Tension and Acceleration - Elevator Cable?

Question

An elevator weighing 2.00 x 10^5 N is supported by a steel cable. What is the tension in the cable when the elevator is accelerated upward at a rate of 3.00 m/s^2?

assume g=9.81 m/s^2

How do I write out this equation correctly? Thanks

Answer 1

I'm going to assume there was a typo and the weight is actually 2.00 * 10^5 Kg.

There are two force on the cable:
1) in the downward direction gravity applies force on the elevator = 2.00*10^5 * 9.81 = 1.96*10^6 N
2) whatever is pulling the elevator upward applies force = 2.00*10^5 *3.00 = 6.00*10^5 N

Total tension is the sum = 1.96*10^6 + 6.00*10^5 = 2.56*10^6 N

Answer 2

m = F/a = 200E5 / 9.81 = 20.4E5 kg (3 sig figs)

T = 200E5 + 20.4E5 * 3.00 = 261E5 N

Conversely...you can add the acceleration of the elevator and the gravity.

Atot = 9.81 + 3.00 = 12.8

T = 20.4 * 12.8 = 261E5 N

Hehe...right Trevor...that's one hefty elevator!