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Block sliding down plane

  1. Dec 19, 2008 #1
    1. The problem statement, all variables and given/known data

    A block of mass 19.0 kg is sliding down an inclined plane. Due to friction between the block and the plane the block is moving at constant speed. If the coefficient of kinetic friction is 0.24), what is the angle of the plane with respect to the horizontal ?

    m = 19kg
    mu = .24
    theta = ?

    2. Relevant equations

    Fk = mu*Fn
    Fn = g sin theta

    3. The attempt at a solution

    I'm confused about the hint of the block moving with a constant speed due to friction.. Because that means that the net force is 0, so how can I calculate this if one of the forces is equal to 0?? I can't get the normal force without the angle and I can't get the friction force without the normal force.. Can anybody help me out with what that clue implies?

  2. jcsd
  3. Dec 19, 2008 #2


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    Hi reaperkid! :smile:

    (have a mu: µ and a theta: θ :wink:)

    There are three forces … normal, friction, and gravitational …

    so just balance them to add to zero

    (either by using components, or a vector triangle ) :smile:
  4. Dec 19, 2008 #3
    Soo does that mean the normal force is equal to the gravitational force?
  5. Dec 19, 2008 #4
    no, draw a FBD
  6. Dec 19, 2008 #5
    Ughhh, I don't know how to get the normal force or friction force. If I could get either of those then it would be easy to balance them. And isn't friction the only negative force here?

    Nevermind I got it, thanks!
    Last edited: Dec 19, 2008
  7. Dec 21, 2008 #6

    I think you should sketched the vector forces. I will give you a hint. There will be two components that make up the forces: The x-component and the y-component. Identify the forces in the x-directions and identify the forces in the y direction. What do you think the value of the acceleration in the y-direction will be?
  8. Dec 22, 2008 #7
    Assume that angle with horizontal is theta.
    mu = 0.24
    You have written Fn = g sin theta
    I think you missed m. Moreover, it should be cos, not sin.
    Draw the diagram. You will find that correct is
    Fn = mg cos(theta)--------------------(1)

    You have written Fk = mu * Fn
    i.e. Fk = 0.24 Fn-----------------------(2)

    You got two equations (1) and (2). By resolving forces parallel to the incline and making net force in this direction as zero, you will get one more equation:-
    Fk = mg sin(theta)-----------------(3)

    You should be able to find theta from these three equations.
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