# Fluid statics with Pascal's barrel

• Undoubtedly0
In summary, Pascal's barrel experiment demonstrates the concept of fluid statics through the use of a cylindrical barrel with a straw inserted. The total weight of the fluid is determined by its density, cross-sectional area, and total height. The normal force exerted on the barrel by a plank is equal to the force due to pressure acting on the bottom of the barrel. However, if the barrel is closed at the top, the pressure at the top will also contribute to the normal force, resulting in a different result than when only considering the force due to pressure at the bottom. This is due to Newton's third law, as the top of the barrel is also pushing down on the water.
Undoubtedly0
I am confused about the some of the concepts behind fluid statics. Here is an example to illustrate this, using Pascal's barrel experiment.

Consider a cylindrical barrel with cross-sectional area A and height H. Now insert a straw of height h and cross-sectional area a so that the total of height of the barrel and straw is h + H. Assuming a fluid of uniform density ρ, the total weight of the fluid is

$$W = m_{tot}g = \rho V g = \rho g (ha + HA)$$

If the barrel is sitting on a plank, then the magnitude of the normal force N of the plank on the barrel is simply N = W.

However, using a different method, I come up with a different result for N. By Pascal's equation, the gauge pressure at the bottom of the tank is

$$P_{bottom} = \rho g (h + H)$$

Thus the force due to pressure acting on the bottom of the barrel is

$$F_P = P_{bottom}A = \rho g A (h + H)$$

In order to balance this force, the normal force N from the plank on the barrel must thus be equal to FP, yet this is a different result then from before: N = W ≠ FP = N.

There must be some error in my reasoning here, but I cannot find it. Thanks a bunch for the help!

Hi Undoubtedly0!

Pascal's barrel was firmly closed at the top (apart from the hole for the straw).

So the water just under the top of the barrel is at higher than atmospheric pressure, pushing upward on the top of the barrel.

By good ol' Newton's third law, the top of the barrel is also pushing down on the water.

Thanks tiny-tim. This accounts for the difference perfectly. The force due to pressure exerted on the top of the barrel is

$$F_{P_{top}} = P_{top}(A-a) = \rho gh(A-a)$$

And so it follows that

$$W = F_P - F_{P_{top}}$$

## 1. What is fluid statics?

Fluid statics is the study of fluids at rest, where there is no relative motion between different parts of the fluid.

## 2. What is Pascal's barrel?

Pascal's barrel is a thought experiment created by the French mathematician and physicist Blaise Pascal. It is a sealed barrel filled with a fluid, with a small opening at the top. According to Pascal's principle, the pressure exerted by the fluid at the bottom of the barrel is transmitted equally throughout the entire barrel.

## 3. How does Pascal's barrel demonstrate fluid statics?

Pascal's barrel demonstrates fluid statics by showing how the pressure exerted by a fluid is transmitted equally in all directions. This means that the pressure at the bottom of the barrel is the same as the pressure at the top, even though the depth and volume of the fluid may be different at different points in the barrel.

## 4. What is the equation for calculating pressure in Pascal's barrel?

The equation for calculating pressure in Pascal's barrel is P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the gravitational acceleration, and h is the height of the fluid column.

## 5. How is Pascal's barrel used in real-world applications?

Pascal's barrel is used in real-world applications to understand and predict the behavior of fluids in different systems, such as hydraulic systems and water towers. It is also used in the design of dams, pipes, and other structures that involve the use of fluids.

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