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Power Screw lifting a load

  1. Sep 10, 2014 #1
    1. The problem statement, all variables and given/known data
    A power screw with a pitch of 1 mm is directly coupled to a stepper motor rotating at 60 rpm. If
    the efficiency of the power screw is 35%, what is the stepper motor torque required to raise a
    4450 N (1000 lb) load?


    2. Relevant equations
    P=τ*ω
    τ=F*r

    3. The attempt at a solution
    60rpm*2∏/60=6.28 rad/s
    τ=4450*(.0005)=2.22 N-m (I assume this is wrong)

    The answer is suppose to be 2.023 N-m
     
  2. jcsd
  3. Sep 10, 2014 #2

    billy_joule

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    Yes, your answer is wrong.

    You need to use all the given data; the pitch to get mechanical advantage and the efficiency to account for torque loss.
     
  4. Sep 10, 2014 #3
    I think torque efficiency is just dividing my torque by .37 once I have solved the torque required? I really have no idea how to get torque though...
     
  5. Sep 10, 2014 #4

    billy_joule

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    Yes, that method will work for the torque loss.



    Where did 0.0005 come from?

    You don't need T=Fr to solve this.

    Do you know how to calc. the velocity of the load? via the velocity ratio of a screw thread?

    Once you know how fast the load is being lifted you calc. the power required (via P=Fv) then finally the torque.
     
  6. Sep 10, 2014 #5
    I thought the pitch was diameter, so I divided by 2 to get radius. And no I don't think I know how to do that, I just started power screws and I mostly only know about how gears work.
     
  7. Sep 10, 2014 #6

    billy_joule

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    No, pitch is not diameter.

    You need to study power screws before attempting exercises on them.
    They are just like an inclined plane which you should have covered in physics.
    Good luck!
     
  8. Sep 10, 2014 #7
    4450=(2pi*0.35*T)/.001
    T=2.028 N-m

    Is this correct?
     
  9. Sep 10, 2014 #8

    billy_joule

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    No.
    It looks like you are plugging numbers into random equations and hoping for the right answer, that calc. doesn't include the motor speed so how could it be right?

    Do you know what pitch is yet? Can you calculate the speed of a power screw from an angular velocity and a pitch?

    Do it step by step:

    Calc velocity of the load.
    calc power required to lift the load at that velocity.
    calc torque required (and account for losses) for the calculated power and given angular velocity.

    You need to include all units.
     
  10. Sep 10, 2014 #9

    CWatters

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    I agree but would suggest breaking as follows...

    Calc. velocity of the load.
    Calc. power required to lift the load at that velocity.
    Calc. the power required after accounting for losses (aka efficiency)
    Calc. torque required for the calculated power and given angular velocity.
     
  11. Sep 10, 2014 #10
    How do I calculate velocity based off pitch and angular velocity? I can't find any information on it.

    Would it just be v=(pitch*angular velocity)/2pi?

    v=(.001*6.28)/2*pi
    =.001
    (.001*4450/.35)/6.28= 2.023
     
    Last edited: Sep 10, 2014
  12. Sep 10, 2014 #11

    CWatters

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    No... or at least it might be but I don't get 2.023 as the velocity.

    I refer you to Billy's post. Specifically....

    and
    The values in the problem (eg 1mm pitch and 60rpm) appear to have been chosen to make things easy.

    For example 60rpm = 1rps. If the pitch is 1mm how far does the load move per revolution or per second?
     
  13. Sep 10, 2014 #12
    That is my calculation for Torque. My calculation for velocity is below. I'm not sure if it's right. I'll write it more clearly.

    Pitch=.001
    60*2pi/60=6.28 angular velocity

    Solving for velocity....

    V=(.001*6.28)/2*pi
    V=.001
     
  14. Sep 10, 2014 #13

    billy_joule

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    We still don't know what any of those numbers represent....You need to include units.

    You have the right number but any professor of mine would give zero marks.
    is it 0.001 Ω? furlongs/fortnight? cubits/afternoon? what about all the other figures?
     
  15. Sep 11, 2014 #14

    CWatters

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    What Billy said. You must state the units.

    In addition...

    If the screw is rotating at one revolution per second and the pitch is 1mm then seems rather obvious (eg no need for equations) that the load will move at 1mm per second.
     
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