# Strain on a slowing elevator

1. Nov 7, 2004

### Tycho

A 1260kg freight elevator is supported by a steel cable of diameter 34.9mm. It is loaded with a crowd of people collectively having a mass of 2850kg and it is descending.
What is the strain in the cable when it is brought to a stop in 0.600s?

I'm not really sure how to figure in the fact that the elevator is slowing in this equation? a little push in the right direction would be GREATLY GREATLY appreciated. thanks!

2. Nov 7, 2004

### Phymath

i believe this is an impluse question, idk what the diameter has to do with anything...

$$\frac{dp}{dt} = F$$
$$-(1260kg + 2850kg)v = F(.6s)$$
solve for F and that is what i would assume "strain" means, however idk what the v would be is it given? idk...sorry...

3. Nov 7, 2004

### Tycho

you need the diameter to find the cross-sectional area of the cable. i understand all that, but i need the strain when it is slowing. i got the strain when it was moving at a constant velocity (i think).
does anyone know how to figure this?