Question

IP The weight of your 1440 kg car is supported equally by its four tires, each inflated to a gauge pressure of 32.0 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? If the gauge pressure is increased, does the area of contact increase, decrease, or stay the same? decrease, increase, or stay the same?

Part C: What gauge pressure is required to give an area of contact of 116 cm2 for each tire?

Answer 1

Part A

Pressure (P)= Force (F)/Area (A)

Force by the car = mass x acceleration due to gravity = mg =1440 kg x 9.8 ms^-2 = 14112 N = 3172.5 lb

Area of contact, A = F/P = 3172.5/32 in² = 99.14 in²

This is the area of contact including all four tires.

So area of contact for each tire = 99.14/4 in² = 24.785 in² = 159.9 cm²

Part B

Pressure is inversely proportional to Area of contact.

So, on increasing Pressure, area of contact decreases

And on Decreasing Pressure, area of contact increases.

Part C

Area of contact for each tire = 116 cm² = 17.98 in²  

Net area of contact = 17.98 x 4 in² =71.92 in².

Force = 3172.5 lb

Pressure Required = 3172.5/ 71.92 lb/in² = 44.11 lb/in²