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Funloving sliding pig

  1. Oct 18, 2007 #1
    A slide loving pig slides down a certain 19° slide in twice the time it would take to slide down a frictionless 19° slide. What is the coefficient of kinetic friction between the pig and the slide?

    here is how i approached the problem:

    since on a slide i changed the x-axis going with the slide with positive heading down and the y-axis facing in the n((Force normal) direction of the pig).

    i split up the weight force into:
    Wy = -(W)cos(19)
    Wx = (W)sin(19)

    and I also know that Fk = n Mk (Mk = coefficient of kinetic friction (mew k))

    since there is no acceleration in the y-direction it is safe to say Wy = n

    so. . . in the x-direction
    Fnetx = ma (acceleration in x-direction)
    break that down to. . .
    Wx - Fk = m a (acceleration in x direction)

    The part I am confused about is what does the time have to do with this problem and since i am not given the mass of the pig I can not go any further. . . any tips to keep me going?
     
  2. jcsd
  3. Oct 18, 2007 #2
    I did this problem a couple of weeks ago, and I got an actual numerical answer for the coefficient of friction in the end, but I'll warn you, it's alot of work, and you have to be careful as you go.

    Here's a clue. To start, you need to figure out the acceleration in the 'x' direction by setting up free body diagrams and getting the forces all down. And instead of using W, you should use mg. Start with the non-friction slide, that'll give you a leg up on the friction slide.

    If you post stuff, I'll help you as you go. It does seem like there's not alot of information, but there really is, so good luck.
     
  4. Oct 18, 2007 #3

    TMM

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    I think you can say that the final velocity will be twice as much for the frictionless ramp because acceleration is constant. From there it is an energy problem that is independant of mass or height of ramp.
     
  5. Oct 18, 2007 #4
    for the frictionless slide there is no friction. . .so the only force acting in the x direction would be Wx which is Wsin(19) . . .and break that down to

    m*g*sin(19) = m*a

    could you then divide both sides by m (which would cancel each other out leaving you with . . .

    g*sin(19) = a . . .so then a = 9.8 * sin(19) . . .a = 3.19 m/s^2 . . .

    since he went down the ramp at twice the time on the frictionless ramp does that mean divide a by 2?
     
  6. Oct 18, 2007 #5
    When I did this, it was a force problem (we had not done energy stuff yet). So if you continue this way, I don't know that you can use the double the time yet. I agree with your acceleration of 3.19.

    Next I did the friction slide. It starts out basically the same as the non-friction slide, but you have to add the fact that there is friction, so now there are two forces on the friction ramp, and you can solve for acceleration again. It'll have the [tex]\mu_k[/tex] in that equation.

    For the last step, since the ramp lengths are the same, I used the equation:
    [tex]x_2 = x_1 v_1t + \frac{1}{2}at^2[/tex]

    Since [tex]v_1 = 0 [/tex] and [tex]x_1 = 0[/tex] those pieces go out, and for the second equation make sure you use [tex]2t^2[/tex], then set the two equations equal to eachother and you can solve for [tex]\mu_k[/tex]
     
  7. Oct 18, 2007 #6
    you forgot a '+' sign inbetween the x1 and v1t but regardless. . .ok so ya i agree with you the final velocity is 0 and the starting postion is 0. but what i do not understand is what are you trying to find? x2 or t? because lets say you set them equal to each other using the 3.19 as acceleration . . .so it looks something like so

    Frictionless: x = 1.595t^2

    Friction: x = 3.19t^2

    and how is that going to help you solve for [tex]\mu_k[/tex] if the formula to find [tex]\mu_k[/tex] is Fk = n [tex]\mu_k[/tex] and i understand that there is no acceleration in the y direction . . .so n = Wy and Wy = -mgcos(19) so. . .

    Fk / (-mgcos19) = [tex]\mu_k[/tex]
     
  8. Oct 18, 2007 #7
    Did you do the forces for the friction slide, then solve for the 'x' acceleration on that slide. Instead of it just being a = 3.19, it'll be a = 3.19 + (or minus) the friction force, [tex]\mu[/tex] or [tex]\mu[/tex]mg(cos19). You can do the calculations.

