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Friction on an inclined plane

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Given an inclined angle of θ = 30° and 10m long. The coefficient of friction is 0,1 and the mas of the block is 1kg.
    a) Find the velocity needed for the block to climb the plane and its velocity be 0 in the top.
    b) Find the time it took to reach the top.
    c) Find the speed of the block when it reaches the bottom after its descent.

    I solved a) and c) already (10,7 m/s and 9m/s)

    2. Relevant equations

    For a) and c) : Wr + Ec1 + Ep1 = Ec2 + Ep2 ===> u.m.g.cos θ.x + 1/2 m v1^2 + mgh1 = 1/2 m v2^2 + mgh2
    For b) : Fr = uN ; Fr = ma ; fv = iv + a.t
    3. The attempt at a solution

    I think i've got to find the acceleration first. But i'm probably messing the equation. Supposedly the friction force (u.N) is equal to mass + acceleration, but im having second thoughts about if i have to add the weight projected on the x plane (m.g.cos ) ?

    ∑F = m.a
    Fr = m.a
    u.m.g.cos210 = m.a
    u.g.cos210 = a
    -0.84m/s^2 = a <===== Thats wrong i guess.

    What am i missing?

    Then i should use :

    fv = iv + a.t
    0 = 10.7m/s + a.t
    (-10.7m/s)/a = t
     
  2. jcsd
  3. May 15, 2015 #2
    Why do you guess it's wrong?
     
  4. May 15, 2015 #3
    We are given the answers, and the time for this item is 1,87s . A number i cant arrive.
     
  5. May 15, 2015 #4
    I can get 9m/s for part c), but why isn't the answer to part a) the same?

    Edit: If you want to get 1.87s, solve a.t=10.7m/s and 10m=0.5a.t^2.
     
    Last edited: May 15, 2015
  6. May 15, 2015 #5

    nrqed

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    Draw a free body diagram. It should be clear that, as you guessed, the friction force is not the only force along the inclined. There is a component of the force of gravity (equal to mg sin(theta) if theta is the angle of the inclined with respect to the horizontal).
     
  7. May 15, 2015 #6

    SammyS

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    Regarding the above line:
    It would help you to get the solution if you can explain what all those terms mean and especially explain how the second equation comes about.

    Did you find the acceleration for going up the ramp or for coming down the ramp?
     
  8. May 15, 2015 #7

    SammyS

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    @insightful ,

    Why should answers (a) and (c) be the same?

    What are you using for acceleration?
     
  9. May 15, 2015 #8
    Ooops, never mind. I forgot the friction force is in the opposite direction when going uphill (red face).
     
  10. May 16, 2015 #9
    Thanks nrqed, you won a cookie. The acceleration was solved by doing:

    Fr + Wx = m.a
    u.m.g.sin240 + m.g.cos240 = m.a ==> cancelling all the m
    0,1 . 9,8 . (-0.86) + 9,8 . (-0.5) = a
    -5.74 = a


    Then :

    fv = iv + a.t
    0 = 10.7 + (-5.74) . t
    -1.86 = t

    That second equation is the law of conservation of energy. English is not my first language so maybe im not translating into the right standars you guys use.

    The acceleration im working with is always the one that goes up.


    I think its because of the friction that you need a higher inicial velocity.





    Thanks everyone.
     
  11. May 16, 2015 #10

    nrqed

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    You are welcome :-)

    Watch out, solving for t you get t= 1.86 second (not -1.86 s).
     
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