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Calculating acceleration on level road/ when road uphill

  1. Oct 22, 2016 #1
    1. The problem statement, all variables and given/known data
    On a level road a motorcycle slows down with an acceleration of -3.40 m/s^2. Calculate the value of its acceleration when the road goes up hill 15 degrees.

    2. Relevant equations
    Fnet = MA



    3. The attempt at a solution
    I tried setting up a system of equations and made friction equal.


    Level road: friction force = ma
    Inclined road: friction + wsin(theta) = ma2

    Ma= ma2-wsin(theta)
    Ma = ma2 - mgsin(theta)
    A = a2 - gsin(theta)
    A2 = A + gsin(theta)
    A2 = -3.40 + (-9.8)(sin15)

    But the answer isn't -5.94, so I don't know what I'm doing wrong. Thanks.
     
  2. jcsd
  3. Oct 22, 2016 #2

    Orodruin

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    Is the friction the same in both cases? What is the limiting case when the inclination is 90 degrees?
     
  4. Oct 22, 2016 #3
    It does not specify. That is the whole question that I posted. I'm in the very beginning of AP physics 1, so I'm not sure what a limiting case is yet.
     
  5. Oct 22, 2016 #4

    Orodruin

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    I know it doesn't, it is a question for you, not for the formulation.

    A limiting case is a corresponding situation in which you can easily reason what the answer should be. In this case replacing the angle with a 90 degree angle.
     
  6. Oct 22, 2016 #5
    Oh okay. Thank you. I would think the friction would be the same because the Natural force and the coefficient of friction wouldn't change, it's the same surface. If the inclination was 90 degrees I suppose it would be zero acceleration, but honestly I'm not quite sure.
     
  7. Oct 22, 2016 #6

    Orodruin

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    Do you mean "normal force"? In that case, why would you think it would be the same?
    You are saying a bike on a vertical wall has zero acceleration?
     
  8. Oct 22, 2016 #7
    Yes, I mean normal force, I apologize. Since it's staying on the road and not floating/sinking, normal force would equal weight and so would be = mg. The mass would stay the same and so would gravity. Coefficient of friction is dependent on the surface, but since the surface is the same it would be equal in both scenarios.

    Oh, the acceleration would be -9.8m/s^2, since gravity would pull it down.
     
  9. Oct 22, 2016 #8

    Orodruin

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    Is this true for an object which is not on a horizontal surface? What does "normal" mean and what is the role of the normal force?

    Yes, but according to the same reasoning you employed for 15 degrees, you would get the result -9.8 sin(90)-3.4 = -13.2 m/s^2. What is the normal force in this case?
     
  10. Oct 22, 2016 #9
    I thought it would be. The normal force is perpendicular to the surface of contact, so I thought that, though normal force and weight may have different directions, their magnitude was equal as long as the object wasn't sinking/floating.

    Would the normal force be -9.8 times mass?
     
  11. Oct 22, 2016 #10

    Orodruin

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    You have then not understood the concept of the normal force. The normal force is the reaction force from the surface that prevents an object from passing through the surface. It ensures force equilibrium in the normal direction (assuming that other forces are not accelerating the object away from the surface in which case the normal force is zero). You can find the normal force by requiring force equilibrium in the normal direction.
     
  12. Oct 22, 2016 #11
    So normal force would be equal to the portion of weight in the opposite direction to normal force?
     
  13. Oct 22, 2016 #12
    The normal force would be equiviliant to weight in the first scenario, correct? In the second scenario, normal force would equal weight in the y direction? I tried drawing a picture and a free body diagram...
     
  14. Oct 22, 2016 #13

    Orodruin

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    If by "portion" you mean "component", yes.

    It is unclear to me what cases you refer to and which direction you have designated as y. Please keep in mind that I cannot see your drawing.
     
  15. Oct 22, 2016 #14
    It is unclear to me what cases you refer to and which direction you have designated as y. Please keep in mind that I cannot see your drawing.[/QUOTE]

    By first scenario I mean when the motorcycle is traveling on a flat road. And second scenario is up the hill. The normal force is along the Y axis and friction is along the X.

    I think I could link the two equations together by coefficient of friction.
     
  16. Oct 22, 2016 #15

    Orodruin

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    You still have not defined your axes. Again, I cannot see your drawing.
     
  17. Oct 22, 2016 #16
    How do I define my axes?
     
  18. Oct 22, 2016 #17

    Orodruin

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    You specify in which direction the x and y axes are pointing. Again, I can guess, but if I guess wrong and you have understood something wrong then I might end up just confusing you. It is therefore very important that you give all of the information and do not just assume I know everything that you can see.
     
  19. Oct 22, 2016 #18
    The positive x axis is pointing up the hill, and the negative x axis is pointing down the hill. The motorcycle is at the origin. The normal force lies on the positive y axis.
     
  20. Oct 22, 2016 #19

    Orodruin

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    Then yes as long as by "portion" you mean "component" in post #12.
     
  21. Oct 22, 2016 #20

    Orodruin

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    I suggest you try to redo the problem with this knowledge.
     
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