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Static Friction, stuck on a problem

  1. Dec 10, 2007 #1
    Sorry to bother with yet another question today, but I'm a little lost with this one.

    QUESTION
    A box with a mass of 25 kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficients of friction between the box and the ramp are us=0.78 and uk=0.65.

    a) Determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest.

    b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.

    MY ANSWER
    a) Fa = Fgh + Ff
    = Fg sin 45 degrees + usFn
    = (22kg)(9.8 m/s^2)(sin 45 degrees) + (0.78)(22kg)(9.8 m/s^2)(cos 45 degrees)
    = 271 N

    Therefore the largest force that can be applied upward if the box is to remain at rest is 271N.

    (I believe this may be incorrect as gravity/normal are balanced and only the force of friction is impeding which would be 118.9N, but I wanted to double check)

    b) I'm really not sure where to start with this as there has not been an example like with with my text :( Could someone give me a hint?
     
    Last edited: Dec 10, 2007
  2. jcsd
  3. Dec 10, 2007 #2

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    a) Correct concept. I've not checked the value.

    b) The friction upward must balance component of weight downward.
    So, mgsin 45 = Fs = (us)*total normal reaction = (us)(mgcos 45 + F). Now you can find F.
     
  4. Dec 10, 2007 #3
    So for part (a) the concept of Fa = Fgh + Ff was correct as to the amount of force I would need to exert on the box?

    Also regarding (b) I believe I am now thinking about the problem correctly, here is my answer for it:

    Fg sin 45 degrees = us(Fg cos 45 degrees + F)
    (25kg)(9.8 m/s^2)(sin 45 degrees) = (0.78)((25kg)(9.8 m/s^2)(cos 45 degrees) + F)
    173.24N / 0.78 = (25kg)(9.8 m/s^2)(cos 45 degrees) + F
    222.10N = 173.24N + F
    F = 48.86N
    F = 48.9N

    Therefore the magnitude of the smallest amount of force that can be applied on top of the box is 48.9N.

    Am I correct with this?

    EDIT: Actually I am quite confused now... as I thought I could also think about it like this:

    Fgh = mg sin 45 degrees
    Ff = us mg cos 45 degrees

    Therefore Fgh = Fa + Ff
    Therefore Fgh - Ff = Fa

    Fa = Fgh - Ff
    = (mg sin 45 degrees) - (us mg cos 45 degrees)
    = ((25kg)(9.8 m/s^2)(sin 45 degrees)) - ((0.78)(25kg)(9.8 m/s^2)(cos 45 degrees))
    = 173.24N - 135.12N
    = 38.11N

    Which of course is different from my original answer, I'm really not sure how to properly think about this one.

    EDIT #2: Ok I think I understand, in my first try I did not apply the friction coefficient to the F as if I did it would yield 38.11N. So am I right in thinking that my second answer is correct?

    So in the end I end up with the following answers:

    a) 271 N (should this be 2.71 x 10^2?)
    b) 38 N

    Thanks for all the help :)
     
    Last edited: Dec 10, 2007
  5. Dec 10, 2007 #4

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    a) That's the correct concept.
    b) 48.8 N is correct.
     
  6. Dec 10, 2007 #5
    Thanks very much! It seems as I worked myself out of understanding when I changed it to 38N. I thought I was wrong the second time applying the friction coefficient to the applied force considering I had already taken friction into account. I just ordered the book you recommended to me on my other thread, so hopefully it comes quick :) For some reason I don't understand why the first way works but why Fa = Fgh - Ff doesn't work.
     
    Last edited: Dec 10, 2007
  7. Dec 10, 2007 #6
    Ah I believe I now understand why Fa = Fgh - Ff does not work, and that is because I messed up the basic mathematics of it. If I would have multiplied out and then moved over I would have had Fa/us = Fgh - Ff which in turn gives me the 48.8N, in error I dropped the us when isolating Fa and therefore had an incorrect force.

    Thanks again for all the help today!
     
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