# Total area of contact between tires and ground?

## Homework Statement

A car has four tires in contact with the ground each of which is inflated to an absolute pressure of 3.1×105Nm^-2. If a person of mass 82 kg gets into the car by how much will the total area of contact between the tyres and the ground increase assuming that the tire pressure remains constant?
m = 82kg
P = 3.1×10^5Nm^-2
g = 9.81ms^-2
P(atmosphere)=10^5Nm^-2

## Homework Equations

Pressure, P = F/A
F = ma

## The Attempt at a Solution

The above equation was used in order to get the solution. P = (9.81ms^-2×82kg)/(3.1x10^5Nm^-2)
= .00259 × 4 (number of tires)
= 0.01m^2

This does not give me the correct answer. Do I have to consider atmospheric pressure? If so, are there any more equations required in this problem?

Chestermiller
Mentor
You need to divide by the number of tires, not multiply. Also, you need to use the gage pressure in the calculation.

Chet

You need to divide by the number of tires, not multiply. Also, you need to use the gage pressure in the calculation.

Chet
I've used gauge pressure which is 2.1x10^5 and divided by the number of tires instead of multiplying and get 0.021m^2. However, this still isn't the correct answer!

Chestermiller
Mentor
I've used gauge pressure which is 2.1x10^5 and divided by the number of tires instead of multiplying and get 0.021m^2. However, this still isn't the correct answer!
There's something wrong with this result. It is not consistent with your previous calculation. It looks high by a factor of about 10.

Chet

gneill
Mentor
I've used gauge pressure which is 2.1x10^5 and divided by the number of tires instead of multiplying and get 0.021m^2. However, this still isn't the correct answer!

Can you show some details of your calculation? I don't see how you've arrived at 0.021m^2 with the given values and your relevant equation.

There's something wrong with this result. It is not consistent with your previous calculation. It looks high by a factor of about 10.

Chet
You're right, I miscalculated. 9.58cm^2 is what I got but it's only partially correct because this is the change in area per tire, but that's not what I've been asked to find apparently?!

Can you show some details of your calculation? I don't see how you've arrived at 0.021m^2 with the given values and your relevant equation.
My mistake! What I actually get is 9.58cm^2 but this is the change in area per tire, which is not what I've been asked to find. Very confusing.

Chestermiller
Mentor
You're right, I miscalculated. 9.58cm^2 is what I got but it's only partially correct because this is the change in area per tire, but that's not what I've been asked to find apparently?!
Ah...I forgot that. So you neither multiply nor divide by 4 to get the answer.
Chet

Ah...I forgot that. So you neither multiply nor divide by 4 to get the answer.
Chet
So since I don't need to multiply nor divide by 4, and just to be completely sure... Is the answer 38.3cm^2?

Chestermiller
Mentor
So since I don't need to multiply nor divide by 4, and just to be completely sure... Is the answer 38.3cm^2?
Yes, if you did the arithmetic correctly.

Chet

To calculate the gauge pressure, is it not the Absolute pressure minus the atmospheric pressure. 3.1x10^5 - 10^5 which would give a negative number?

Chestermiller
Mentor
To calculate the gauge pressure, is it not the Absolute pressure minus the atmospheric pressure. 3.1x10^5 - 10^5 which would give a negative number?
How do you figure that this gives a negative number?

Chet

Well 3.1 - 10 = -6.9 does it not?

Chestermiller
Mentor
Well 3.1 - 10 = -6.9 does it not?
You're having a little bit of a notation issue. 3.1x10^5 is supposed to be 310000, and 10^5 is supposed to be 100000. So 310000 - 100000=210000=2.1x10^5.

haruspex