 # How to figure force of water

lets supposes it didn’t know or have access to the formulas used for moving water. What confuses me is how do you determine the weight of rushing water. If I knew the speed and I measured the force on 1 square inch I could come up with a weight figure but applying weight to moving water just doesn’t seem right, its not like a solid moving object. What terms are used to describe the force of moving water??

I am going to slightly change my question, how long is the chain of molecules that affects the front molecule with force? Is it shaped like a cone???

Get a bucket. Measure the weight of the water that comes out of your system in one second. (or, measure the water flow in volume per time, say gallons per minute, and use the volume to weight conversion of your choice). That will tell you the literal weight of the moving water.

Pressure is force or weight per unit area. Which, thinking about it, is more of a static quantity. If you have a compressed air tank and open a hole, the pressure will be going down as long as air is flowing. If you knew the rate of pressure change, the hole area, and time, that might be enough to let you calculate how much water moved, and hence the weight. delta P, delta T, A - hm, I guess you would probably also need the total initial volume. Maybe?

Would this be the same for an open system like a creek or flood channel?

Are you trying to find the discharge (volume of water conveyed per unit time) of an open channel? That’s complicated a bit by the fact that it’s non-uniform, but here’s a pretty good (if not brief) overview of standard methods (warning: pdf).

Most often you’ll break the channel into subsections and then measure the water velocity in each subsection. The subsection discharge then is the (section width) times (section depth) times (velocity). For the total discharge, you simply add all the subsection discharges together. Figure 1 of that pdf (p. 19) gives a pretty good graphical representation of that.

Or do you want to estimate the force of moving water exerted on something in the channel?

Yes per square inch force it would exert on something.

How the water develops its force is what I am trying to figure out, if the channel were 10 miles long or 1 mile long and the water was the same depth and the same speed it would have the same force behind it, so I guess i am trying to figure out is what the force field behind the object being pushed looks like

The way it’s usually described is as a drag force; it applies to the force experienced by any object moving relative to the flow of a fluid (liquid or gas).

In general, it’s not just a force per square inch, but instead depends on the three-dimensional shape of the object. For a given shape, the force is proportional to the density of the fluid, the square of the velocity of the fluid, and the cross-sectional area. But the proportionality factor (called the drag coefficient) differs from shape to shape. For example, a sphere, a hemisphere, a cylinder, a cone pointing “upstream”, and a cone pointing “downstream” will all experience different drag forces in the same water flow, even if they all have the same circular cross-section.

I thought of this XKCD strip.

Mike not how the object experiences the force but how the water itself collects the force

Spherical. But then you have to consider how they add together. For an object much bigger than a molecule, the assumption its spherical goes away.

Anyway, there’s no point considering individual molecules and then summing them.
One molecule at one instant might be pushing 100 times more than the average molecule.
Here’s how its done.
youtube video of a physics tutor explaining it.

Consider a metre of water inside the pipe. It has velocity v m/s. So it takes 1/v seconds to turn at the thing its hitting.

Assume the turn is 90 degree. spreading out to form a disc… a plane.
So the water flow in each direction is constant, and the net momentum is zero.

But the metre of water had mass * velocity momentum.

Now momentum = m v, and force = mA , and so Force = change in momentum / time taken to lose that momentum.

So you’ve got the original momentum of that metre of water, the end momentum is zero, and you have the time taken to lose it… So there’s the force.
Now if the water has a net momentum, you can easily fix the maths.
Use any test length you want, it doesn’t have to be a metre. Use a variable for it , it will cancel.
If energy is “real”, there is no such real thing as force (apart from the 5 fields… eg electric and magnetic fields, but we aren’t going there.) or momentum, or pressure. So your desire to sum forces is just as invalid as doing calculations on momentum, or on pressure … I hope not invalid.

If I’m interpreting the question correctly, water is supported against its own weight by the pressure of the water below it. And this is true regardless of whether the water is still or flowing. The only case in which the pressure of the water at the bottom of the channel would be anything other than density times g times the depth of the water is when the water has some nonzero vertical component of acceleration. Now, water flowing in a channel probably will have some nonzero vertical component of acceleration, but in situations like a river, it’s going to be very, very small.

Not enough coffee in my belly yet for mathing …

Could we put a clear tube into the water at a 45º angle to the horizon … one end in the water upstream and the other above the water … for any velocity of the water, a certain amount of water will push up the tube above water line … however much water is pushed up will be acted upon by gravity equal to the force of water entering the tube in the stream …

Help me mathmagicians … a one square meter opening normal to the flow that pushes one kilogram of water up the tube is 9.8 Newtons of force? …

ETA: We’re neglecting friction, but we have something that fits in our pocket …

This question however has stumped me in the sense that I can’t figure out what you are asking.

Think of a long gentle slide (like a kids slide but with a gentler slope). Imagine a bunch of marbles falling continuously (like seeping down from a mountain) on the high end and rolling down to the low end of the slide.

The force of gravity gets the marble to start rolling. If you stop a marble mid-way, the force you feel is the impact of something moving being stopped.

• The above is a simplification. Water flows in sheets in slow/laminar flow (analytical expressions work in laminar flow only). At fast speeds,(high Reynolds number) water flow is turbulent and analytical methods work poorly. (

I think the marbles may be a fair example. Once the object has been placed to block the water and becomes static then I am wondering how the marbles behind it would react? How far back would the stream of marbles be affected? Would it be different if I were to dam up the channel completely and just let the water start rising?

If I’m interpreting the question correctly the flow pattern would be similar to how air flow around objects is calculated; there should be a ton of material describing this wrt aviation. This picture of water flow around a pier shows the velocity (I think) which would be proportional to force around a bridge pier. In terms of how far upstream the water flow is changed, it appears very little.

And yes damming the river would profoundly affect the flow pattern for a long distance upstream since you be creating an impoundment where the water would slow down.

The force felt when water hits a surface is due to the acceleration of water at the surface. In order to change velocity there has to be a force on the water.

This is what I was looking for, I just couldn’t figure out how to pose the question. Kind of what i had imagined it would be.

If the channel is enclosed (a pipe), the flow is affected ‘all the way back’: water is incompressible for ordinary calculations.

If the channel is open and you have super-critical flow, you get a hydraulic jump at an obstruction. The dynamic pressure lifts some of the water, and the energy is dissipated in turbulence. You can calculate the length of the turbulent region, but it depends on how fast the water is flowing, and where the water goes: there is no simple number you can put to it, because fluid flow equations are not simple. Even when you have a simple situation with clearly defined edges.

If the channel is open and you have only turbulent flow coming to an obstruction, you get, I think (?) a much simpler situation where the flow away from the obstruction depends on the angle of the surface of the water (?). and, (?) I think (?) the water behind the obstruction will eventually just flatten out at the correct level.