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Homework Help: Block sliding Down; find kinetic coefficient

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 25.7°, the block starts to slide down the incline, traveling 3.50 m down the incline in 1.50 s. Calculate the coefficient of static friction between the block and the plank.
    Motion of a block:


    Part 2:
    Calculate the kinetic coefficent of friction between the block and the plank.

    2. Relevant equations

    3. The attempt at a solution
    not sure how but i keep getting: .21766
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 20, 2011 #2
    So, first, you have a FBD correct?

    I'll assume that you do.

    You should have a weight force going down.
    A normal force perpendicular to the plank's surface.
    And a friction force going the opposite direction of motion, correct?
  4. Sep 20, 2011 #3
    IDk what an FBD even is :(
  5. Sep 20, 2011 #4
    Google it real quick and you should get some good notes on it. :smile:
  6. Sep 20, 2011 #5
    Frozen Beverage Dispenser?!
    just kidding. bhaha
    Okay a Free Body Diagram. Well mmy professor told me to make one but I don't know how.
    But okay, what you said makes sense. A normal force, a weight force and a friction force and then the motion.
  7. Sep 20, 2011 #6
    Right. Now, when you draw your free body diagram, you'll notice that N is slanted in a particular direction.

    You can picture the x-y axis as being slightly rotated, that way the N is now directly on the y-axis, and the f force is now directly on the x-axis.

    W is now a hypotenuse going down, with an angle between the W and y-axis of 25.7 degrees.

    Before I go on, does the way I (tried) to explain the FBD make sense?
  8. Sep 20, 2011 #7
    Everything makes sense excpet where the W goes.
  9. Sep 20, 2011 #8
    Alright, so let me try it a different way.

    Consider a normal x-y axis with N going upward directly on the y-axis and the f force going directly on the x-axis.

    Now, your W force will be a diagonal line in the 4th quadrant (that is, negative y, positive x).

    The angle between W and the y-axis will be 25.7 degrees. Does that make any more sense that way? :smile:
  10. Sep 20, 2011 #9
    That makes much more sense. :)
    You're my new best friend.
  11. Sep 20, 2011 #10
    But now what? :(
  12. Sep 20, 2011 #11
    Haha. Thanks, just here to try to help.

    Now, you need to find the sum of the forces in the x-direction.

    Well, let's start by figuring out what N will be.

    Since N is pointing straight upwards, and by Newton's 3rd law, N - Wcos(theta) = 0, right?

    Now, N = Wcos(theta) -> W = mg, so...

    N = mg*cos(theta)

    f=[itex]\mu[/itex]N -> [itex]\mu[/itex]mg*cos(theta)....

    [itex]\Sigma[/itex]F[itex]_{x}[/itex] = mg*sin(theta) - [itex]\mu[/itex]mg*cos(theta) = ma[itex]_{x}[/itex]


    If you see an error in my reasoning, correct me.
  13. Sep 20, 2011 #12
    Alright, most of that makes sense. i don't have the acceleration tho to find u?
  14. Sep 20, 2011 #13
    And all three masses cancel out right?
  15. Sep 20, 2011 #14
    Alright, well, that's what we need to do next. :smile:

    And yes, all three masses cancel out.
  16. Sep 20, 2011 #15
    Okay. So now we have:
    gSin(Theta) - u*g*cos(theta) = a
    4.25 - 8.83u = a
  17. Sep 20, 2011 #16
    Consider that the final velocity will be the Delta x / Delta t.

    So, Vf = 3.50 m / 1.50 s

    Vf = 2.333 m/s

    Now, use your kinematic equation...


    V[itex]_{i}[/itex]=0 m/s

    Now, solve for a.....

    a = [itex]\frac{V_{f}}{t}[/itex]

    a = [itex]\frac{2.333}{1.50}[/itex]

    a = ?

    After that, you can solve for uk.
  18. Sep 20, 2011 #17
    That looks to be correct.
  19. Sep 20, 2011 #18
    it says i'm wrong....
    i got a=1.553
    so, u should equal .305
  20. Sep 20, 2011 #19
    That should be correct. (Unless we did something wrong in the procedure).
  21. Sep 20, 2011 #20
    we must have done something wrong...cuz it says i'm incorrect.
  22. Sep 20, 2011 #21
    g is always 9.8 ?
    and theta is always 25.7 ?
  23. Sep 20, 2011 #22
    g is approximately 9.8 m/s^2

    Theta will be 25.7 degrees in this problem.

    uk may be .322.... or .323. I'm not quite sure right now.
  24. Sep 20, 2011 #23
    Then i have no idea what to do. i relooked my math, and it makes sense to me.
  25. Sep 20, 2011 #24
    Maybe you could multiply the 1.553 m/s^2 by cos(25.7). Since a would be a vector, and that would, technically, give you ax.

    Try that and see if that works for you. I'm honestly confused.
  26. Sep 20, 2011 #25
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