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Ramp Problem

  1. Aug 13, 2014 #1
    1. The problem statement, all variables and given/known data
    A Truck is going around a circular track of radius 72m and is banked at 60 deg. A spider is at rest on the side of the truck and the coeffiecient of friction is 0.91, what is the maximum velocity of the truck so that the spider is at rest.


    2. Relevant equations

    [tex]F = \frac{mv^2}{r}[/tex]

    3. The attempt at a solution
    What I did is take the coordinate axes along the wall of the spider, then made -mgsin(%)+ N = -mv^2/2cos(%) for the vertical axes (where (%)is 60 deg ). -mgcos%+ musN = mv^2/2sin% for the horizontal axis. then substitute in N for the horizontal equation and solve for v. (The solution was (mv^2/r)sin60 - mus((mv^2/r)cos60 + mgsin60) = mgcos60, v = 47 m/s When I look at the solution I see that for the vertical axis the centripetal force and gravitational force have the same sign when it equal to Normal force which implies that when you subtract it to from both sides, the normal force is in the same direction as the centripetal force. Isn't the centripetal force just another way of writing the resultant force and should therefore be a result of other forces, instead of contributing itself.
     
    Last edited: Aug 13, 2014
  2. jcsd
  3. Aug 13, 2014 #2

    haruspex

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    Please explain how you get the right hand side of that equation.
    (I would recommend using vertical and horizontal axes instead, since the centripetal force is horizontal.)
     
  4. Aug 14, 2014 #3
    if you mean -mv^2/rcos60 I reason after breaking the centripetal force on the wall of the truck the direction you get is opposite to the normal force of the spider, making it in the same directin of the component gravitational force
     
  5. Aug 14, 2014 #4

    haruspex

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    It isn't stated, but I gather the spider is on the outside of the bend wrt the truck.
    Oh, that's ok (presuming you mean (mv^2/r)cos60), but you wrote mv^2/2cos60.

    Sorry, I cannot follow what you are saying there. In the vertical direction, the centripetal force has no component. In the horizontal, the gravitational has no component.
    To get them with the same sign, you must be referring to the direction parallel to the road surface. But then I don't know what you mean by "it equal to the normal force". What is?
     
  6. Aug 15, 2014 #5
    Actually from the image the spider is on the wall closer to the ground, that is to say it doesn't have the wall of the truck pushing on it in the centripetal motion. I have uploaded an image


    Thats because I broke up the radial acceleration on the wall, but I'll use the axis along the radial acceleration. here what I get

    fsin(60) - Ncos(60) = (mv^2/r) for the horizontal axis f for friction force

    fcos(60) + Nsin(60) = mg for the vertical axis

    then I get v = sqrt( (musin(60) - cos(60)/( mucos(60) + sin(60)))*Rg R for radius

    V= 12.4 m/s , but the answer is 40.7 m/s, I upload how the solution was done.
     

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    Last edited: Aug 15, 2014
  7. Aug 15, 2014 #6

    haruspex

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    OK, I was thinking it was on the outside of the truck, outside of the bend, but inside-inside is the same.
    I agree with your answer. The book answer seems to have some signs wrong (the cos 60 and sin 30 terms) Intuitively, the spider wasn't that far off slipping in a stationary truck, so it is not going to take much speed to dislodge it.
     
  8. Aug 15, 2014 #7
    alright, I shall move on thanks
     
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