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Statics: Friction Problem

  1. Nov 14, 2005 #1
    Ok this problem seems way too simple. I have worked it out without using any of the dimensions given.

    The problem is here

    My work is here

    I dont think the solution could be that easy. Anyone want to try and see if this is more complicated than I made it? Thanks
  2. jcsd
  3. Nov 14, 2005 #2


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    Staff: Mentor

    What if the friction coefficient at the end of the tongs goes to zero? Youre solution doesn't take that into account. As the mu at the tong tips gets lower, you will be limited in how much weight you can pull, or the tongs will slip off.
  4. Nov 14, 2005 #3
    Thanks for noticing Berkeman. I am going to attempt to sum the moments about C to solve for the normal force applied at the tips of the tongs. feel free to stop me if this is incorrect! thanks
  5. Nov 14, 2005 #4
    P.s. That is hibbeler and the problem in the book is wrong. I found that out the hard way myself. There is no solution.
    Last edited: Nov 14, 2005
  6. Nov 14, 2005 #5

    It is Hibbeler, and the answer in the back of the book is incorrect. I am very confident there is a correct solution however.
  7. Nov 14, 2005 #6


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    Staff: Mentor

    Really? It sure seems like there should be a solution -- the situation seems pretty physical. There probably is no solution if you have to build up the pulling force from zero, because zero initial normal force at the squeeze point would mean that the jaws would always slip off. But if the jaws could be held on the wood as the pulling force was built up to some number, and then they are released, it seems like you could solve for what force you could apply before they finally slipped off. Definitely let us know what you find. What's the wrong answer in the book, BTW?

    Edit -- Hmm, maybe with the angles given and the mu, the jaws would always pull off. Maybe that's what's going on. Maybe it takes at least some minimum mechanical advantage on the squeeze point and some minimum mu for the jaws not to slip off?
  8. Nov 14, 2005 #7
    you got it. Solve for the friction force given the value of mu. It will be
    2F> P, sorry that wont work, as it is not in equilibrium anymore. The values are bogus. ;-)
  9. Nov 15, 2005 #8
    the answer in the back is m = 54.9 kg

    i will back back a little later to read your response a little more thoroughly. Thanks
  10. Nov 15, 2005 #9
    Dont pay attention to what the book has, its wrong. To get a solution you will have to solve to find what the value of mu can be. Its really quite trivial, if you have a force of P to the left, then at each tong you must have a force of 1/2 P in the other direction, which means you have a force of P acting on the box. So you need to solve for what minimum value of mu will allow you to have a value of 1/2 P at each tong, or change the angle of the chains.
    Last edited: Nov 15, 2005
  11. Nov 15, 2005 #10

    So you are saying that I should disregard the given value of mu between the tongs and the pallet, and solve for the normal force, thus giving me the correct value of mu? So was my work correct for given value of mu?

    I am confused because this problem was assigned by my professor. I don't understand why he would assign it if he knew it was a trick question. I am positive that he is aware of the problems inaccuracies.
    Last edited: Nov 15, 2005
  12. Nov 15, 2005 #11
    Yep, and your work seems good too.

    Nope, the book is wrong. Its not a trick question, its an improperly written problem in the book. Just bring it to his attention and ask for extra credit because you spotted a mistake, and then thank him that you do not use the 9th edition that had TONS of mistakes in it.
  13. Nov 16, 2005 #12
    lol thanks a lot for helping me out!
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