Total area of contact between tires and ground?

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Homework Help Overview

The problem involves calculating the increase in the total area of contact between a car's tires and the ground when an additional mass is added, while maintaining constant tire pressure. The subject area includes concepts of pressure, force, and area in the context of physics and mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of pressure equations, questioning whether to use absolute or gauge pressure. There are attempts to clarify the calculations and the correct interpretation of the problem statement.

Discussion Status

The discussion is ongoing with participants providing guidance on the calculations and questioning the assumptions made regarding pressure. Some participants have identified discrepancies in previous calculations and are seeking clarification on the correct approach to find the total area of contact.

Contextual Notes

There is confusion regarding the use of absolute versus gauge pressure, as well as the proper handling of the number of tires in the calculations. Participants are also addressing notation issues related to the values used in the calculations.

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

Chet
 
Chestermiller said:
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!
 
Dennydont said:
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
 
Dennydont said:
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.
 
Chestermiller said:
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?!
 
gneill said:
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.
 
Dennydont said:
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
 
Chestermiller said:
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
Dennydont said:
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
Sam Fielder said:
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
Sam Fielder said:
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.
 
  • #15
Sam Fielder said:
Well 3.1 - 10 = -6.9 does it not?
It's 105 = 1 *105 , not 10*105.
 
  • #16
haruspex said:
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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