T}{m}-g=\frac{0.2EA}{m}-g=5.2\text{ m/s}^2.$$

The tension force in the cable for 6 m/s² is

$$T=m(a+g)=1000(6+9.8)=15800\text{ N}.$$