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Friction: Block on Ramp

  1. Jan 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A block slides down a slope from point O with initial speed V. THe sliding coefficient, mu, brings the block to rest at time T. Using the ramp as the x-axis and the perpendicular as the y-axis, find T.

    2. Relevant equations

    F = ma
    mu = mg sin(theta)
    a=dv/dt

    3. The attempt at a solution

    I'm taking modern physics, and this is a problem to demonstrate how picking different frames of reference can change a problem. I have to take the axes as the vertical and horizontal after solving this part, butI haven't had general physics in about 3 years, so I'm just not sure how to do the easy part. I think if I get this part I can get the second, I just need a small refresher.

    I was thinking somewhere along the lines of this:

    ma = mg sin(theta) - mu
    dv/dt = g sin(theta) - mu/m
    dt = dv/(g sin(theta) - mu/m)

    Maybe? But, I'm not sure how I should do that derivative.
    Thanks!
     
  2. jcsd
  3. Jan 21, 2008 #2

    Doc Al

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    That's the one.
    mu is the coefficient of friction, so this makes no sense. How does friction relate to normal force?

    What you need to do is identify all the forces acting on the block (draw a free body diagram!) and then apply Newton's 2nd law.
     
  4. Jan 21, 2008 #3
    You're making it too complicated, just find the acceleration and use those old constant acceleration motion equations.
     
  5. Jan 21, 2008 #4
    I did draw a diagram. F = mu N yeah?
     
  6. Jan 21, 2008 #5

    Doc Al

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    Yep. What's the normal force? Now find the net force parallel to the ramp.

    As Feldoh says, keep it simple. (No need for calculus.)
     
  7. Jan 21, 2008 #6
    N = mg cos(theta)

    So,
    a = g sin(theta) - mu g cos(theta)?

    Then,
    v = v_0 +at

    Since we're looking at when it comes to rest, v=0

    T = -v_0/a

    Look right?
     
  8. Jan 21, 2008 #7

    Doc Al

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    Looks good.
     
  9. Jan 21, 2008 #8
    Excellent. Thank you.

    For the second part, with the axes being the vertical and horizontal, I'm going to have a multi-component acceleration, yeah?

    In the y-direction, the forces working is the weight/gravity which is pointing in the -y-direction, and components of the Normal and Friction pointing in the +y-direction.
    For the x-direction, I'm going to have the normal in the positive x-direction, and the friction in the -x-direction.
    Does that sound right?
     
  10. Jan 21, 2008 #9

    Doc Al

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    Sounds good.
     
  11. Jan 21, 2008 #10
    For part a, I got
    T = -v0/(g sin(theta) - mu g cos(theta)]

    Obviously since it's the same situation, I should be able to get the same answer for part b, But I'm having difficulty.

    x-direction
    Fx = N cos() - mu N cos()
    ax = [N cos() - mu N cos()]/m
    T = -[v0x cos() m]/[N cos() - mu N cos()]

    y-direction
    Fy = mu N sin() + N sin() - mg sin()
    ay = [mu N sin() + N sin() - mg sin()]\m
    T = -[v0y sin() m]/[mu N sin() + N sin() - mg sin()]

    theta is in all the empty (), of course.

    And then, I'm stuck. I don't see how I could get the original answer from that, so either it's something I just haven't seen or I did something wrong.
     
  12. Jan 21, 2008 #11

    Doc Al

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    Check those components. They can't both be cosine. :wink:
     
  13. Jan 21, 2008 #12
    No? I always thought that if it was an x component it was cosine, and y component was sine. Those are both the x components, so...
     
  14. Jan 21, 2008 #13

    Doc Al

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    Sure, if the angle is with respect to the x-axis. But theta is the angle of the ramp. What's the angle of the normal force?
     
  15. Jan 21, 2008 #14
    wrt the ramp 90º or wrt x-axis 90º+theta

    How is the angle not wrt the x-axis? It's the angle made by the ramp with the x-axis
     
  16. Jan 21, 2008 #15

    Doc Al

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    That's with respect with the -x-axis, but OK. (What's the angle with the +x-axis?) In any case, you wrote cos(theta).
     
  17. Jan 21, 2008 #16
    How is it the -x-axis?

    Although, I now see that you're saying that the normal is theta from the positive y-axis, and needs to be sin(theta), right?
     
  18. Jan 21, 2008 #17

    Doc Al

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    Right.
     
  19. Jan 21, 2008 #18
    So now I get

    x-direction
    Fx = N sin() - mu N cos()
    ax = [N sin() - mu N cos()]/m
    T = -[v0x cos() m]/[N sin() - mu N cos()]

    Substitute N = mg cos()

    T = -[v0x cos() m]/[mg cos() sin() - mu mg cos() cos()]

    cancel m cos() to arrive at original answer, right?

    Thanks for all of your help.
     
  20. Jan 21, 2008 #19

    Doc Al

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    Looks good to me. (And you're welcome.)
     
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