Gauge pressure

[#H34N1]

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.

[tiny-tim]

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.

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:

[#H34N1]

[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.

[pixel01]

You can not add pressure in this case.

[tiny-tim]

You can not add pressure in this case.

Hi H34N1! :smile:

Yes, pixel01 is right:

pressure = total mass / total area.

You must only use the pressure once! :smile:

[tiny-tim]

You must only use the pressure once! :smile:

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: