Understanding the Power of Elevator Lift Systems

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SUMMARY

The discussion focuses on the mechanics of elevator lift systems, specifically analyzing the power dynamics between the elevator, its counterweight, and the motor. The key equation used is P=Fv, where the power is derived from the force exerted by the weights of the elevator (m1) and counterweight (m2) and their velocities. The participants clarify that the energy difference in the system during a finite time Δt is represented as (m1-m2)gΔh, and they emphasize the importance of understanding the signs in work done by the motor and gravity. The conversation concludes with a clear understanding that power can have directional implications despite being a scalar quantity.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of force, velocity, and power
  • Basic knowledge of gravitational force and its effects on mass
  • Ability to interpret energy equations in mechanical systems
NEXT STEPS
  • Study the principles of mechanical advantage in elevator systems
  • Learn about the role of counterweights in lift efficiency
  • Explore the implications of power transfer in mechanical systems
  • Investigate the effects of varying speeds on elevator performance
USEFUL FOR

Mechanical engineers, physics students, and professionals involved in elevator design and maintenance will benefit from this discussion, as it provides insights into the fundamental mechanics of lift systems and power dynamics.

Janiceleong26
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Homework Statement


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Homework Equations


P=Fv

The Attempt at a Solution


Since the elevator is rising at a steady speed, a=0, thus, the tension on the string equals to the weight of the elevator. I chose C as I thought just for the elevator case, it would be m1gv, but how does m2 affects the power of the system?
 
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Answer C is not correct.

Imagine the movement within a finite time Δt (and therefore a change of position of Δh for both m1 and m2). What is the energy difference of the whole system during Δt?
 
Janiceleong26 said:
Since the elevator is rising at a steady speed, a=0, thus, the tension on the string equals to the weight of the elevator. I chose C as I thought just for the elevator case, it would be m1gv, but how does m2 affects the power of the system?
Yes, the motor is working at that rate in raising the elevator, but what is happening on the other side? How much power is being transferred, and which way, between the counterweight and the motor?
 
stockzahn said:
Answer C is not correct.

Imagine the movement within a finite time Δt (and therefore a change of position of Δh for both m1 and m2). What is the energy difference of the whole system during Δt?
Energy difference would be (m1-m2)gΔh ?
 
haruspex said:
Yes, the motor is working at that rate in raising the elevator, but what is happening on the other side? How much power is being transferred, and which way, between the counterweight and the motor?
m2gv, and anti-clockwise? But I thought power is scalar? Does the minus sign comes from the velocity?
 
Janiceleong26 said:
Energy difference would be (m1-m2)gΔh ?
Correct. And instead of distance and energy you need velocity and power - so what to do?
 
stockzahn said:
Correct. And instead of distance and energy you need velocity and power - so what to do?
Divide it by Δt and you will get velocity and (m1-m2)g is the force, so force times velocity gives power. I got it! Thanks!
 
Janiceleong26 said:
m2gv, and anti-clockwise? But I thought power is scalar? Does the minus sign comes from the velocity?
Even scalar values can have sign. If A does positive work on B then B does negative work on A.
If you take up as positive and gravity as -g then the motor does work at rate -m1(-g)v on the elevator and at rate -m2(-g)(-v) on m2.
 
haruspex said:
Even scalar values can have sign. If A does positive work on B then B does negative work on A.
If you take up as positive and gravity as -g then the motor does work at rate -m1(-g)v on the elevator and at rate -m2(-g)(-v) on m2.
Oh I see, thanks. But why is there a negative sign before the mass?
 
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Janiceleong26 said:
Oh I see, thanks. But why is there a negative sign before the mass?
Because m(-g) is the force from gravity. The motor has to supply an equal and opposite force, -m(-g).
 
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