# Gauge Pressure of car tires

1. Nov 17, 2007

### Nghi

1. The problem statement, all variables and given/known data

The weight of your 1205 kg car is supported equally by its four tires, each inflated to a gauge pressure of 35.7 lb/in.^2.

a) What is the area of contact each tire makes with the road? (in m^2)
b) What gauge pressure is required to give an area of contact of 113 cm^2 for each tire? (in lb/in^2)

2. Relevant equations

Pgauge = P - Patmosphere

Pressure = Force/Area

3. The attempt at a solution

I actually went to my professor for help on this one, hehe. He told me that since the weight is equally supported, I just had to make sure that the force is divided by four. So I did that. First I found the real 'P'.

35.7 = P - 14.7
P = 50.4 lb/in^2

Since they wanted the area of contact in meters, I decided to convert the pressure into N/m^2.

50.4 lb/in^2 x (1.01e5 N/m^2 / 14.7 lb/in^2) = 346285.7143 N/m^2 (we're going to let that equal to X to make it simpler)

Now that I had the pressure and the force, all I should do is plug it into the definition of pressure.

P = F/A
X = (1205 x 9.81 / 4) / A
A = 2955.2625 / X
A = 0.00853 m^2

This isn't the answer, though, and I'm confused as heck. This was how my professor explained it to me, and it's wrong. o_o

For part b, I should theoretically just work backwards. First, I convert cm^2 to m^2.

113 cm^2 x (1 m^2/10000 cm^2) = 0.0113 m^2

P = F/A
P = 2955.2625/0.0113
P = 261527.6549 N/m^2 x (14.7 lb/in^2 / 1.01e5 N/m^2) = 38.064 lb/in^2

Pgauge = P - Patmosphere
Pgauge = 38.064 - 14.7
Pgauge = 23.364 lb/in^2

This is also the wrong answer. :/

Last edited: Nov 17, 2007
2. Nov 17, 2007

### katchum

A thought on this, I may be completely wrong:

You use the absolute pressure P for your calculation? But the atmosphere is also pressing on the outside of your tires. Maybe the solution is to just use gauge pressure? I don't know.

3. Nov 17, 2007

### dynamicsolo

katchum beat me to it: your tires already have one atmosphere of pressure inside them when they're "flat". It's the difference between the pressure inside and outside of the tire wall that matters; the gauge pressure is what supports the car's weight.

4. Nov 17, 2007

### Nghi

asd;falsflasldf;

I hate myself so much sometimes. :( But thank you so much for the help. Seriously. I was ready to punch myself in the eyeball from this problem.

Last edited: Nov 17, 2007
5. Nov 17, 2007

### dynamicsolo

Don't beat yourself up over it: the students I work with had a similar problem a week or so ago and I had to stop and think a moment as to whether it was the absolute or the gauge pressure that mattered. Different problems may require the use of one or the other...