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

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SUMMARY

The tension in the supporting cable of an elevator with a combined mass of 1500 kg, moving downward at 11 m/s and brought to rest with constant acceleration over a distance of 43 m, can be calculated using the equations of motion. The time to decelerate was determined to be 0.78 seconds, leading to an acceleration of 14.1 m/s². The total force acting on the cable is calculated using F = ma, resulting in a force of 35,850 N when considering both gravitational and deceleration forces. The initial calculation contained an error due to a decimal mistake.

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  • Familiarity with kinematic equations
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  • Study kinematic equations in detail, focusing on deceleration scenarios
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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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