# Homework Help: Hydrostatics and Cylindrical Tank

1. May 4, 2012

### Shoelace Thm.

1. The problem statement, all variables and given/known data
I fill a cylindrical tank of length 50 ft and diameter 90 ft with water (to the brim). What is the force the bottom 1 ft band of water exerts on the lining?

2. Relevant equations

3. The attempt at a solution
My main doubt is whether i should integrate from 0 to 1 and add the weight of the water above or simply integrate from 49 to 50 and be done. Both yield significantly different results.

2. May 4, 2012

### Dick

Depends on what variable you integrating with respect to. Is it distance from the top or distance from the bottom? Show your setup.

3. May 5, 2012

### Shoelace Thm.

I have one representation being $\vec{F} = 62.4 * 2 * \pi*45 \int^1_0 x\,dx + \pi*45^2*49$, where dx is a differential movement from the top of the 1 ft band to the base of the tank, and the other being $\vec{F} = 62.4 * 2 * \pi*45 \int^{50}_{49} x\,dx$, where dx represents the same. Both expressions take into account the fact that the force is greater because of there being water above, but nevertheless yield significantly different results.

4. May 5, 2012

### Dick

The pi*45^2*49 you are adding in the first case has nothing to do with the force exerted by the first 49 feet of water. What should it be? If you'd put units on things, you'd see it doesn't even have correct units.

5. May 5, 2012

### Shoelace Thm.

Would $62.4lbs/ft^3*\pi*45^2ft^2*49ft$ be correct?

6. May 5, 2012

### Shoelace Thm.

And if so the answers are still inconsistent.

7. May 5, 2012

### Dick

You want to add the force due to the pressure of the first 49 feet of water on the area around the side of the bottom foot of the pool. What would that be?

8. May 5, 2012

### Shoelace Thm.

I can't think of an expression because the 49 ft of water in question is not even in contact with the bottom 1 ft of the tank, in which case I can't find a reasonable way to incorporate the fact that the bottom 1 ft of water is "submerged" beneath 49 ft of water.

9. May 5, 2012

### Dick

The pressure of the 49 ft of water adds to the pressure of the last foot of water. It doesn't have to be in direct contact. What's wrong with doing it with your 49 to 50 integral?