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Homework Help: Acceleration,Time,Velocity with Friction Problem?

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data

    Oscar the otter is having fun sliding down a mud bank.

    Oscar has a mass m=13.2kg
    The length of the mud bank is L=4m
    The coefficient of kinetic friction between Oscar and the bank is MU=.365
    The angle the mud bank makes W.R.T. the horizontal is theta=26.2 degrees

    A) Calculate Oscar’s acceleration down hill? A=
    B) Starting from rest at the top of the mud bank, how long will it take Oscar to slide the length of the hill and go for a swim? T=
    C) Starting from rest from the top of the hill, what is Oscar’s velocity as he hits the water? V=

    After work, Oscar decides to calculate what he could about the day’s fun. He found that taking the +x axis to lie along the bank and pointing down hill seemed to make solutions a little easier.

    2. Relevant equations
    A)Not really sure about this equation EFx=mg sin theta
    B)Delta x=1/2at^2 solve for t

    3. The attempt at a solution
    A)Is my equation correct for acceleration and how do i equate my friction of MU into the correct equation?
    B)To solve for time would the equation stated above be correct?
    C)Would my velocity equation be correct for this problem?

    Please help I need some direction and it starts with my equations because i really have no good examples of this problem. Thank you
  2. jcsd
  3. Oct 5, 2007 #2


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    Draw a freebody diagram... take the x-axis along the plane... y-axis perpendicular to the plane...

    use the equations: [tex]\Sigma{F_x} = ma_x[/tex] and [tex]\Sigma{F_y} = ma_y[/tex].

    here ay = 0, hence [tex]\Sigma{F_y} = 0[/tex]... because the otter can't travel into the plane...
  4. Oct 5, 2007 #3
    Thank you but still alittle confused,

    So I should just use the EFX=max equation only?
    Will my EFx equal my friction .365 correct?
    Im sort of confused would my equation look lke this:


    and what importance is my theta of 26.2 degrees?
  5. Oct 5, 2007 #4


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    0.365 is not the friction, but the coefficient of friction... to get friction you need the normal force... then friction = 0.365*normal force.

    it is important to draw the freebody diagram so that you see all the forces involved...

    The theta is necessary to get the components of gravity along the plane (x-axis), and perpendicular to the plane (y-axis).

    For [tex]\Sigma{F_x} = ma_x[/tex]... friction is not the only force. you also have the component of gravity along the plane.
  6. Oct 5, 2007 #5

    So my Normal force would be 13.2kg(9.8m/s^2)=129.36N correct?

    Then I take my friction formula and do 0.365(129.36N)=47.21 correct?
  7. Oct 6, 2007 #6


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    no, you need to take into account the angle... have a look at this page:

    http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Dynamics/InclinePlanePhys.html [Broken]

    it explains how to deal with inclined planes etc...
    Last edited by a moderator: May 3, 2017
  8. Oct 6, 2007 #7
    Thanx again Learn, the website was alot of help.
    What I was able to get from it was that the overal equation i need for acceleration is:

    a=Fapp-un-mg(sin theta)/m

    Is this the correct formula? If so what is my Fapp and un?
  9. Oct 6, 2007 #8


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    Yes, that's exactly the right formula. In this case there is no applied force.. Fapp = 0.

    [tex]\mu*N[/tex] = 0.365*N

    For N have a look at the website...

    Have a look at:

    "Incline Plane with Parallel Applied Force and Friction" in the website... this is exactly like your problem. only difference is in this case Fapp = 0... everything else is exactly the same as for your problem.
  10. Oct 6, 2007 #9
    Ok I might just have it:

    This is what i did

    N=mgcos theta


    Then I use
    a=Fapp-uN-mg(sin theta)/m

    Are my numbers mostly right?
    Im concerned about the -(minus) signs because im getting negative outcomes,should I turn the minus signs positive because of my positice numbers?
  11. Oct 6, 2007 #10


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    Oh... I made a mistake.

    This situation is not exactly the same as the one I pointed out in the websiste because friction is in the opposite direction...

    In this case the block is moving down the plane, but friction opposes motion so it is up the plane...

    if we take down the plane as negative, and up the plane as positive:

    [tex]a = \frac{\mu*N - mgsin(\theta)}{m}[/tex]

    so a = [(0.365)116 - 13.2kg(9.8m/s)(0.4415)]/13.2

    a = -1.12m/s^2

    the minus sign just means the acceleration is down the plane (it just indicates direction)... if we took our initial convention to be down the plane as positive and up the plane as negative, it would come out as +1.12m/s^2...
  12. Oct 6, 2007 #11
    Ok I gotcha thanx a whole lot, so for my part B I just use this equation
    Delta x=1/2at^2 solve for t
    For part C i use V=at

    Are they correct? If so what is my Delta x
  13. Oct 6, 2007 #12


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    Yup. that all looks right. you're delta x is the length of the mudbank which is 4m
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