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Homework Help: Friction help, please

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the magnitude of the horizontal force Fp required to push a block of a mass 30 kg, up along an inclined plane, with a constant acceleration of 0.8m/s2. The plane makes an angle of 25° to the horizontal and the coefficient of kinetic friction between the plane and the block is 0.21. (Hint: The solution involves setting up simultaneous equations in terms of x and y components of the unknown force, i.e. Fpcos25° and Fpsin25°, and then solving them for Fp.)

    2. Relevant equations

    I asked my teacher how to do it but i didn't really understand him. He gave me this formula to find Fp. The answer given by the book is 250 N.

    Fpcosθ = Fg +μ(mgcosθ + Fpsinθ) + Fa

    where Fp= Force applied/of pull
    Fa= Acceleration Force
    Fg= Force due to gravity/weight

    3. The attempt at a solution
    I have plugged in the given but i still get the wrong answer (400+ N).

    Can someone please explain this to me? Help me please. Please.
    Last edited: Mar 4, 2012
  2. jcsd
  3. Mar 5, 2012 #2
    First of all, you need to learn trigonometry to have these types of problems easier. I have attached an image where I have pointed out the most important parts of the problem.
    At the top left corner you have the chosen directions for x and for y.

    [itex]f[/itex] => force of friction (the red vector). The force of friction always is on the opposite side of the force being applied to the object.

    [itex]G[/itex] => force of gravity (the orange vector). It is then divided into two components. The [itex]G[/itex][itex]x[/itex] component (the yellow vector) and [itex]G[/itex][itex]y[/itex] (the black vector). If you add those vectors you would get [itex]G[/itex] so it is fair to break it up in two components as it makes it easier to solve the problem. By trigonometry, I have written [itex]G[/itex][itex]x[/itex] = [itex]mgsin25[/itex] and [itex]G[/itex][itex]y[/itex] = [itex]mgcos25[/itex].

    [itex]F[/itex]P => the force you apply (purple vector)

    [itex]F[/itex]N => the normal force. (light green vector)

    Now go on and try to solve it. Even though I got a result of [itex]200N[/itex], I believe it has been mistakenly put on the book [itex]250N[/itex].
    Last edited: Jun 16, 2012
  4. Mar 5, 2012 #3


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    welcome to pf!

    hi leianne! welcome to pf! :smile:
    that Fg (which = mg) shouldn't be there

    try again without it :smile:

    (if you still don't understand why that is the formula, let us know)
  5. Mar 5, 2012 #4

    Thank you so much! If my memory doesn't fail me, I've actually gotten 200N and 250 N at one time. It's just that i didn't understand how i got it. :rofl:

    Anyway i've plugged it in in the formula (without Fg) but my answer still comes up to 356.25 N :confused: Why is that? I don't get it im sorry. The formula i used is:

    Fpcosθ = μ(mgcosθ + Fpsinθ) + Fa

    Another question is that isn't friction acting in the opposite direction of the pull? Shouldn't that be negative then? Please help
  6. Mar 5, 2012 #5
    And if possible can you/anyone also explain to me how my teacher got that formula? Please bear with my little understanding. :( Thank you very much.
  7. Mar 5, 2012 #6
    did you see the image I've attached?
  8. Mar 5, 2012 #7
    Yes, I did. But isn't the question asking for the magnitude of the horizontal force? So shouldn't Fp also be broken down into components?

    Anyway is your diagram showing that F = Fp - Gx - f? I've tried it and it gave me 204 N. yay am i annoying you? im really sorry.
  9. Mar 5, 2012 #8


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    (tonit, your diagram is clearly wrong, that is not being helpful :redface:)

    hi leianne! :smile:
    can you show us exactly how you got from that formula to your answer? :smile:
    your teacher did F = ma along the slope

    really, all the forces should be on the left, and only ma on the right

    (in my opinion, it is really confusing to refer to "acceleration force" :frown:)

    so, on the left, it is obvious that Fp is positive, but the other two are negative :smile:
  10. Mar 5, 2012 #9

    Fpcosθ = μ(mgcosθ + Fpsinθ) + Fa
    Fpcos(25) = 0.21[(30x9.81xcos25) + Fpsin(25)] + 30x0.8
    0.9Fp = 0.21(266.7 + 0.4Fp) + 24
    0.9Fp = 56 + 0.084Fp + 24
    0.9Fp - 0.084Fp = 56 + 24
    0.816Fp/0.816 = 80/0.816
    Fp = 98 N (what?!:eek:)

    yay i think i got 356 N from inventing/mixing up formulas. wah what should i do?

    anyway with what you last said, will the formula then be :

    Fp - Fgx - f = ma?
  11. Mar 5, 2012 #10


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    oops! :redface:

    i didn't draw a diagram, and i left out an mgsinθ (the x component of the weight)

    try that :smile:
  12. Mar 5, 2012 #11

    Well actually the Fgx in my Fp - Fgx - f = ma is equal to mgsinθ. I just called it Fgx since it's the horizontal component of the weight. And the answer i kept getting using that formula is 204 N which means im still missing 46 N (according from the answer at the back of my book. oh yes, my book is Engineering Mechanics by val ivanoff)
  13. Mar 5, 2012 #12


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    no, that should be 204/.816, shouldn't it? :wink:
  14. Mar 5, 2012 #13
    WOAH it's 250 N!!!!!! finally!!!! that is awesome yay thank you so much!!! but what is the right formula then? how should i show that? again thank you very very much yay!!!

    is it simply :
    ma = Fp - mgsin(theta) - f
    ma = (Fpcostheta - Fpsintheta) - mgsintheta - f ?
  15. Mar 5, 2012 #14
    yay i get it now!!! thank you very very very much!!! yay im just genuinely ecstatic right now and really really grateful!!! thank you thank you thank you!!!
  16. Mar 5, 2012 #15


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    Pcosθ - mgsinθ - μ(mgcosθ + Psinθ) = ma :smile:

    (ie, all the x-components on the LHS, and ma on the RHS)
  17. Mar 5, 2012 #16
    yes i get it now yay and it's all thanks to you!!!! thank you very very very much!!!! i feel kind of embarrassed now that i've figured out what my silly mistake was :redface: i understand everything now and i better burn it to my memory yay!!! really thank you very very very much!!!
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