Max and min forces on an elevator.

[yeyspaz]

" A elevator with a mass of 4850 kg is to be designed so that the maximum acceleration is 0.0600 g's. What are the max and min forces the motor should exert on the cable?"
I converted the .0600 gs to .558 m/s/s, but when iplugged it into the F=ma formula i got the wrong answer? am i supposed to use another formula?

 

[kirbykirbykirby]

Yes.

I think you use Fnet = Ft - Fg

Where Ft is the forces the motor exerts on cable.

^
|
| Ft
^ |
| ______
Fnet | |
_______
|
| Fg
V


For the min Fnet = 0 so Fg = Ft so Ft = 4850 x 9.8

and the max Fnet = Ft - Fg
would be (4850)(0.06)(9.8) + (9.8)(4850) = Ft

I could be wrong because I'm crap at Physics but, eh, at least I moved your question to the top again.

 

[daniel_i_l]

You used the right equation but with the wrong variables. Think, on the way down gravity makes the elevater to accelerate at 9.8m/s/s so to cancel that out the moter needs to pull enough to accelerate the elevator upwards with the acceleration of 9.8 + .588 m/s/s. (this is what kirby said but I didn't see him until I posted)