    After that, you need to solve for [tex]\mu[/tex]. I used the equation below. as you start plugging stuff in, more erroneous stuff (like 't') will cancel, and you can solve for [tex]\mu[/tex].
     
  9. Oct 18, 2007 #8
    alright i will try to comprend everything and see if i can come up with an answer and I will let you know if I got an answer chocokat by tomorrow (20 hours by now) thanks for all the help . . again ill let you know if i came up with an answer 20 hours from now.
     
  10. Oct 18, 2007 #9
    O.k., goodluck kiddo.

    As I mentioned earlier, just be careful with your calculations and + or - signs, because you don't want to get messed up by a little mistake, they're so annoying.
     
  11. Oct 19, 2007 #10
    ok sorry chocokat but im still lost. . .here is what i have so far (this is only acceleration for x because in the y direction is 0) . . .

    acceleration for the frictionless slide is a = 3.19 so we plug that into the equation
    [tex]x_2 = x_1 + v_1t + \frac{1}{2}at^2[/tex]
    and i came up with x = 1/2(3.19)t^2 and i solved that for t so

    t = square root of (x/1.595)

    then for the friction slide i got a = 3.19 - (mu k)mgcos19
    it is minus because friction is acting the opposite way of motion so if i plug it into the equation of
    [tex]x_2 = x_1 + v_1t + \frac{1}{2}at^2[/tex]

    i get x = 3.19- (mu k) mgcos19(t^2)
    and i solved that for t to get

    t = square root of (x/3.19- (mu k) mgcos19)
    now i set those two equations equal to each other. . .but i do not know where to go from there. . .
     
    Last edited: Oct 19, 2007
  12. Oct 19, 2007 #11
    O.k., this is what you do. For the frictionless slide you have

    [tex]a = 3.19 m \backslash s^2 [/tex]

    For the slide with friction you have:

    [tex]a = 3.19 m \backslash s^2 - \mu_k 9.267 m \backslash s^2[/tex]

    1. Now, put those into two equations. On the equation with the slide with friction, remember to put [tex]2t^2[/tex]
    Don't do any further calculations at this point, just leave stuff as is, it will make it easier in a couple of steps. For example

    [tex]\frac{1}{2} 3.19t^2[/tex]

    leave the [tex]\frac{1}{2}[/tex] alone, don't multiply it by the 3.19.

    2. Set the equations equal to eachother (since the distance 'x' is the same for both slides, the first equation equals the second equation) If you don't understand this, please let me know and I'll try to explain it further. So now, there are no x's.

    3. If you haven't multiplied anything through, such as the [tex]\frac{1}{2}[/tex], you should notice that you can easily multiply stuff out. If you get stuck here, let me know. After this, you should be able to solve for [tex]\mu_k[/tex]
     
  13. Oct 19, 2007 #12
    ok so now i take

    (1/2)3.19t^2 = (1/2)(3.19 - [tex]\mu_k[/tex](9.269))2t^2 . . .

    you can cancel out the 1/2 andt^2 and that leaves you alone with

    3.19 = 2(3.19 - [tex]\mu_k[/tex](9.267)

    so shouldnt [tex]\mu_k[/tex] = 0.17212
     
  14. Oct 19, 2007 #13
    and i understand about the x's
     
  15. Oct 19, 2007 #14
    You're so close. For the right side, the 't' is [tex](2t)^2[/tex] so the '2' becomes '4' [tex](2t)^2 = 4t^2[/tex]. Then when you finish the calculations:

    3.19 = 4(3.19 - [tex]\mu_k[/tex]9.267)

    so the final answer is [tex]\mu_k = .258[/tex]

    This problem was alot of work, but one of the key things to remember is that even though you weren't given alot of information, you were able to calculate alot, and if you run into similar problems, sometimes it's easier not to 'solve' for variables for awhile, as they end up cancelling out making the solution easier to calculate.

    And be careful as you go. I make so many dumb mistakes missing - signs, or with simple math errors, for a big problem just do it step by step to avoid those types of problems, you'll end up saving time and frustration in the long run.

    Good work!
     
  16. Oct 19, 2007 #15
    man choco. . .forgot about that squared part. . .amazing work thanks for helping me out. . .God bless
     
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