Given max tension, find max upward acceleration

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Homework Help Overview

The discussion revolves around calculating the maximum upward acceleration of an elevator with worn cables, given its mass and the maximum tension the cables can withstand. The subject area includes dynamics and forces acting on objects in motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between tension, mass, and acceleration, with some suggesting the use of the equation Ef = ma as a starting point. There are repeated requests for confirmation of a specific calculation regarding maximum acceleration.

Discussion Status

The discussion includes attempts to verify calculations and clarify the setup of the problem. Some participants have provided feedback on the notation used, indicating a focus on ensuring clarity and correctness in the mathematical representation.

Contextual Notes

There are indications of confusion regarding the mass value due to a typographical error, as well as preferences for using proper units and notation in calculations. The original poster's assumptions about the problem setup are also being examined.

bigman8424
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a elevator with very worn cables has mass of 1600 kb, cables can withstand max tension of 24000 N, what is max upward acceleration if the cables are not to break

please confirm my answer to be amax = 24000/16000 -9.8 = 5.2 m/s2
 
Last edited:
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bigman8424 said:
a elevator with very worn cables has mass of 1600 kb, cables can withstand max tension of 24000 N, what is max upward acceleration if the cables are not to break

start this problem with Ef = ma?

Show us your attempt to solve this problem.
 
please confirm 24000/16000 -9.8 = 5.2 m/s2

please confirm 24000/16000 -9.8 = 5.2 m/s2
 
bigman8424 said:
please confirm 24000/16000 -9.8 = 5.2 m/s2

Too many zeros in 16000, but the result is OK. It's better to write your exponents as s^2 rather than s2. It's even better to use the built in LaTex and keep track of units to write

[tex]a = \frac{24000N}{1600kg} - g = 5.2 m/s^2[/tex]
 

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