Finding the weight of a car.. with just pressure.?

Question

I am working on this problem for physics. But I frankly think that this is impossible.. even though it wont be. Here is the problem:

The four tires of an automobile are inflated to a gauge pressure of 2.0x10^5 Pa. Each tire as an area of 0.024 m^2 in contact with the ground. Determine the weight of the automobile.

The first thing that came to my mind was using P=F/A and fin F so I could find the force and then the mass.. but that didn't worked. I would gladly appreciate if anybody can answer and explain how the process work. Thanks :)

Answer 1

The total force of the car on the ground is just

F = (2.0 x 10^5 N/m²) (4) (0.024 m²) = 19200 N = 19200 kg m / sec²

The mass of the car is (19200 kg m / sec²) / (9.81 m / sec²) = 1957.19 kg

I'm guessing that you needed to divide the force by acceleration to find the mass. Recall that F = m a = m g, where g is 9.81 m / sec².

Addendum: I read Possum's answer with interest. When you measure tire pressure, a reading of 0 does not mean that there's a vacuum, it actually still does mean about 1.01325 x 10^5 Pa. A reading of of 2.0 x 10^5 Pa is the difference between tire pressure and atmospheric pressure, and it's this difference that's what holds up the vehicle. Archimede's principle and all that.

2.0 x 10^5 Pa is roughly 29 psi, which is slighly less than the standard 34 psi for most tires.

Answer 2

Not sure how tricky your professor is. The following are things to watch for:

Don't forget to multiply by four, for four tires.

Make sure it is PSIG (PSI Gauged, which means the pressure is that in excess of the atmospheric pressure). If it is PSIA (PSI Absolute) you need to subtract 1 atm from it.

Make sure the pressure was measured with the tires mounted on the car and the car parked on the ground with no temperature change. That is not the pressure of a dismounted tire or a pressure when the tires were hot off of a highway and had since cooled down before the contact areas were measured.

The elastic forces of the steel radial belts and the nylon-rubber sidewalls of the tires are hard to model. But they may be significant. And you can estimate them by the contact area change vs. the pressure change.

Answer 3

Pa = kg/m*s^2
2.0 * 10^5 Pa = 200,000kgs / m*s^2
200,000 kgs/ m*s^2 * 0.024 m^2 = 4800 kgm /s^2
multiplied by 4 tyres:

19,200kgm/s^2

divided by acceleration due to gravity : 9.8 m/s^2

1959.18 kgs

the units of measure are important so you know if you're doing the right thing. hope this helps.

Answer 4

P = F/A
F=mg

then P = mg/A
m=PA/g

You've written P= 200,000 pg(g), so absolute pressure = 301,325 Pa. (You have to use absolute pressure)

m = 301325*(4x0.024) / 9.8
m= 2,952 kg.