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Power required to lift something vertically

  1. Dec 4, 2017 at 3:26 AM #1
    1. The problem statement, all variables and given/known data
    I'm using a force of 392 pound feet of force to move an object that weighs 250 lb a distance of 22 in in 1.3 seconds.. how do I determine how much power was consumed in this move?


    2. Relevant equations
    Power = force * distance/ time

    3. The attempt at a solution
    749 watts
     
  2. jcsd
  3. Dec 4, 2017 at 3:37 AM #2
    Isn't "pound-foot" the unit of a torque?
     
  4. Dec 4, 2017 at 3:39 AM #3
    Force is 392 pound force .....disregard the "foot"
     
  5. Dec 4, 2017 at 3:48 AM #4
    OK. Given the force, the distance and the time, your formula in 2) seems to be correct. Just wondering, why the weight of the object is stated. Are there more questions to answer you need the weight for or is it for confusion?
     
  6. Dec 4, 2017 at 10:52 AM #5
    No, the reason I included the 250 was to illustrate the difference between weight being lifted in force required to lift weight. If I had put 250 pounds into the equation, it would have given me an incorrect figure, correct?
     
  7. Dec 4, 2017 at 12:08 PM #6

    CWatters

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    If the applied force is constant for the full 1.3 seconds then yes that is the average power consumed during the movement.

    The instantaneous power at any time during the 1.3 seconds will vary because the mass is accelerating due to the applied force being greater than the weight.

    I calculated that the 250lb mass is moving upwards at about 2.6 meters per second by the time it gets to the top. If it's not then either I made a mistake or there is an error in the data somewhere.
     
  8. Dec 4, 2017 at 2:42 PM #7
    Yes, there is a difference between the force you apply and the weight force of the body. But like CWatters did, you could try to calculate what the lifting force does with the body and calculate its kinetic state at the end of the acceleration.
     
  9. Dec 4, 2017 at 3:07 PM #8

    haruspex

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    Are you sure of those numbers? The acceleration would be ((392-250)/250)g. The distance is 22 inches, right? I calculate it should achieve that height in under half a second.
     
  10. Dec 4, 2017 at 6:08 PM #9

    CWatters

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    Good point. I never checked if the data was consistent.
     
  11. Dec 4, 2017 at 6:20 PM #10
    Yes. There are centripetal forces at work here as well. So as the the weight is being lifted vertically, its also rotating at a rate of 1.57 rad/sec around a common center point or axle. I know its a little confusing. See attachment please.
     

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  12. Dec 4, 2017 at 7:02 PM #11

    haruspex

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    You'll need to add some description to that.
    It looks like two masses of 100lb on an arm rotating at a steady rate in ... a vertical plane?
    One mass is 22" closer to the axis than the other? But we do not know how long the arm is?
    Two positions are shown, but no angle specified.
    How did the mass become 250lb? Does the arm have mass 50lb?

    This being so, I do not understand how the distance moved becomes 22", nor why the thread title says a mass is being lifted vertically.

    Please provide a clear statement of the problem.
     
  13. Dec 4, 2017 at 7:11 PM #12

    CWatters

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    That drawing appears to show two 100lb weights = 200lbs not 250lbs.

    It also doesn't show the 392lb applied force, where it acts or the direction in which it acts.
     
  14. Dec 4, 2017 at 7:12 PM #13
    Yes, you are correct, it is rotating 360 degrees in a vertical plane. Much like a windmill or ferris wheel. The weight became 250 because the connecting rods that join the 100 pound weights are around 50 lb. So you have the two 100-pound masses and then another 50 lb in the connecting rods. The overall length of the arm is 10 ft. The two, 100lb masses make a vertical transition 22 in between the 11 o clock position and the 1 o'clock position. Takes around one second to make that part of the rotation.
     
  15. Dec 4, 2017 at 7:15 PM #14
    The 392 pound force is created by a electric linear actuator which is hard mounted to the back side of one of the 100-pound Masses.
     
  16. Dec 4, 2017 at 7:30 PM #15

    haruspex

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    Then the wrong correction was made to your original post:
    You meant torque of 392 foot pounds.
     
  17. Dec 4, 2017 at 7:37 PM #16

    haruspex

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    I cannot relate that to the diagram. 11 o'clock to 1 o'clock would only be a rotation of 60 degrees. The diagram seems to show (depending on which end of the arm is considered the "hour hand") either 11 o'clock to 7 o'clock or 5 o'clock to 1 o'clock, a rotation of 240 degrees.
    Either way, I do not see how that equates to lifting 200 pounds (certainly not 250 pounds since the arms do not lift) through 22". That would involve a rotation from, say, 6 o'clock to 12 o'clock.
     
  18. Dec 4, 2017 at 7:38 PM #17

    CWatters

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    They can't move vertically because the rig rotates while they move. Do you mean the weights slide along the central rod 22" while the angle of the rod changes?
     
  19. Dec 5, 2017 at 2:58 PM #18
    I am going to post a Solidworks drawing for reference. Can you guys delete the drawing from the thread once this discussion is over?
     
  20. Dec 5, 2017 at 5:29 PM #19

    CWatters

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    Don't know. I suggest you ask the moderators by reporting your own post with a comment.

    I think you also need to reword the question because it's clear the weights aren't simply being lifted vertically. I think it will be difficult to analyse what's going on. You would need to know things like....

    The exact force profile produced by the solenoid as I doubt it is constant. The figure you quoted is just the makers figure right?

    The instantaneous angular velocity immediately before and after the "lift".

    I suppose then you could calculate the energy input via the solenoid and compare it with the energy added to the wheel. However I would need a very good reason why I should put in a lot of effort to do so.
     
  21. Dec 5, 2017 at 6:25 PM #20
    cwatters. A gentleman by the name Merlin 3189 wrote me a message. I responded to him and informed him of my situation. I don't know if you can correspond with him directly or not, but I believe we've come to an agreement on what needs to happen before we pursue this any further. I'm going to show you guys the full scope of what I'm talking about complete with video animation simulation Etc. I just don't know where the best place to post this information would be. Hopefully he will tell me and we can go from there. I think you've been with me since my first post several months ago. I appreciate your patience and all the knowledge that you've given to me. It's been a real blessing.
     
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