Total area of contact between tires and ground?

In summary: Yes I realize that now, not sure why it did not click in my head earlier, I should be better at this :p
  • #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?
 
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  • #2
You need to divide by the number of tires, not multiply. Also, you need to use the gage pressure in the calculation.

Chet
 
  • #3
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!
 
  • #4
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
 
  • #5
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.
 
  • #6
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?!
 
  • #7
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.
 
  • #8
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
 
  • #9
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
 

1. How is the total area of contact between tires and ground measured?

The total area of contact between tires and ground is typically measured by using a pressure-sensitive film or electronic pressure sensors placed between the tire and ground. These tools can accurately measure the distribution of pressure across the contact patch and calculate the total area.

2. Does the total area of contact differ between different types of tires?

Yes, the total area of contact can vary between different types of tires due to differences in tire design, tread pattern, and inflation pressure. For example, wider tires may have a larger contact patch than narrower tires, and off-road tires may have a larger contact patch than street tires.

3. How does the total area of contact affect a vehicle's performance?

The total area of contact between tires and ground has a significant impact on a vehicle's performance. A larger contact patch can provide better traction and handling, but it also increases rolling resistance and can decrease fuel efficiency. A smaller contact patch may improve fuel efficiency but can decrease overall grip and handling.

4. Can the total area of contact change while driving?

Yes, the total area of contact can change while driving. Factors such as vehicle speed, tire pressure, and weight distribution can affect the size and shape of the contact patch. For example, when braking, the weight of the vehicle shifts to the front, causing the front tires to have a larger contact patch than the rear tires.

5. How does the total area of contact impact tire wear?

The total area of contact between tires and ground can impact tire wear. A larger contact patch can distribute the vehicle's weight more evenly, reducing the wear on individual parts of the tire. However, a larger contact patch also means more friction, which can cause the tire to wear out faster. Proper tire maintenance, including regular rotation and inflation, can help optimize the contact patch and extend the life of the tire.

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