How Many People Can an Elevator Hold Without Breaking the Cable?

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Homework Help Overview

The problem involves determining how many people an elevator can safely hold without exceeding the maximum tension in the cable. The context includes the mass of the elevator, the maximum allowable acceleration, and safety standards regarding cable tension.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the elevator, including tension and gravity, and how these relate to the maximum allowable tension. There are attempts to calculate the number of people based on different interpretations of the forces involved.

Discussion Status

Several participants have provided calculations and reasoning regarding the forces acting on the elevator. There is acknowledgment of the need to consider both the weight of the elevator and the additional weight of the passengers. Some participants express understanding and appreciation for the guidance received.

Contextual Notes

Participants are working under the constraint of safety standards that dictate the maximum tension in the cable, and there is a focus on ensuring that calculations account for both the elevator's mass and the mass of the passengers.

Voltrical
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Homework Statement



1) An elevator cable can withstand a maximum tension of 19,500N before breaking. The elevator has a mass of 490 kg and a maximum acceleration of 2.24 m/s2. Engineering safety standards require that the cable tension never exceed two-thirds of the breaking tension. How may 65-kg people can the elevator safely accommodate?



Homework Equations




F = ma


The Attempt at a Solution



F = ma

F = (2/3)*(19,500 N)
F = 13,000 N

Then:

n = { [(13,000 N) / (2.24 m/s^2)] - 490 kg } / 65 kg
n = 81.7

n = 81 people
 
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Do not forget gravity. The cable exerts the upward tension T and gravity exerts the downward force mg. The resultant of these forces accelerate the elevator upward.

ehild
 
Voltrical said:
n = { [(13,000 N) / (2.24 m/s^2)] - 490 kg } / 65 kg
In addition to what ehild said, don't forget that the entire elevator, people plus car, are accelerating.
 
This is what I've gotten so far.

Since "require that the cable tension never exceed two-thirds of the breaking tension" :
F = ma
F = (2/3)*(19500)
F = 13000

Max Acceleration = 2.24 m/s^2
Acceleration due to gravity = 9.81 m/s^2
Acceleration = 2.24 + 9.81 = 12.05 m/s^2

So,

F = ma
13000 = ((490kg)+(65kgx))(12.05m/s^2)
x = 9.059 people

Rounded it to 9 people.
 
Voltrical said:
This is what I've gotten so far.

Since "require that the cable tension never exceed two-thirds of the breaking tension" :
F = ma
F = (2/3)*(19500)
F = 13000

Max Acceleration = 2.24 m/s^2
Acceleration due to gravity = 9.81 m/s^2
Acceleration = 2.24 + 9.81 = 12.05 m/s^2

So,

F = ma
13000 = ((490kg)+(65kgx))(12.05m/s^2)
x = 9.059 people

Rounded it to 9 people.

Your "F" is the tension in the cable.

The elevator is accelerated by the sum of both forces: tension and gravity. The net force on the elevator is T-mg upward. So ma=T-mg.

Rearranging the equation, it becomes the same you used. T=m(a+g). The result is correct.

ehild
 
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Oh that makes sense now ehild.

Thanks a lot guys, the help is much appreciated.
 
Voltrical said:
Oh that makes sense now ehild.

Thanks a lot guys, the help is much appreciated.

You are welcome:smile:

ehild
 

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