Maximum tension a cable can withstand.

[cdx]

Homework Statement

If the maximum tension that a cable can safely withstand is 9000N and the maximum upward acceleration of the elevator is 1.00 ms², how many 80kg passengers may be safely carried in the elevator?

(The mass of the cable is negligible)

Homework Equations

F=ma
T=mg

The Attempt at a Solution

If I let a_y=1.00 ms² and F=9000N then I have:
9000N = m * a_y and m = 9000. The number of 80kg people the elevator can withstand is 9000 / 80 = 112 people.

Is this correct?

 

[PhanthomJay]

No. You first need to know the mass of the elevator. Is it given? Then note that weight and mass are not the same, and that net force = ma. You need to identify all the forces acting on the elevator/people system. Hint: don't forget the weight acting down.

 

[cdx]

The mass of the elevator is 520 kg. Sorry I forgot to include this. Correct me here: There are two forces acting on the elevator, F_G and F_N.
F_G = mg = 520 kg * (-10) = -5200N.

Help?

 

[NBAJam100]

The mass of the elevator is 520 kg. Sorry I forgot to include this. Correct me here: There are two forces acting on the elevator, F_G and F_N.
F_G = mg = 520 kg * (-10) = -5200N.

Help?

I think the 2 forces you want to analyze in the Y direction are Fw and Ft, which oppose each other. Fw is acting in the negative Y direction and Ft is acting in the positive Y direction. Ft of course being the force of tension

So id look at it as the max force of tension it can hold is 9000N=Ft,

Ft-Fw=ma ----> Ft=ma+(Fw)