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Homework Help: Coefficient of Friction / Speed on incline.

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A waterslide ride consists of a straight guided ramp, at an angle theta (to the horizontal plane), with a transition to a short horizontal section at the bottom of the slide (1m long), just above the surface of a splace pool of length L. Patrons queue on a platform of height h (edge of platform is 18m from start of horizontal section of slide) before climbing onto a 10kg toboggan and sliding down slide under the force of gravity until they leave the slide and skip along surface of splash pool.

    Water slide is lubricated by 12L/s of water

    When theta = 53degrees
    h = 7.6m

    1. Determine Rider speed
    2. and stopping distance if;
    average weight of rider is 100kg and toboggan is 10kg
    coefficient of friction between toboggan and slide (us) is;
    0.2 (Q 20)/50 (Where Q is in L/s)
    coefficient of friction between toboggan and pool water uw is;
    0.6
    ****Taking into account dynamic friction only*****
    3. Determine length of splash pool needed L




    2. Relevant equations




    3. The attempt at a solution

    I have been trying to figure this out using various equations, none of which seems correct... someone please help guide me onto the right track?!

    I think i need to use the equation a= netforce / mass = g(sin theta - cos theta)
    then use the formula V(final)^2 = V(initial)^2 + 2 as
    to find net force I just used netforce= mass x g <<<<<<< I think this is incorrect because I am supposed to factor in the angle?! but how?

    then use formula Stopping distance (D) = (V(final)^2 / 2x us x g) - (V(final)^2 / 2x uw x g)

    i cannot figure out where to use the flow rate in this at all... i think i have totally stuffed up.... i really need help
     
    Last edited: May 23, 2010
  2. jcsd
  3. May 24, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi NoobertNoClue! Welcome to PF! :smile:

    (have a mu: µ and a theta: θ and try using the X2 and X2 tags just above the Reply box :wink:)

    Don't use acceleration.

    Use ∆(KE) + ∆(PE) = work done by friction. :wink:

    (The only relevance of the flow rate is that you need it in the formula for the friction on the slide)
     
  4. May 24, 2010 #3
    ∆KE = 1/2mV(final)² - 1/2mV(initial)² ???
    ∆PE = mg∆h
    ∆EE is N/A
    ∆KE + ∆PE = 0

    ∆PE = 110kg x 9.81 x 7.6m
    =8201.16

    .·. Should ∆KE = -8201.16 ???

    V(initial) = 0
    ∆KE = 0 - 1/2mV(final)²
    V(final) = √(2x∆KE/ -m)
    V(final) = 12.2114.... m/s
    at the end of the angular part of slide...

    velocity at staright part of slide would be;
    ∆PE = 110kg x 9.81 x 0
    =0
    so ∆KE = 0 ???

    0 = 12.21 - 1/2mV(final)²
    -12.21 = -1/2mV(final)²

    V(final) = 0.4712 m/s

    am i correct so far??

    Then velocity on pool =
    0 = 0.4712 - 1/2mV(final)²
    -0.4712 = -1/2mV(final)²
    V(final)= 0.092559....m/s

    how do i determine stopping distance from all of this?
    do i use the coefficient of friction to determine force?
    then divide?
    if so do i use the formula

    F= µR where R is normal force = mg <<<<<<<<<At this point was I supposed to use F= µN where N = mg(sin[tex]\vartheta-cos\vartheta[/tex])

    F on slide = 0.2 x (110 x 9.81)
    = 0.2 x 1079.1
    = 215.82 N

    then on the pool
    = 0.6 x 1079.1
    = 647.46

    Stopping distance =
    work = distance x force

    so d = w/f
    Total work (∆KE) = 1/2mV(final)² - 1/2mV(initial)²
    = 1/2x110x0² - 1/2x110x0.09255²
    =-0.4711026...

    at this point i know i have gone terribly wrong...
    should i just be using the formula
    d=V²/2µg
    where V would be the velocity at the end of the straight part of the slide
    and µ would be the coefficient of friction of the pool water.?????
     
    Last edited: May 24, 2010
  5. May 25, 2010 #4

    tiny-tim

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    Hi NoobertNoClue! :smile:

    (just got up :zzz: …)
    No!! As I said …

    Start again. :smile:
     
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