Calculating Tension in an Elevator Cable During Upward Acceleration

[alwysnforevr002]

An elevator weighing 2.00 x 105 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/s2? (g = 9.81 m/s2)

for this equation i think i may have found the answer but i am really not sure. I do not know if there is an equation for tension. Here is what i thought:

Weight = mass x acceleration
mass = weight in N divided by gravity
2 x 10(to the fifth) / 9.81 m/s
= 20387.4 x 3
=61162.1N

Also what i don't understand is if their is an equation specifically designed for Tension

 

[neutrino]

Draw a free-body diagram. What are the forces acting on the elevator? What you've calculated is the net force acting on the elevator, not the one due to the cable.

In this case, there aren't any equations "designed" for tension. Just consider the tension to be 'T'.

 

[denni89627]

well the tension varies in the cable. it would be a very complex equation i would think. as the elevator goes up, more weight (the weight of the cable) goes over a drive sheave and hangs down on the other (counterweight) side of that drive sheave. if you neglect the weight of the cable itself i would imagine you could add the upward acceleration to g, calculate the new g force (somewhere in the order of 1.3g) and multiply the original mass by that figure. the tension would still vary in the cable but you can at least find out the weight it is now supporting.

ps. I'm an elevator constructor so i had to jump on this question, although I'm sure someone more qualified in physics can be of more help

edit to say: the practical accelerations of modern elevators don't exceed 1.5m/s^2
More than that and you might need those air sickness bags!