Buoyant force or water pressure?

[jibes138]

Sum of the forces

So Bill my original response on the stability thread that described the sum of the forces equal to zero in the vertical direction as defining equilibrium was correct and therefore we did not need another 150 posts or so about it? I think people are failing to realize that the water pressure on the hull of a floating boat varies with the depth of every point of the submerged hull and that it is a calculus problem to sum up all those infinitely small points to get the sum of the forces. Another way to think about it is exactly the same forces would be applied to the water that would normally be where the boat is.

 

[MoonSailer]

http://www.howstuffworks.com/question254.htm

Maybe this will help.

 

[Ross]

My sister said when she started physics 101 in college on the first day they encountered a prof standing in front of a chalk board covered with symbols and equations. When the class was seated and the usual proceedures gotten out of the way the prof asked if anyone knew the meaning of the work on the chalk board. When no one answered he said simply, "that is the way the ball bounces."
Then he said now let's get started we have a lot to cover.

 

[Brian M H23]

This thread is interesting..yet kind of silly.
Take a hydraulic jack and try to change a tire.
When the piston hits the frame of the car, the weight of the car exterts a downward force. The next pump raises the pressure in the cylinder, and that pressure exerts a force that lifts the car.

A claim that "force doesn't lift your boat, pressure does" is absurd and silly, because forces are naturally applied through an area (unless you know how to have an infitely small contact), and pressure = force/area

 

[Bill Roosa]

Forces, pressures, densitys, ......

Hi All
This is one of my pet peeves if you couldn't tell. I hear folks say that its the density that makes a boat float or the displaced water makes the boat float and I know that all those things are true but irrelevant. You can't take the fact the the a boat displaces a volume of water who's weight is equal to the boats weight and then use that to determine anything. If the boat was a barge the I could use the formula for the volume of a prism to calculate how deep the boat will ride in the water but for the most part the shape of a boats hull is not a mathmaticaly definable surface. It can be broken down into a set of surfaces that are but there is no one equation that we can work with. To work with all those different surfaces requires a pretty good grasp of the basics. That is what I've been trying to accomplish.

I apologize if I've upset anyone or bored you folks. Such was not my intent.

 

[Ross]

Bill , Let me see if I have this right water floats a boat. Any floating object displaces a quantity of water equal to the weight of the floating object. That displaced water is spread out over the entire surface of the aforesaid body of water and therefore becomes negligible in the process of floating the object. The rest of the water is what does the floating part. Or in the case of my bowls the displaced water returns to the ocean and from there to the sky where eventually it falls as rain and gets into the ground and is pumped out of the well and into the pipes that lead to my faucet so that I can refill the bowl after I lift the inner one from the outer one.

 

[jibes138]

Finite Element Analysis

While an equation for the hull of a boat is indeterminate, unless the boat is a barge or similar simple structure, we should be able to use a finite element analysis program to determine the net force on the hull based on a solid model of the hull shape and a definition of the water pressure versus depth. Does anyone have access to a finitie element program and can show a proof that the water pressure applied over the hull sums up to a total force that equals the weight of the boat?

Someone said we are arguing over force versus pressure. No, we are arguing that it is water pressure that is applying a force in PSI normal to the hull at any point and the sum of the vertical components of those force vectors yields a total upward force that balances the force in the downward direction due to gravity, or in short "the sum of the forces in the vertical direction equal zero" this is a mandatory condition for equilibrium or in other words the boat is floating, not sinking.

 

[Jim Kolstoe]

Bill, try your bowl/hemisphere experiment backwards. Put the smaller bowl into the larger bowl, then trickle water in between them until the smaller bowl floats (as opposed to being wetted - which would be the result of applying an earlier suggestion that a ship would be said to float in a few tablespoons of water if the basin containing the water were a tight enough fit). Then compare the weight of the water and the small bowl.
Jim Kolstoe, h23 Kara's Boo

 

[DanW9494]

http://www.howstuffworks.com/question254.htm

Maybe this will help.

Actually this is wrong because the water pressure one foot below the surface is not 0.44 PSI but 0.587 PSI, because they forgot to add the air pressure on the surface of the water.

You start at 0.147 PSI at the surface and then add 0.44 PSI at one foot depth for a total of .587 PSI, and 0.44 PSI for every additional foot of water depth.

Water pressure is due to the density and depth of the measured water in the presence of earth's gravity.

Consider what it would be like for a scuba diver if water pressure was the buoyant force. At 70 feet down the water pressure is about 31 PSI. Why would you need to empty air from your BC to go down or add air to your BC to go up if you had that much water pressure trying to eject you from the water?

Archimede's principle states that the buoyant force on an object is the weight of the volume of water that it displaces. He didn't say that the buoyant force on an object increases at 70 feet down where the water pressure is 31 PSI. And the weight of a cubic foot of seawater at 70 feet down is still 64 lbs.

 

[Roger Long]

Braaaap!

(The BS alarm goes off and Roger comes running back into the room from where he's been dealing with having a root canal and other distracting stuff.)

Actually this is wrong because the water pressure one foot below the surface is not 0.44 PSI but 0.587 PSI, because they forgot to add the air pressure on the surface of the water.

