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What is the direction and magnitude of the frictional force?

  1. Oct 10, 2013 #1
    1. The problem statement, all variables and given/known data
    A 2,200-kg car is moving down a road with a slope (grade) of 14% at a constant speed of 11 m/s.

    2. Relevant equations
    The equation that I'm using is ƩF=ma and friction=μN


    3. The attempt at a solution
    the mass is 2200kg
    the acceleration is zero because of the constant speed.
    I really cannot think of anyway to solve this solution.
     
  2. jcsd
  3. Oct 10, 2013 #2
    The frictional force always opposes the motion, it "eats up" the total energy in the system. To do any calculations with the frictional force we need at least the coefficient of friction μ.
     
  4. Oct 11, 2013 #3
    And that the problem, mu is not given. Do I have to find mu in order to move on?
     
  5. Oct 11, 2013 #4
    Well for this problem you do not need to find mu. Here are some steps to get you started:

    1. Draw a free body diagram on the car.
    2. Utilize Newtons Second Law in the x and y directions respectively
    3. Algebraic Wizardry, manipulate the equations until you can find the frictional force using values you already know.
     
  6. Oct 11, 2013 #5
    ok, but what about the 11m/s, does that velocity come into play?
     
  7. Oct 11, 2013 #6
    your question mentions that velocity is constant, and when velocity is constant the net force is always zero. so, here friction is equal to the component of gravitational force along the slope.

    P.S. - like Legaldose already mentioned drawing free body diagram always clears out doubt.
     
  8. Oct 11, 2013 #7
    ok so I solved the problem, and this is what I did: m*g*sin(7.96) the 7.96 is the reverse tangent.

    so 2,200kg*9.81m/s^2*sin(7.96)= 2,989N

    would that be considered my magnitude and direction?
     
  9. Oct 11, 2013 #8

    rcgldr

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    Not a big deal, but arctan(.14) ~= 7.97.
     
  10. Oct 11, 2013 #9
    thanks, but what about the solution that I solved 2,989N is that solution the magnitude and direction, or if there is there still more?
     
  11. Oct 12, 2013 #10

    haruspex

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    That formulation can confuse people when thinking about e.g. a car being driven up a slope. It leads them to think that the friction between road and tyre would act down the slope. Better always to write that friction opposes relative motion of the surfaces in contact.
     
  12. Oct 12, 2013 #11
    it's the magnitude you have calculated. For direction refer to haruspex's post above.
     
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