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Frictional Force and µk problem

  1. Nov 6, 2007 #1
    1. The problem statement, all variables and given/known data
    A brick is on a 121 cm plank, and the plank is lifted gradually until the brick begins to slide, when the plank is lifted 48cm in the air at an angle of 30º. I need to find the Frictional Force, and µ kinetic.
    How do I find these things knowing only these three things?

    2. Relevant equations
    Ff=(µk)(Fn) let Fn= Normal Force
    g=9.81 m/s/s

    3. The attempt at a solution
    I tried using the first equation I mentioned, but was unable to figure how to get both values.
    Help is much appreciated.
  2. jcsd
  3. Nov 6, 2007 #2

    Doc Al

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    Staff: Mentor

    Draw a diagram showing all the forces acting on the brick at the moment just before it begins to slide. What must the net force be in any direction?
  4. Nov 6, 2007 #3
    You already have the force normal to the surface. You need the other component as well.
  5. Nov 6, 2007 #4
    I think I should do this working purely without numbers, so I can just plug my figures in, I suppose.

    I have the height of the end of the plank when the brick started sliding, the length of the plank, Speed of the brick, Mass, Weight, Fn, and the Fp [the force in the plane], I also have µ static. How do I find Ff and µk?
  6. Nov 6, 2007 #5

    Doc Al

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    Staff: Mentor

    Identify the forces acting on the brick. Apply Newton's 2nd law to the forces parallel to the incline.
  7. Nov 6, 2007 #6
    Er - speed of brick? no. The question is "What is the frictional force?"

    As the plane is tipped steeper, there is a force component in the direction down the plane. There is another force component in the direction normal to the plane. They both derive from the force (from gravity) acting vertically downwards.

    The solution is very easy once you take the major step of understanding the forces. You already have the component normal to the slope. This component, multiplied by the coefficient of friction usually called [tex] \mu[/tex] will be the force of friction resisting movement down that slope. It is the bit you do not have!

    You do have the way to get at the frictional force. It has to equal the force down that slope, (because the brick stays put!!) Eventually, you discover its maximum at the un-stick angle. Get that, and you have one of the required answers. Getting at the (unknown) coefficient of friction then follows.

    The static coefficient is the one that decides where the brick un-sticks. The dynamic coefficient a lesser value, (not so steep slope required). This is when the brick is moving. A special case is when the speed is steady. If accelerating, the available force to do that accelerating is what is left over after subtracting the friction.

    Now we ask that you actually get to grips with the forces diagram, and demonstrate that you can take that step. It really is the one necessary and key bit of knowledge this question requires.
    Last edited: Nov 6, 2007
  8. Nov 6, 2007 #7
    Thanks a lot, man. I get it now.
  9. Nov 7, 2007 #8
    OK then - so post the answer of what you think the force is down the slope, or better, the formula expression that you used to figure it.
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