# Under PRESSURE

1. Aug 7, 2011

### recreated

Hypothetical question to help me in design:

A PVC air tight balloon like object is used to suspend a weight just above the ground. It can be thought of as being very similar to a car tyre's function. The weight is known, the part of the balloons area in contact with the ground is known.
Thus, pressure is found by using ((Pressure = Force/ground contact area) = N/m^2).

From this, how do I find the PSI on the walls of the balloon? It has a cylindrical shape.

Pressure/Surface area of balloon?

Cylinder is thin walled, other than top and bottom which have half inch plates attached, how would they effect the spread of pressure?

Surely the analogy of a car tyre could be used and give near perfect results as it is basically a balloon like object with differing wall thickness? Any relavent links about this in relation to car tyres would be appreciated, I have already searched quite a bit on them.

Thank You.

2. Relevant equations

3. The attempt at a solution

2. Aug 16, 2011

### nvn

recreated: Is this a homework question? You say the cylinder wall is made of PVC, but then you say the cylinder wall acts like a balloon, which would be like an elastomer, such as polychloroprene.

Air pressure, p, inside the cylinder will be the same, throughout the cylinder, on the cylinder wall and end caps. Assuming the cylinder wall is a thin elastomer, and acts like a balloon, I currently think air pressure p = po + (W/A), where po = initial gauge air pressure inside the cylinder (before weight W is applied), W = suspended object weight, A = 0.25*pi*d^2, and d = current cylinder midspan diameter (after weight W is applied).