What is the equation for calculating tension in an elevator rope?

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Homework Help Overview

The discussion revolves around calculating the tension in an elevator rope when the elevator is decelerating. The problem involves a combined mass of 1500 kg and a specific scenario where the elevator is brought to rest from a downward motion.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of time required for deceleration and the application of kinematic equations. There are attempts to clarify the method used to find the time and the resulting acceleration.

Discussion Status

Some participants are questioning the calculations presented, particularly regarding the time to decelerate. There is a recognition of potential errors in the original calculations, and guidance is being sought to clarify the approach taken.

Contextual Notes

Participants note the importance of correctly applying kinematic equations and the implications of the initial conditions provided in the problem. There is an acknowledgment of the age of the thread, which may affect the relevance of the discussion.

B-80
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An elevator and its load have a combined mass of 1500 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 11 m/s, is brought to rest with constant acceleration in a distance of 43 m.

First I found the time it to to decelerate
D=(V+Vo)/2 T
gave me a value of .78seconds
plugged it into V=Vo+at
a=14.1m/s^2

F=ma
so the force is 1500Kg * (9.8+14.1) which is 35850
thats wrong, thanks in advance
 
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B-80 said:
First I found the time it to to decelerate
D=(V+Vo)/2 T
gave me a value of .78seconds
Redo this calculation.
 
thanks, stupid decimal.
 
How did you figure out the time?
 
SolidTwinz said:
How did you figure out the time?
Read the first post. (Distance = average speed X time)

This thread is 4 years old.
 

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