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Coverting Amp to pull force?

  1. Aug 24, 2012 #1
    Coverting Amp to pull force?!?!

    I am trying to figure out a way to covert amps to pull force(lbs). I have a known amp that I would like to find out the pull force in lbs.

    I have 2.8 amp @ 120v and would like to know the pull force that is needed to turn this type of amperage.
     
  2. jcsd
  3. Aug 24, 2012 #2

    Drakkith

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    Re: Coverting Amp to pull force?!?!

    Are you referring to the force from an electromagnet?
     
  4. Aug 24, 2012 #3
    Re: Coverting Amp to pull force?!?!

    Yes and no.....

    No in the case that I am working with neodymium magnets...and yes that I believe that the strengh of pull are both rated the same (pull force lbs)
     
  5. Aug 24, 2012 #4

    Drakkith

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    Re: Coverting Amp to pull force?!?!

    Amp is a measure of electrical current and does not directly relate to a pull force, so I don't understand your question. Are you trying to generate electrical power or something? Please give us some details of what you are trying to do.
     
  6. Aug 24, 2012 #5
    Re: Coverting Amp to pull force?!?!

    I have an electric motor that I'm "modifying". The motor is rated at 2.8amp @ 115v. Given simplified version of calculating horsepower. (a * v = w)*.0013=HP, my motor is .43hp. Where I get stuck is converting HP to pull force, I think this would be the easiest way. I am trying to make a system to convert magnet pull force into a rated HP to replace said motor. In other words, replace electrically created magnetic force into just magnet force ie: electricless motor :) I think what I need is a formula to calculate two like forces (pull force and hp) ...I hope this helps! It's been hindering me for a week
     
  7. Aug 24, 2012 #6

    Drakkith

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    Re: Coverting Amp to pull force?!?!

    Why don't you just find the torque generated by the motor and use that instead of "pull force". I'm not even sure what you mean by pull force. What is pulling what?
     
  8. Aug 24, 2012 #7
    Re: Coverting Amp to pull force?!?!

    Are you saying you are trying to create a motor powered by magnets only? If so I have some bad news for you.
     
  9. Aug 24, 2012 #8

    Drakkith

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    Re: Coverting Amp to pull force?!?!

    Ah, I missed that part of his post I guess. You are correct, replacing the electrical components with permanent magnets will not work.
     
  10. Aug 24, 2012 #9
    Re: Coverting Amp to pull force?!?!

    hp is without crank/arm of wrench to resist

    resistance is proportional to the length of the arm the wrench
    torqe = force * length of wrench
     
  11. Aug 24, 2012 #10
    Re: Coverting Amp to pull force?!?!

    possibly.... Given my original amps and volts, is there a way to calculate the pull the the winding are generating?
     
  12. Aug 25, 2012 #11

    Drakkith

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  13. Aug 25, 2012 #12

    K^2

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    Re: Coverting Amp to pull force?!?!

    Power can't give you the pull force. It depends on how fast you are pulling something. To pull something twice as fast at the same pull force, you need twice as much power. And vice versa. To pull something with twice as much force at the same speed requires the same doubling of power.

    Alternatively, but following the same logic as above, you can only get torque of an electric motor if you know at which RPM you are going to need that torque. Power divided by angular velocity will give you torque.

    The current times voltage is, indeed, power consumed by the motor. You do have to subtract the power wasted to heat, however. So the full formula is IV - IR², where R is the resistance of the coil.

    Finally, keep in mind that simply because the motor is rated at 2.8A and 115V, it doesn't mean it will be drawing 2.8A at 115V. The amount of current the motor draws will depend on the RPMs the motor is going at and applied voltage. The formula for current is I=(V-kω)/R, where R is aforementioned resistance, V is applied voltage, ω is angular velocity, and k is a constant unique to the motor. If you know the maximum RPM the motor reaches under no load and given voltage, you can estimate that constant by k=V/ωmax.

    Replacing magnets in a motor will effectively alter the constant k. R will remain the same, unless you change the coils as well. And these two constants effectively determine performance of an electric motor under ideal conditions. Naturally, real world tends to be slightly more complicated, but it's a very good estimate to start off with.

    Hope some of that helps.
     
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