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

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SUMMARY

The maximum tension a cable can withstand in an elevator is 19,500 N, with engineering standards requiring that it never exceed two-thirds of this limit, resulting in a safe tension of 13,000 N. Given the elevator's mass of 490 kg and a maximum acceleration of 2.24 m/s², calculations show that the elevator can safely accommodate 9 people, each weighing 65 kg. This is derived from the equation F = ma, considering both the upward tension and the downward gravitational force. The final formula used is T = m(a + g), confirming the calculations are accurate.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Basic knowledge of forces, including tension and gravity
  • Familiarity with units of mass (kg) and force (N)
  • Concept of acceleration in physics
NEXT STEPS
  • Study the implications of safety factors in engineering design
  • Learn about the dynamics of elevators and cable systems
  • Explore the effects of varying weights and acceleration on cable tension
  • Investigate other safety standards in mechanical engineering
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Engineering students, mechanical engineers, safety compliance professionals, and anyone involved in the design and operation of elevator systems.

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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