Calculating Strain in a Slowing Elevator Cable

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    Elevator Strain
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To calculate the strain in the elevator cable when it is brought to a stop, the total mass supported by the cable is 4110 kg (1260 kg for the elevator plus 2850 kg for the crowd). The force exerted on the cable can be determined using the impulse-momentum principle, considering the change in momentum over the stopping time of 0.600 seconds. The diameter of the cable is relevant for calculating its cross-sectional area, which is necessary for determining stress and subsequently strain. The user is seeking clarification on how to incorporate the deceleration of the elevator into the strain calculation, as they have only calculated strain for constant velocity. Understanding the initial velocity and applying the correct equations of motion will be essential for solving this problem.
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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!
 
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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...
 
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?
 
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