1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gauge pressure

  1. Jul 27, 2008 #1
    The gauge pressure in each of the four tires of an automobile is 240 kPa. If each tire has a "footprint" of 220 cm2, estimate the mass of the car.

    @Mods: please move if this is too easy for this forum.
     
  2. jcsd
  3. Jul 27, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi #H34N1! Welcome to PF! :smile:

    Show us how far you've got, and where you're stuck, and then we'll know how to help you! (same for your other thread, of course) :smile:
     
  4. Jul 27, 2008 #3
    [hide=My work]Pressure = F/A.

    There are four tires and four footprints so the total pressure of all four tires is 240*4+101.3 = 1061.3.

    I am not sure if I need to add the atmospheric pressure to the gauge pressure. There could be reasons for both. Adding atmospheric pressure is correct because it is the pressure against which the tire is inflated. On the other hand, the atmospheric pressure also adds to the mass of the car so I am not sure how to proceed here.

    The total area is 220*4 = 880cm^2. We now have to convert to m^2 because the initial footprint is given in terms of cm^2 and the pressure is given in kPa. $=0.088$.


    F = mg, so we have

    1061.3 = mg/0.088

    Solving for m, we have m=9.53004082 kg, which does not seem right.
     
  5. Jul 27, 2008 #4
    You can not add pressure in this case.
     
  6. Jul 28, 2008 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi H34N1! :smile:

    Yes, pixel01 is right:

    pressure = total mass / total area.

    You must only use the pressure once! :smile:
     
  7. Jul 28, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hmm … on second thoughts, although that statement is correct, it only works because all the tyres have the same pressure (which they need not), and so in the general case you should consider each tyre separately.

    The amount of weight supported by each tyre (equal to the reaction force on that tyre from the ground) divided by the contact area for that tyre equals the pressure for that tyre:

    W1 = P1A1 etc

    In this case, the Ps are all the same.

    So total weight = W1 + W2 + W3 + W4 = P(A1 + A2 + A3 + A4) :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gauge pressure
  1. Gauge Pressure (Replies: 1)

  2. Gauge Pressure (Replies: 2)

  3. Pressure gauge (Replies: 2)

  4. Gauge pressure (Replies: 8)

Loading...