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Lever Question

  1. Sep 1, 2005 #1
    A lever 2 meters long is used to lift up a load of 1000 Newtons. The lever gives a mechanical advantage of 4. Sketch the lever and show the forces at each end of the lever and at the pivot point.

    Steer me in the right direction. What class of lever are we talking about?
     
  2. jcsd
  3. Sep 1, 2005 #2

    Fermat

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    You're talking about a straight bar. With a load of 1000N at one end. And a force of P newtons, say, at the other end which you have to work out, The lever/bar will have a fulcrum, or pivot point, at some point along it to give the MA of 4. You should also mark on the reaction at the fulcrum.
     
  4. Sep 1, 2005 #3

    quasar987

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    It's the one kind Archimedes was refering to when he said 'Give me a rod long enough and I shall lift the earth.'

    I believe that the mention of Mechanical Advantage imediately tells you what kind of lever we're talking about...
     
  5. Sep 1, 2005 #4

    Chi Meson

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    As for which class lever... it could be any one of the three. Just get the ratio of lever arms correct for whichever you pick.
     
  6. Sep 1, 2005 #5

    HallsofIvy

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    Not quite true. With a class 3 lever, the mechanical advantage is always less than 1.

    The basic concept for a lever is "conservation of energy" together with "work = force times distance". To have a mechanical advantage of 4, the load must be 4 times as great as the force applied. Set up your equation with "force applied*applied lever arm= (4*force applied)*load lever arm" to determine the relation between the two lever arms.
     
  7. Sep 3, 2005 #6
    --------------------------------------------------------------------------

    A force of P newtons equals

    1000 Newtons
    ------------
    4

    = 250 Newtons

    So what we have so far is a load of 1000 Newtons on one end and a force of 250 Newtons on the other.
     
  8. Sep 3, 2005 #7

    HallsofIvy

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    Since this has been here for a while now, with a class 1 lever (you have the fulcrum between the applied force and the load), the mechanical advantage is the length of the "application arm" (the distance between the application point and the fulcrum) and the length of the "load arm" (the distance between the load and the fulcrum). Here that must be 4. Draw a picture. label the length between the application and the fulcrum "4" and the length between the fulcrum and the load "1". What is the length of the entire lever?
     
  9. Sep 3, 2005 #8

    Chi Meson

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    I knew that. Don't you think I knew that? I knew that. Of course I knew that. I did. :grumpy:
     
  10. Sep 4, 2005 #9
    -----------------------------------------------------
    The length of the entire lever is 4m.
    The ma=4 x 1
     
  11. Sep 4, 2005 #10

    Doc Al

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    Then why does the problem state "A lever 2 meters long..."
     
  12. Sep 4, 2005 #11
    i'd say, by logics, that the fulcrum is at 1/5th of the length, starting at the end which lifts, am i right?
     
  13. Sep 4, 2005 #12
    formula

    -------------------------------------------------------------

    Is there a formula I use to calculate the reaction at the fulcrum?
     
  14. Sep 5, 2005 #13

    Fermat

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    All the forces - load, effort and reaction (at the fulcrum) are considered to be in the same direction. Assuming they are all vertical, do a balance of Vertical Forces.
     
  15. Sep 5, 2005 #14
    ------------------------------------------------------------
    Ok then,
    the load = 1000 Newtons + force of 250 Newtons = 1250 Newtons, reaction force at the fulcrum.
    the effort = load
    ------
    MA

    = 1000N
    -----
    4

    = 250 Newtons

    ---------------------------------------------

    Taking moments about the fulcrum:
    Would clockwise moments = anticlockwise moments apply in this question?
    This is what my book says.

    Please post your reply and correct my answer.
    scientist
     
  16. Sep 5, 2005 #15

    Fermat

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    This is what you mean to post, yes?

    You haven't actually said that the force at the fulcrum = 1250N but, if so, then that is correct. Your downward forces (load plus effort) = 1250N, so your single upward force (fulcrum reaction) is also 1250N.
    Since the lever is assumed to be in static equilibrium, then
    Σ Fv = 0
    Σ Fh = 0
    Σ T = 0
    where Σ Fv is the sum of Vertical forces, Σ Fh is the sum of Horizontal forces, and Σ T is the sum of the Torques.
    There are no horizontal forces and you have dealt with the vertical forces when finding the reaction at the fulcrum.
    They could, yes.
    Now you can do Σ T = 0, as one way of evaluating the position of the fulcrum
    Take moments about the fulcrum. Call the distance from the fulcrum to the load as x. So what is the distance fornt the fulcrum to the effort?
    Then the moment of the Load about the fulcrum is equal to the moment of the Effort about the fulcrum. Solve for x.
     
  17. Sep 6, 2005 #16
    x solved

    ----------------------------------------------------------------
    Yes, I mentioned at the beginning of my post that the reaction force at the fulcrum is 1250 Newtons.
    The length between the application and the fulcrum is 4 meters. The length between the fulcrum and the load is 1 meter.
    This is taken from the MA =4:1.
    I added the 4 + 1 = 5 meters
    The total lever length is 2 meters as stated in the question.
    I divided the lever length of 2 meters by 5 meters.
    2/5=0.4 meters.

    X = 0.4 meters as the final answer.
    So the pivot point is 0.4 meters from the load arm of the lever.
    This is a first class lever.
    The lever is said to be in a static equilibrium because the lever has no movement and the lever is balanced. So in other words the lever is perfectly straight across and there are no outside forces acting upon this lever.

    Please post a reply.
     
  18. Sep 6, 2005 #17

    Doc Al

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    Careful. The ratio of the lengths is 4 to 1; this does not mean that the lengths are 4 meters and 1 meter! (Since the total length is only 2 meters, you know that can't make sense.)
    If you call the load-fulcrum distance X, then the two distances are X and 4X. Their sum must equal 2 meters, so X + 4X = 5X = 2 meters. Thus X = 2/5 meters.
     
  19. Sep 6, 2005 #18

    HallsofIvy

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    Yes, and 2/5= 0.4 meters as he said.
     
  20. Sep 6, 2005 #19
    Correct

    -------------------------------------------
    You are correct! Because the total length of 2 meters divide by
    the MA of 4:1 equals 0.4 meters.

    2 meters
    -------
    4+1=5

    =0.4 meters
     
  21. Sep 6, 2005 #20
    From scientist

    I would like to thank all my physics tutors for the outstanding information I received during this lengthly post! I have learned a lot! I will study and learn these principles.
     
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