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Tire pressure problem

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Your right rear tire normally has to support a weight of 3000N. The contact area of your tire with the road is 200 square cm. If the pressure in your tire is suddenly reduced from 32 pounds psi to 16 pounds psi, what must be the new contact area to support the car?


    2. Relevant equations
    Classical mechanics, pressures in Pascal etc.


    3. The attempt at a solution

    Not quite sure where to go with this... My intuition tells me that the area should be larger but I'm not sure what equation to start plugging stuff into to prove it.
     
  2. jcsd
  3. Oct 22, 2012 #2

    SteamKing

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    What is the relationship between force and pressure?
     
  4. Oct 22, 2012 #3
    Hmm lets see here.

    P=F/A

    So the initial pressure is 1500 Pa, correct? How does that help me?
     
  5. Oct 22, 2012 #4

    SteamKing

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    Read your problem carefully.

    In the initial instance, you are given a force, a pressure and an area of contact. Are all of these quantities consistent, when using consistent units?

    In the second instance, the pressure in the tire is reduced. Does the force on the tire change? What happens to the contact area when the tire pressure is reduced?
     
  6. Oct 23, 2012 #5
    Back when I used to work heavy equipment, civil engineers would often ask me how much pressure a piece of equipment applied to the ground, so that they could design a concrete slab to park it on. I've seen all sorts of schemes to measure contact area to divide the axle weight by. But that always works out to a number equal to tire pressure. So I would tell them to figure 180 psi because none of the regular equipment is higher than that. Some aircraft is much higher, and that results in some very thick heavily reinforced concrete.
     
  7. Oct 24, 2012 #6
    Hmm the 3000N remains the same, that's for sure. If the pressure is decreased the contact area must increase... The pressure decreases by 1/2 so the contact area increases by a factor of 2?
     
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