# Does a falling column of fluid exert the same pressure as a static column?

Suppose you have a $$5\,\text{m}$$ high waterfall. The water will be accelerated downwards by gravity and impart a force, proportional to the cross sectional area of the falling column of water, on the surface at the bottom. So the falling water exerts a constant hydrodynamic pressure (force/area) as a function of the drop height and the fluid density, ignoring atmospheric drag.

Now suppose you are at the bottom of a pool of water $$5\,\text{m}$$ deep. You will experience the hydrostatic pressure of just the weight of all the water above you. The water at the bottom will be compressed, and under pressure, and exert a constant hydrostatic pressure as a function of depth and density.

The falling water is not being compressed by gravity, since the water is allowed to fall (Falling observer experiences $$0g$$), so there is no hydrostatic pressure. But the same water is moving at a velocity enough to exert a given hydrodynamic pressure.

My question is, given the same fluid density, height/depth, and gravitational acceleration, is the hydrodynamic pressure at the bottom of a falling stream always equal to the hydrostatic pressure at the bottom of a static column?

• Does a falling rock exert the same pressure on you as the same rock sitting on you? Commented Apr 27 at 23:49
• @mmesser314 No. But water is a fluid and rock is a solid. A rock will exert a force dependent on the rigidity of the surface on impact. A less rigid surface will decelerate the rock slower resulting in a lower force. Commented Apr 27 at 23:50
• Suppose you want to keep the rock or water from accerating you. Whatever the surface area the rock or water hits, you must exert a certain force to oppose its weight. If it is moving, you must also add a force to decelerate it. Yes there are differences in this additional force for a rigid rock and fluid water. Commented Apr 28 at 3:10

This is very similar to the falling chain problem.

Assume a column of water with a cross-sectional area A and column length L, is suspended directly above a scale.

The weight of water is $$Mg = \rho A L g$$

The total weight measured on the scale after all the water has fallen will be:

$$W_T = W_{water} + F_{impact}$$ where $$F_{impact}$$ is the force due to the momentum change that occurs due to each each falling drop dm that hits the scale.

The velocity the dm achieves after falling a distance L is: $$v = \sqrt{2gL}$$

EDIT: The water ground collision is not perfectly inelastic like a falling chain. After collision, the water spreads with a velocity that is a fraction of original:

$$v_s = \gamma v$$

If the water column is uniform, then dm can be stated as a function of dx by using the linear density $$\frac{M}{L}$$

$$dm = \frac{M}{L}dx$$

$$F = \frac {dp}{dt} = \frac{M}{L}dv\frac{dx}{dt}$$

Since $$dv = \sqrt{2gL}(1-\gamma)$$ and $$\frac{dx}{dt} = \sqrt{2gL}$$

$$F_{impact} = 2Mg(1- \gamma)$$ and $$F_T = Mg + 2Mg(1-\gamma)$$

When $$\gamma = 1$$ then $$F_T = Mg$$ which is the answer given by Vincent Thacker using Bernoulli's equation. Otherwise, the water imparts a fraction of its momentum to the ground.

My question is, given the same fluid density, height/depth, and gravitational acceleration, is the hydrodynamic pressure at the bottom of a falling stream always equal to the hydrostatic pressure at the bottom of a static column?

Unless the water collision is perfectly elastic, then No. Water is highly incompressible and at 5m any additional pressure will be small compared to the momentum imparted to the ground if the collision is inelastic.

Falling chain problem: https://www.feynmanlectures.caltech.edu/info/solutions/falling_chain_sol_1.pdf

• The water doesn't stop moving after reaching the bottom, unlike the chain. It spreads out and flows radially outwards. I indeed made the same mistake as you (see the earlier revisions of my answer). The correct equation to use is Bernoulli's equation for a jet of water hitting a surface. Commented Apr 27 at 23:01
• @VincentThacker Doesn't that just remove the static weight (left on the scale?) If I let the water roll off the scale, that still leaves 2Mg from the momentum change due to the drop. I will edit my post. Commented Apr 27 at 23:08
• No, the water continues to flow with nonzero velocity. It is known as a stagnation point. It's described in figure 16 here. If you look at the previous version of my answer, I got exactly the same (wrong) 2Mg result as you did. Commented Apr 27 at 23:12
• @VincentThacker OK. I get it. I have assumed a completely inelastic collision of the water with the ground. If I assume a perfectly elastic collision where $v_i^2 = v_f^2$ for the water, then I get your answer where F = Mg since none of the momentum is transferred to the ground This assumes all the vertical velocity is converted to a horizontal spread velocity. Commented Apr 28 at 0:59

If the water is assumed to start from rest at height $$h$$, using Bernoulli's equation, we get $$P_0 + \rho gh = P_1$$ so the additional pressure at the central point of the bottom is $$P_1-P_0$$ which is indeed equal to $$\rho gh$$. The velocity of the cylindrical jet of water just before hitting the bottom is $$v=\sqrt{2gh}$$.

If the water breaks up into droplets, the pressure can vary and becomes much more difficult to calculate, cf this post.

• The higher the water falls from, the faster, and thus more force imparted, on the bottom. So I do not understand. The flow rate defines how much water there is, thus the cross-sectional area of the stream. That defines the force, but not the pressure. Less flow rate is less cross-sectional area of stream, and less force, but same pressure overall, right? Commented Apr 27 at 21:35
• @CPlus See the updated answer. Commented Apr 27 at 22:02
• @CPlus If the water that hits the ground with velocity v spreads away at the same velocity then momentum of the water is conserved without imparting any momentum to the ground so F = Mg Commented Apr 28 at 1:04

At the center of the falling fluid contact, the pressure is going to be equal to the stagnation pressure $$p_s=\frac{1}{2}\rho v^2$$. For the velocity $$v=\sqrt{2gh}$$, that means that, at the center of contact, $$p=\rho g h$$ This is the same as the hydrostatic pressure in the static case. Further from the center of contact at the surface, the pressure will be less than this.

• Please do not tactically downvote other answers just to post a nearly identical one. This isn't the first time you've done this. Commented Apr 28 at 11:50
• @VincentThacker My answer isn't the same at all. The gauge pressure within the falling jet is not hydrostatic; it is zero along the path of the jet, and only rises to $\rho g h$ at the center of the contact with the ground. Commented Apr 28 at 14:07
• This is the correct answer! Bernoulli's equation holds along a stream line, and the stream line which ends in the stagnation point, where the water velocity becomes zero, gives the stagnation pressure, which is simply equal to the kinetic energy of the water element gained by from the potential energy of the fall. Away from the stagnation point, where the water flow parts, the velocity doesn't become zero and thus the pressure will be less. As you correctly state in your comment, the pressure is zero in the falling jet before contact. Commented Apr 28 at 15:19
• @freecharly Thank you. Well said! Commented Apr 28 at 15:30
• @ChetMiller Your answer is essentially the same as mine sans the last sentence. None of what you said in your comment (pressure being zero in the column) is even in your answer. Commented Apr 28 at 16:04