# Force question on elevator cable

1. Nov 7, 2007

### get_rekd

A 700 kg elevator suspended by a cable accelerates upwards at 3 m/s^2. The force exerted by the cable on the elevator is?
(use g=10 m/s^2)

m=700 kg
a= 3.0 m/s^2

Fnet= 700 kg * 3.0 m/s^2
= 2,100.0 N Upward force

Fnet= 700 kg * 10 m/s^2
= 7,000.0 N Downward force

Fnet= 2,100.0 N + 7,000.0 N = 9,100.0 N

So is 9,100.0 N the total force exerted on the cable?

2. Nov 7, 2007

### Dick

Yes, but you didn't express that very well. All of those different numbers can't be Fnet. One of them is Fnet=ma, one of them is gravitational force and one is the cable tension. Can you label them correctly?

3. Nov 7, 2007

### get_rekd

Fnet= 700 kg * 3 m/s^2
= 2,100.0 N

Fg= 700 kg * 10 m/s^2
= 7,000.0 N

T= 2,100.0 N + 7,000.0 N
= 9,100.0 N

What is the variable for tension. I am not sure so I just used T.
Would this be a more suitable answer?

4. Nov 7, 2007

### Dick

Better. Fnet is the sum of all forces. Fnet=T-Fg. So yes, T=Fnet+Fg.

5. Nov 7, 2007

### get_rekd

Thanks a lot. Appreciate your help.