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Lever and Torque questions

  1. Feb 17, 2012 #1
    If I have a 10 foot long lever and the fulcrum is 1 foot from the end, if I place 2000 lbs. of force on the lever, what will be my "force" or torque at the end of the lever? And, very important, how much distance will the tip of the short end of the lever move up if the long end of the lever can only go down 1" at its end?

    Also, can I somehow translate the torque developed at the end of the lever into a horse-power calculation? Say I'm driving a crankshaft or piston into a cylinder (air-compressor) at the short end of the lever. What kind of HP and CFM can I generate?

    Thank you!
     
  2. jcsd
  3. Feb 17, 2012 #2
    The first part of your post is not concise. The static torque (I assume static) about the [itex] x-axis [/itex] is [itex] 2000 lb \times 10 ft[/itex] and about the [itex] y-axis [/itex] is [itex] 2000 lb \times 1 ft [/itex]. I assumed what you were trying to indicate by fulcrum as a bend of 90 deg. in the lever.

    You comment about "horsepower". Power is Torque[itex]\times[/itex]angular velocity. So you need some kind of movement (Kinetics/Dynamics) for there to be any power developed.
    Cheers,
     
  4. Feb 17, 2012 #3
    OK. I have a "lever", a 10 ft long piece of steel. It will be placed horizontally level with a "fulcrum" 1 or 2 ft from one end. 2000 lbs of kinetic energy (a vehicle) will roll over the other end of the lever thus applying the force. I want the lever to drive a crankshaft or piston into a cylinder and compress air. The "long" end of the lever cannot depress more than 1".

    If I can determine the foot pounds of pressure I can develop at the short end of the lever I can likely calculate the HP equivalency I will have to drive a piston.
     
  5. Feb 17, 2012 #4
    You are still not clear on the setup. A diagram would help.
    Comments:
    1) lbs is NOT kinetic energy.
    2) you cannot calculate HP from pressure.

    You need much more information to be able to do what I think you are doing.
     
  6. Feb 17, 2012 #5

    russ_watters

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    The car drives onto the "long" end and depresses it 1"? Well, if you apply 2000 lb of force over a distance of 1", that's 2000*1/12= 166.7 ft-lb of work (energy). If this happens in 1 second, you've applied 166.7/550 = 0.3 hp of power to the lever (for 1 second).

    Note, the torque and power are equal on each side of the lever.
     
  7. Feb 17, 2012 #6
    I agree. But you assumed both distance and time which are not specified in the OP.
    The OP also uses incorrect terminology.
    I believe the time factor is the most difficult to determine in a situation like this as it would require a dynamic simulation of the car-lever interaction.
     
  8. Feb 17, 2012 #7

    russ_watters

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    Distance is in the OP's second post - I only assumed the time.

    You're right that the OP butchered the units.
     
  9. Feb 17, 2012 #8
    Thank you for your patience!

    Simple diagram attached. Simply put: if a 4000 lb vehicle drives over a "bump" in the road, which in essence is a lever, and I have a fulcrum point 1 or 2 ft from the other end of the lever which, on its upward movement drives a crankshaft to an air compressor piston.
     

    Attached Files:

  10. Feb 17, 2012 #9

    russ_watters

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    My calculation was correct then - though the time is irrelevant, it is actually the frequency of cars driving over that matters.

    And FYI, we actually have had several threads about harvesting energy from highways in this manner. A forum search should turn them up.
     
  11. Feb 17, 2012 #10
    Guys! I'm on here because I don't know the answers to these questions and it's even difficult to phrase the questions properly! Sorry! Trying not to be a "BUTCHER"!
     
  12. Feb 17, 2012 #11
    Excellent! I will do a search. Thank you!
     
  13. Feb 17, 2012 #12
    Fair enough russ.
    Although for the OP. I think this calculation is much to simple. Approximating the force of a moving vehicle riding over a lever requires a kinetic analysis. The force will not be static.

    This is a good start however. Usually, it takes some practical application to convince people their calculations are far too simplistic and ideal.
    There are soo many unanswered questions. Vehicle speed, suspension kinematics, tire inflation and specs, etc.... these all will change the end result.
     
  14. Feb 17, 2012 #13
    Very interesting. Thank you very much.
     
  15. Feb 17, 2012 #14
    I guess I'm just looking for a Yay or Nay that I could move a crankshaft at the end of the lever with sufficient force to drive a piston and compress air. I have extensive uses for CA and solutions for compressed air storage and distribution systems. I'd like to aerate polluted bodies of water adjacent to a highway with the compressed air, for one example.
     
  16. Feb 17, 2012 #15
    Yes you can do what you are thinking to do. But 'engineering' a design that you can correlate properly or predict the output can't be done with one simple formula. You can try to estimate possible power but it will be very inaccurate like I mentioned above. Even if all the factors are taken into account there will still be some error. If you really want to attempt this I suggest contact modeling using a dynamic simulator like MSC ADAMS. By "hand" this problem can also be attempted (then coded to solve) but it needs high expertise in the filed of multibody dynamics.

    In short, your only problem in just trying it is designing the lever-crank interface properly.

    Cheers,
     
  17. Feb 20, 2012 #16
    Thank you very much, everyone. I am going to leave the kinematics to our mechanical engineering school partners, when the time comes. I will only be trying to pull together a proto-type model of my concept.

    Could someone please direct me to the proper forum for answering an initial question I had? I need to know the formula for calculating how far a lever moves up at one end of the lever depending upon how far it can move down at the other end and where the fulcrum is located.

    Thank you.
     
  18. Feb 20, 2012 #17

    russ_watters

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    It is just the ratio of the lengths of the two sides.
     
  19. Feb 20, 2012 #18
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