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Coke In a Car: Pressure Problem

  1. Aug 21, 2006 #1
    Suppose somebody put a coke can in the cup holder. The can is a perfect cylinder with radius R and height H. There is a very small airpocket since the can isn't completly full with a pressure, [tex]P_o[/tex]. The car accelerates horizonatlly with a value of A. The soda has a density of [tex]\rho[/tex]. What is the total pressure half way down the coke can?

    Here is my attempt:

    [tex]\SigmaF_x = mA = P_x*h*2R[/tex]
    [tex]\SigmaF_y = 0 = P_o + \rho*g*/frac{H}{2} - P_y[/tex]
    [tex]P=\sqrt{P_x^2 + P_y^2}[/tex]

    Somethings seems wrong. Does anyone know a better way to do it? Or why this method is wrong?
     
    Last edited: Aug 21, 2006
  2. jcsd
  3. Aug 21, 2006 #2

    Andrew Mason

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    It seems to me that the pressure is not uniform around the can. Where exactly are you measuring the pressure?

    Without the acceleration, the pressure at a point h below the top is P0 + [itex]\rho gh[/itex]

    Divide the can into discs of thickness dh and determine the force required to accelerate each disc. You can see that the pressure varies depending on the horizontal position.

    AM
     
  4. Aug 22, 2006 #3
    When a liquid accelerates there is a pressure differential in the direction of the acceleration.
     
  5. Aug 22, 2006 #4
    Thanks I see what to do now.

    [tex]A=\sqrt{A_x^2+A_y^2}[/tex]
    [tex] P=P_o+A*\rho*\sqrt{(\frac{H}{2})^2+R^2}[/tex]
     
  6. Aug 22, 2006 #5

    Andrew Mason

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    The pressure depends on the point at which it is measured. The pressure on the side in the direction of acceleration would be less than the pressure on the other side. How does your formula show this?

    The force on the contents is:

    [tex]f = mA = \pi r^2\rho HA[/tex] where H is the height of the can.

    This force is distributed over an area that is 1/2 the area of the cylinder.

    If you slice the can vertically in planes parallel to the direction of acceleration, the maximum force is on the largest slice (middle) of width dx. That force would be:

    [tex]f = dmA = \rho dVA = \rho 2rHdxA[/tex]

    That force is distributed over area dxH so the maximum pressure due to acceleration is:

    [tex]P = f/area = \rho 2rHdxA/dxH = \rho 2rA[/tex]

    If you add that to the pressure in the can and weight of the liquid, the maximum pressure at a point h from the top is:

    [tex]P = P_0 + \rho gh + \rho 2rA[/tex]

    AM
     
  7. Aug 23, 2006 #6
    That is a good way to do it. Your right that the formula I put out was wrong. I was about to fall asleep and then I realized that following revisions would have to be made. The idea i had was to make the axis along the vector resultant of the accelleration, A. Follow the line to the end of the can to give us distance d. The gives us the [tex]\rho*A*d[/tex] Then for the initial pressure if it is on the top edge it is just [tex]P_o+\rho*A_x*x[/tex] where x is the horizontal distance from the edge of the can. Similar methods could be used if the accelleration vector that is intersecting the center of the can hits the side instead of the top. I made this problem up and maybe it doesn't work out that well. The angle that the accelleration vector line makes with the horizontal is the arctangent of [tex]\frac{g}{a_x}[/tex]
     
  8. Aug 24, 2006 #7
    Can you just add the pressure resulting form the horizontal and vertical accellerations together of do you have to find a new accelleration?
     
  9. Aug 24, 2006 #8

    Andrew Mason

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    What vertical acceleration is there?

    AM
     
  10. Aug 30, 2006 #9
    the gravity of earth, i mean the coke molecules feel gravity

    It just seems like all the forces need to be computed in a vector fashion. Then see how far from edge of the can that the spot of question is following the vector resultant of the acceleration. Calculating the initial pressure at the edge of the can I don't think is to difficult. Does this make sense?
     
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