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Cushion seat angle : Radius of curvature

  1. Apr 11, 2014 #1
    The problem is stated in the attachment.

    I would include my attempt at the question if I got anywhere.
    I'm really only looking for a hint as to how I set up the solution.

    PS, I understand how to work out the angle if the car wasn't moving.

    Thanks




    1. The problem statement, all variables and given/known data
    2. Relevant equations
    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 11, 2014 #2
    It's question 11 by the way
     
  4. Apr 11, 2014 #3

    berkeman

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    Staff: Mentor

    Start with a free body diagram (FBD).
     
  5. Apr 11, 2014 #4
    Is the radius of curvature and the speed of the car of any relevance?
     
  6. Apr 11, 2014 #5

    berkeman

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    Staff: Mentor

    At first glance, I don't think so. Your FBD will include forces, which will sum and be related to the acceleration...
     
  7. Apr 12, 2014 #6
    So I don't think I did it correct. I got an answer of 10.873°.

    I started off with the forces on the box being mgsinθ parallel down the plain and mgcosθ perpendicular into the plain.

    I broke up m x 4 (... mass x acceleration) into two components, 4mcosθ acting parallel to and up the plain, and 4msinθ acting into the plain.

    Angle for slip...

    mgsinθ + (mgcosθ + masinθ) x μ = macosθ

    (g+μa)sinθ = (a - μg)cosθ

    θ = ArcTan [(a-μg)/(g+μa)]

    θ = 10.87°
     
  8. Apr 12, 2014 #7

    berkeman

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    Staff: Mentor

    Could you attach your FBD? That would be a help in figuring this out...
     
  9. Apr 14, 2014 #8
  10. Apr 14, 2014 #9

    berkeman

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    Staff: Mentor

    Sorry, maybe I confused you with my reply. The motion of the car on the curved surface adds an additional acceleration beyond just gravity. That term should also be added in.
     
  11. Apr 14, 2014 #10
    Is that an additional tangential and normal acceleration?
     
  12. Apr 14, 2014 #11

    berkeman

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    Staff: Mentor

    The braking force would be tangential and the centripital force would be normal, right?
     
  13. Apr 15, 2014 #12
    Hey Berkeman, does this look okay to you?
    I followed it through and I got an answer of 3.8° which is isn't right, but I think the FBD looks okay.
    Thanks,
    cmcd
     

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