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Force required to move a crate

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data
    You drag a trunk of mass m across a level floor using a massless rope that makes an angle with the horizontal (figure below). Given a kinetic-friction coefficient (mu).
    RW-05-64.jpg
    Find the minimum force needed to move the trunk with constant speed.


    2. Relevant equations
    F=ma
    Fk=µk(Fn)


    3. The attempt at a solution
    What's really getting me is that the site says the answer doesn't depend on theta.

    I found the normal force being Fn=mg-FsinΘ

    the total friction force being µk(mg-FsinΘ)

    For the force equation,
    F=ma
    F=0
    FcosΘ=µk(mg-FsinΘ)

    Solving for F, I came up with:

    F= (-µkmg)/(µksinΘ-cosΘ)

    Again, says no theta in the answer though. Any pointers?
     
  2. jcsd
  3. Feb 16, 2009 #2

    tiny-tim

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    Hi bakin! :smile:
    Yes that's the correct formula …

    now find the value of Θ for which it's a minimum. :wink:
     
  4. Feb 16, 2009 #3
    Well, to get the smallest force value, you want the largest denominator, correct? That would be when cosΘ=0, and sinΘ=1, being pi/2. That would make the answer just -mg. Doesn't make sense to me :(

    edit: actually, no. Because if sinΘ=1, and cosΘ=0, the answer would be less than one on the denominator, because it would be µk times sinΘ, giving an answer of 0.XX . I guess then it would be when Θ=0. then, the denominator would be equal to -1, and the answer would be just µkmg?

    edit2: I tried putting in µkmg in as an answer before, though, and it wasn't correct. Could I have entered it in wrong or something?

    edit3omg: or maybe the angle is pi/4, giving you less pull, but also making the normal force less. not 100% sure.
     
    Last edited: Feb 16, 2009
  5. Feb 16, 2009 #4

    tiny-tim

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    oops!

    Sorry, I misread it :redface:

    should be a plus on the bottom …

    can you see why?
     
  6. Feb 16, 2009 #5
    Aw boo, looks like I need to review the distributive property :blushing:

    Ok, so redoing my formula I have the same thing, except all the negatives are now positive.

    But, making the denominator equal to one, I'm still left with µkmg, which it said was wrong..
     
  7. Feb 16, 2009 #6

    tiny-tim

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    uhhh? why? :confused:

    Differentiate the bottom, and put it equal to zero. :smile:
     
  8. Feb 16, 2009 #7
    Doing that gives me tan-1(µk)=Θ ????

    But, how would you find this out without differentiating? I'm working on this with a friend of mine, and I don't think her class uses any calculus.
     
  9. Feb 16, 2009 #8

    tiny-tim

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    without differentiating …

    you can always rewrite asinΘ + cosΘ in the form b(sinφsinΘ + cosφcosΘ), = b cos(Θ-φ), which is easy to minimise. :wink:
     
  10. Feb 16, 2009 #9
    Would plugging in tan-1(µk) in for theta be it, or is there another step to that method?

    Where did you get φ? Not exactly sure how you went from asinΘ+cosΘ to the other form :uhh:
     
    Last edited: Feb 16, 2009
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