Question

The weight of your 1430 kg car is supported equally by its four tires, each inflated to a gauge pressure of 33.0 lb/in2

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

Part B: If the gauge pressure is increased, does the area of contact increase, decrease, or stay the same?

Part C: What gauge pressure is required to give an area of contact of 111 cm2cm2 for each tire?

Answer 1

Part A.

We know that Pressure is given by:

Pressure = Force/Contact Area = F/A

A = F/P

F = Force on each tire = W/4 = m*g/4 = 1430*9.81/4 = 3507.075 N

P = Pressure inside each tire = 33.0 lb/in^2

Remember: 1 atm = 1.01*10^5 Pa = 14.7 lb/in^2, So

P = (33.0 lb/in^2)*(1.01*10^5 Pa/14.7 lb/in^2) = 33.0*1.01*10^5/14.7 = 226734.7 Pa

So,

A = 3507.075/226734.7 = 0.0155 m^2

Since 1 cm^2 = 10^-4 m^2, So

A = 155*10^-4 m^2 = 155 cm^2

Part B.

From above we can see that area of contact is inversely proportional to the gauge pressure, So if gauge pressure is increased than area of contact will decrease.

Part C.

Now when area of contact is 111 cm^2, So

A = 111 cm^2 = 111*10^-4 m^2

F = Force applied on each wire = 3507.075 N

So,

P = gauge pressure required = F/A

P = 3507.075/(111*10^-4)

P = 315952.7 Pa = (315952.7 Pa)*(14.7 lb/in^2/(1.01*10^5 Pa))

P = 315952.7*14.7/(1.01*10^5)

P = 45.985 lb.in^2

P = 46.0 lb/in^2 = gauge pressure required

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