Help with Physics?

Question

The weight of your 1255 kg car is supported equally by its four tires, each inflated to a gauge pressure of 34.1 lb/in2.

What is the area of contact each tire makes with the road?

I keep getting confused on this one. Help is much appreciated. Thanks.

Answer 1

Divide single wheel load by tire pressure inflation:

1255 kg = 2 766.80 lbs (convert your units)

2766.8/4 = 691.7 lbs (the weight on each tire)

691.7 lbs /34.1 lbs/in^2 = 20.28 in^2

You have to remember to keep your units straight - if you have the weight in kg, your pressure should be in kg/cm^2 or convert the metric (Kg) to imperial (Lbs) and then do the math. I could have also converted 34.1 lbs/in^2 to Kg/cm^2 (about 2.4 Kg/cm^2) and then done the math, but it is easier to convert the simple unit rather than the complex.

Once you do that, you are basically dividing out all of the mass units so that you are left with the area units.

Answer 2

1255*2.2 = 2,761 lb
2,761 lb / 34.1 psi ≈ 80.96774 in^2
80.96774 in^2 / 4 ≈ 20.241935 in^2 ≈ 20.2 in^2

Answer 3

pressure = weight /(area of contact)

pressure of the car = 34.1 lb/in2 = 15.47kg / in2.
it's not pressure but pressure /(acceleration of free falling)

pressure = 154.7/in2
12550/A=154.7

A=4.(area of contact of each tire)=4.At

12550/4.(At)= 154.7
=>At= 20.24 in2