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Derivation of Applied Force in a Grade 12 Force Question

  1. Oct 14, 2016 #1
    1. The problem statement, all variables and given/known data
    When an applied force elevated at angle θ is used to drag mass m along a surface with μ as the coefficient of friction, the mass accelerates horizontally with some value, a. Derive the cleanest and most simplified formula for the applied force, Fapplied, that will give the required acceleration.

    Since the Fapplied is elevated at angle θ one could break it down into components FaCosθ
    [Horizontal]and FaSinθ[Vertical]. Additionally, FN will be mg-FaSinθ because FaSinθ is alleviating some of the force that FN normally has to counteract Fg, meaning that Ff will be μ(mg-FaSinθ).
    I'm just so sure my derivation is correct but when I used sample numbers it didn't work at all, I'd really appreciate any advice or help, thanks!


    2. Relevant equations
    a=Fnet/m
    Fg=mg
    Ff=μFn



    3. The attempt at a solution
    a=Fnet/m
    a=FaCosθ- μ(mg-FaSinθ/m
    a=FaCosθ-μmg-μFaSinθ/m
    am+μmg=Fa(cosθ-μsinθ)
    am+μmg/cosθ-μsinθ=Fa
     
    Last edited by a moderator: Oct 14, 2016
  2. jcsd
  3. Oct 14, 2016 #2
    Hi troutkibe, you're reasoning seems sound. Your equations seem sound too (although the bracketing is a little funny in part "3. The attempt at a solution", but I'm guessing that's a typo?). Maybe give us the numbers you need to plug in and we can try plugging them in too (might just be the way you've inputted those values into the calculator you're using.)
     
  4. Oct 14, 2016 #3
    I withdraw my comment. See SammyS' post below.
     
  5. Oct 14, 2016 #4

    SammyS

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    Hello Troutkibe. Welcome to PF !

    You are somewhat cavalier regarding parentheses. They're important.

    I've corrected a few in the above quote of you OP.

    I'll leave fixing parentheses up to you in that last line. If that is what you are literally using for plugging in your numbers, then you will definitely get erroneous results.
     
  6. Oct 14, 2016 #5
    Yeah, that's my bad on not using brackets here, but I was using them in the actual calculation. I think I figured it out though, when I multiplied everything by -μ I neglected its negative value and that really threw everything off.
     
  7. Oct 14, 2016 #6

    SammyS

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    (am+μmg)/(cosθ-μsinθ) = Fa
     
  8. Oct 14, 2016 #7
    Shouldn't that be a + sign in the denominator?
     
  9. Oct 14, 2016 #8
    Yeah thats what I realize I messed up on. Dumb mistake too :/
     
  10. Oct 14, 2016 #9
    The hard part was expressing the physics of the problem in the form of equations. You did masterfully at that. Then you messed up on the simple part: solving the equations algebraically.
     
  11. Oct 14, 2016 #10

    gneill

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    Bonus question: What would be the minimum force you can apply to achieve a given acceleration? In other words, if the force is applied at an optimum angle to minimize the force required, what is the magnitude of that force? :smile:
     
  12. Oct 16, 2016 #11
    Set dF/dθ=0
    get θ and substitute back to get F
     
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