Calculating Strain in a Slowing Elevator Cable

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SUMMARY

The discussion centers on calculating the strain in a steel cable supporting a 1260kg freight elevator loaded with an additional 2850kg of people, as it descends and is brought to a stop in 0.600 seconds. The relevant formula involves impulse, where the change in momentum is equal to the force applied over time. The diameter of the cable, 34.9mm, is crucial for determining the cable's cross-sectional area, which is necessary for calculating strain. The user seeks clarification on how to incorporate the deceleration into the strain calculation.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with the concept of impulse and momentum
  • Knowledge of strain and stress in materials
  • Ability to calculate cross-sectional area from diameter
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  • Study the relationship between force, mass, and acceleration using Newton's laws
  • Learn how to calculate impulse and its application in real-world scenarios
  • Explore material properties related to strain and stress in engineering contexts
  • Investigate the formulas for calculating cross-sectional area from diameter in circular sections
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Mechanical engineers, physics students, and anyone involved in elevator design or safety assessments will benefit from this discussion.

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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...

[tex]\frac{dp}{dt} = F[/tex]
[tex]-(1260kg + 2850kg)v = F(.6s)[/tex]
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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