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I Simple Fulcrum Force Calculation Question

  1. Apr 5, 2016 #1
    Hello all,

    I have a (at what first seemed) fairly simple question about calculating the forces on this fulcrum design I have.

    Here is an image of the sketch and free body diagram that I drew:


    Basically, the pivot point in my real life application is actually a cam, but for the sake of this question, imagine it as a simple fulcrum.

    On the right hand side of the fulcrum bar, I have a 30lb force pushing down. On the left hand side, I have a the fulcrum bar mounted to a bearing that only allows rotation.

    For my FBD, I labeled there being a reaction force at the fulcrum pivot (obviously), which I labeled F_A. I also put a reaction force at the bearing, labeled F_N. This is the force I am slightly confused by. I know there has to be a reaction force here, but I am worried I am calculating it wrong, therefore calculating F_A wrong.

    I am trying to figure out the force on the fulcrum pivot point, which I termed F_A in my drawing. Please look through my calculations. According to this, F_A is equal to 102LB, more than three times the original, single force in my system! Does this really mean that I just translated 30LB into 102LB, simply by fixing the far left end of the fulcrum bar with that bearing? Did I do the calculations correct?

    This just seems a little goofy to me. I am having trouble grasping the fact that F_N, which is PURELY a reaction force, is contributing to the downward force on the fulcrum pivot, F_A. I was under the impression that since my singlular force of 30LB is the only force originally there, my F_A couldn't possibly be more than that. My calculations say otherwise.

    Let me know if you think I am correct in my calculations. If I am not, how can I rectify this? Where did I go wrong?

    Thank you for your time.
  2. jcsd
  3. Apr 5, 2016 #2

    Simon Bridge

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    You will expect force amplification because the setup is basically a lever. That is what levers do.
    The arrangement you have there is how nutcrackers work.
    I got ##F_A= (17/5)F_L## by equating moments about the bearing.
  4. Apr 6, 2016 #3


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    This is why I don't like physically meaningless qualifiers of forces like "reaction". They just confuse people. The forces you analyze all act simultaneously, and for the force balance at that time point it's completely irrelevant which force "was originally there".
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