Tire inflation to determine weight

[BooGTS]

Tire inflation to determine weight - solved

The problem goes as follows: he four tires of an automobile are inflated to a gauge pressure of 1.85E+5 Pa. Each tire has an area of 0.0237 m2 in contact with the ground. Determine the weight of the automobile.

But I've applied the Pfluid = F/A and it doesn't seem to pan out. I come up with 447 N per tire, which is wrong.

 

[Hootenanny]

Please show your working. It's hard to see where you've gone wrong if you don't show your working.

 

[HallsofIvy]

Then you done the arithmetic wrong: 2 x 10^5 times 2x 10^(-2)= 4 x 10^3 or 4000, not 400.

 

[BooGTS]

I worked on this problem forever before posting and figured it out in this little window on this site of all places. I'm sure it won't be my last tonight though... thanks for the response.

1. 1.85 x10^5 N/m^2 = (x/.0237m^2)

2. 1.85 x10^5 N x .0237 = 4384.5 N = x

3. 4384.5 N x 4 = 17538N =x

 

[Hootenanny]

4384.5 N x 4 = 17538N =x

Look's good to me

 

[Curious3141]

The problem goes as follows: he four tires of an automobile are inflated to a gauge pressure of 1.85E+5 Pa. Each tire has an area of 0.0237 m2 in contact with the ground. Determine the weight of the automobile.

But I've applied the Pfluid = F/A and it doesn't seem to pan out. I come up with 447 N per tire, which is wrong.

As Halls of Ivy said, that looks like a simple mathematical error.

Don't forget each tyre only bears a fraction of the weight of the car. So after getting the force exerted by each tyre on the ground, you need to do something else to find the weight of the car.

EDIT : Ah, OK, you've already posted it, that's correct.