Total area of contact between tires and ground?

  • #1
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0

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?
 

Answers and Replies

  • #2
20,870
4,546
You need to divide by the number of tires, not multiply. Also, you need to use the gage pressure in the calculation.

Chet
 
  • #3
45
0
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!
 
  • #4
20,870
4,546
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
 
  • #5
gneill
Mentor
20,913
2,862
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.
 
  • #6
45
0
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?!
 
  • #7
45
0
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.
 
  • #8
20,870
4,546
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
 
  • #9
45
0
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?
 
  • #10
20,870
4,546
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
 
  • #11
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?
 
  • #12
20,870
4,546
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
 
  • #13
Well 3.1 - 10 = -6.9 does it not?
 
  • #14
20,870
4,546
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.
 
  • #16
It's 105 = 1 *105 , not 10*105.
Yes I realize that now, not sure why it did not click in my head earlier, I should be better at this :p
 

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