What is the power of an elevator?

Click For Summary
SUMMARY

The discussion centers on calculating the power required for an elevator system, specifically addressing the relationship between force, weight, and acceleration. Participants clarify that power is defined as the product of force and velocity, and that the net force acting on the elevator system must account for both the elevator's weight and the counterweight. The correct approach involves using the equation Power = Force x Velocity, where the force is derived from the net weight being lifted, which is 3300 lbs at a velocity of 15 ft/s. Additionally, the effects of acceleration on power requirements are discussed, emphasizing that the motor must provide additional force to achieve the desired acceleration.

PREREQUISITES
  • Understanding of Newton's laws of motion, particularly F = ma
  • Familiarity with the concepts of force, weight, and tension in mechanical systems
  • Knowledge of power calculations in physics, specifically Power = Work/Time
  • Ability to analyze free body diagrams for mechanical systems
NEXT STEPS
  • Study the principles of mechanical advantage in pulley systems
  • Learn about the dynamics of elevator systems and counterweights
  • Explore advanced power calculations involving acceleration and velocity
  • Investigate the implications of gravitational forces on moving systems
USEFUL FOR

Students studying physics, engineers designing elevator systems, and anyone interested in the mechanics of force and power in mechanical applications.

CGI
Messages
74
Reaction score
1

Homework Statement


media%2Fe4c%2Fe4c463dd-a209-4ffa-8091-5bfcc0fffbc0%2Fphp1kc8uI.png

Homework Equations


So I've been thinking about this one for a while and I've tried to read up about Power in general.
I think Power = Work/ Time = (Force x Distance)/Time or Force x Speed.

The Attempt at a Solution


F = ma, so if velocity is constant, a = 0. So that would mean that the force is zero, so would that make the motor have zero power? That doesn't really sound right. If someone could help me out, I would really appreciate it. Thanks in advance!
 
Physics news on Phys.org
F = ma, but you're forgetting about gravity working more on one side than the other.

Gravitational acceleration works on both the elevator and counterweight.. if they're the same mass, it cancels out, so it won't accelerate.. If they're a different mass, the motor will have to do (or absorb) work.

Draw out free body diagrams for it all to start with.
 
Alright. I got for the counter weight, that the sum of forces in the y direction with up as positive, is T - W = 0 or T = W for the FBD of the counter weight.
For the Elevator, I have that 3T - W = 0 or that 3T = W for the FBD at the elevator.

Would these be correct?
 
I'm a little lost.. What is T and W here, and why 3T?

If it's not accelerating, the motor has to do work on the difference of the two weights... so moving the elevator 5000 lb elevator up is moving the 2200 lb counterweight down, The net weight you're lifting will be 3300 lbs @ 15ft/s
 
Oh wow, I'm sorry. T in my FBD stands for tension in the cables and W stands for weight. In the elevator, I thought that it would be 3T - W = 0 because there were three cables lifting that elevator up. As for my counterweight, I thought it would T - W = 0 since there is only one cable so the tension would have to be equal to the weight.

But okay, if the net force of gravity on the system is 3300lbs, could I just multiply 3300 by 15 and have the Power generated by the motor?
 
Forget about the tension in the cables.. simplify it... if you look closely at the diagram, you'll notice it's a 1:1 movement ratio for the elevator : counterweight

The tension on the cable to the motor is half the weight of the elevator minus counterweight, but the tension in the cable is irrelevant to work.. the motor needs to move twice as far at half the load which means the work stays the same.

Yes, the motor is making 3300lbs * 15ft/s of work :)
 
Oh okay that makes things so simple! So when it says that there is acceleration upward, how would that change the power from the motor?
 
The motor will have to do more work.. and it'll need to apply more force to accelerate both masses.

Just as an exercise (and it'll give you a hint), how fast would the elevator accelerate if the cable to the motor broke?
 
It would just fall so wouldn't it just be the acceleration due to gravity?
 
  • #10
Nope... there's gravity acting on both weights, so the net force will be the difference of gravitational forces, but both weights need to accelerate at the same speed (one up, one down evidently)
 
  • #11
Well, yeah, OK, the acceleration will be due to gravity, but the acceleration will not be 32ft/s^2
 
  • #12
Rx7man said:
moving the elevator 5000 lb elevator up is moving the 2200 lb counterweight down, The net weight you're lifting will be 3300 lbs @ 15ft/s
3300?
 
  • #13
Oh that's right, the net weight would be 2800lbs.

And I don't think I'm understanding the way the acceleration would affect the elevator. I am so sorry :(
 
  • #14
CGI said:
Oh that's right, the net weight would be 2800lbs.

And I don't think I'm understanding the way the acceleration would affect the elevator. I am so sorry :(
You cannot simply add an extra power term to handle the acceleration. The power required for an acceleration depends on the current speed.
Easiest is to use power = force x velocity. What force would be required from the motor to achieve the desired acceleration?
 
Last edited:
  • #15
If I used F = ma with (a) as 3, would the force needed just be 3 x 5000?
 
  • #16
CGI said:
If I used F = ma with (a) as 3, would the force needed just be 3 x 5000?
In the equation F=ma, it is important to remember exactly what F represents.
 
  • #17
F in this case is just the applied amount of force on the elevator, no?
 
  • #19
Okay so F stands for the vector sum of forces on an object. So if I used F = ma, I would have to add up all the forces acting upon the elevator, which would not just be the force of gravity. Is that headed somewhere in the right direction?
 
  • #20
CGI said:
Okay so F stands for the vector sum of forces on an object. So if I used F = ma, I would have to add up all the forces acting upon the elevator, which would not just be the force of gravity. Is that headed somewhere in the right direction?
Yes. F=ma will tell you the net force, and from that you can deduce the force provided by the motor.
 
  • #21
the total power needed will be the power needed to lift it at it's current rate, plus additional power to accelerate it... Don't forget the effects of the counterweight either.
 
  • #22
Rx7man said:
the total power needed will be the power needed to lift it at it's current rate, plus additional power to accelerate it... Don't forget the effects of the counterweight either.
As I wrote in post #14, that doesn't really help. The power required to accelerate depends on current speed, not just on the desired acceleration rate. Finding the force that would give the required acceleration is as simple a route as any.
 
  • #23
You are right, but sometimes separating things helps me visualize how to solve the problem (it works for me)
 

Similar threads

Find the power needed to accelerate this elevator downward
  • · Replies 4 ·
Replies
4
Views
1K
What is the Average Power of an Elevator in Physics?
  • · Replies 17 ·
Replies
17
Views
2K
When is power constant in this scenario?
  • · Replies 25 ·
Replies
25
Views
2K
A Physics 11 Elevator Work/Power Problem
  • · Replies 3 ·
Replies
3
Views
4K
Power To Accelerate 1000kg Elevator Cab Upward
  • · Replies 31 ·
2
Replies
31
Views
3K
What is the power input of a motor pulling up an elevator?
  • · Replies 3 ·
Replies
3
Views
3K
Questioning the Solution for Elevator vs Rock Problem
  • · Replies 7 ·
Replies
7
Views
2K
Elevator Motion and Bathroom Scale Readings
  • · Replies 2 ·
Replies
2
Views
2K
Conservation of Energy Problem (Power)
  • · Replies 14 ·
Replies
14
Views
3K
Space elevator minimum initial speed
  • · Replies 3 ·
Replies
3
Views
2K