Sorry, but this is completely wrong. Trust me, I've been using water head pressures to design marine structures for 36 years. If the Society of Naval Architects, the naval architecture schools, and everyone I've ever worked for were off by .147 psi on this most basic factor, I'm sure someone would have noticed before this.

It is true that the weight of the atmosphere has a pressure effect on the surface of the water. In fact, if you were to remove the atmosphere and its pressure, the water would start boiling away because boiling temperature drops with pressure. Propeller cavitation is actually the pressure on the front of the prop blades getting so low that the water boils into steam at the temperature of the water. When the bubbles collapse, they do so with enough force to cause wear on the blades.

In practice though, you can just ignore air pressure. The reason is the root misconception that has given me several ROFL moments reading down this thread. This is not all stuff stacked up like a pile of books. It is happening in a fluid and gas which are free to move sideways and upwards as well as responding to the force of gravity.

Sea water weighs 64 pounds per cubic foot. Suspend a foot of the stuff in air on a scale (adjusted for the weight of the container) and you get 64 pounds. What happened to the air pressure? It is pushing up on the bottom and sides of the container virtually as hard as it is pushing down. Since air pressure decreases with altitude, there is slightly more pressure towards the bottom of our cube of seawater and this actually creates a tiny buoyancy or lightening of the cube. The effect in the real world is so small however, that we simply ignore it (except for balloonists). Convention is to consider all weights as being in air. When you buy a pound of bananas at a farmer's market in New York, don't try telling the vendor that he needs to consider the air density.

Now lower the cube of seawater in to water. We’ll imagine that it is in a one pound container made of plastic that had a density of 64 pounds per cubic foot, exactly the same as the seawater. The scale would have read 65 pounds in the air exposed to the air pressure. In the water, it reads zero, because of buoyancy but it’s no longer exposed to the air pressure. What happened to the air pressure? Whatever effect it has on the water is distributed around the cube exactly as it was in air. Air pressure is already factored into the weight, just as it is in air.

Air also has no effect on a vessel’s draft because the net effect of it pushing on the decks (in the case of an airtight boat) or hull and its contribution to the water pressure pushing up is always zero. In 36 years of naval architecture, this is the first I’ve heard it mentioned and I doubt any discoveries are being made here.

(Apologies Dan, I’m not intending to pick on you but it was a good opening to jump back in here.)

 

[Hermit Scott]

Jibes138-if your boat is new enough, it was probably designed in a solid modeling program. You don't need FEA to tell you the volume displaced. Solidworks or what ever program you use will give you volumes for sections.
Of course you could just take the weight of the boat. That will tell you how much water weight is displaced. lol
Dan makes a good point that's hard to rebut.
You could alway put your boat in a pool and measure the level of the water before and after to determine how much water is displaced.
Pressure is nomal to the surface. But if you put a cone, point up in water, the normal force along the surface area is actually represented as a downward angle minus the area on the bottom, but the cone will still float.

 

[Hermit Scott]

Roger "Since air pressure decreases with altitude, there is slightly more pressure towards the bottom of our cube of seawater and this actually creates a tiny buoyancy or lightening of the cube"
I don't think that's true. The cubic foot of water is boyant in the air becasuse it displaces that amount of air. So in a vacuum the cubic foot of water would weigh more by exactly weight of the amount of air it displaces.
A sinking object in water weighs less exactly by the amount of water that it displaces.
To prove that point with a thought experiment, change the shape of the water to be very thin and parallel to the surface of the earth, then weigh it. It will weigh the same. no matter what the shape.

 

[Roger Long]

I don't think that's true.

Then explain the hot air or helium balloons.

In the case of the wide thin sheet of water (or anything else subject to buoyancy forces in either air or water) the pressure difference due to altitude or depth is very small but the surface area over which it acts is very large. The net sum of force is the same.

Even a cubic foot of lead has buoyancy in water. The weight in air is 708 pounds in air and 643 pounds in sea water. A cubic foot of anything in seawater will have 64 pounds of buoyancy. Whether it goes rises or sinks depends on the weight. If the top is just level with the surface, there will be zero pressure on the top. Pressure on the sides will vary but not push up or down. Pressure on the bottom will be 64 pounds. Now change the shape so that it is a rectangle of the same volume but two feet from top to bottom. Pressure per square foot on the bottom will be twice as much because the depth has increased but the area of the bottom will only be half as much so buoyancy will still be 64 pounds. You can keep doing this until the object is as deep as the ocean and just a tiny rectangular rod. The enormous force on the bottom will still be creating 64 pounds of upwards force.

I'll be back after I go spend some time working on my boat.

 

[Bill Roosa]

Ross Graduating to the bathtub

No Ross, it is still water pressure acting upward on the inner bowl that buoys it up.
Consider the bathtub, when I float something in it the water it displaces causes the water level in the tub to rise somewhat. The bowl now floats at that higher level. The energy liberated in the bowl being lowered (weight of the bowl * distance dropped till it floats starting at water contact is transferred to raise the rest of the water. Just like the energy on your master cylinder in your car brakes is transferred to the slave cylinders. That energy is conserved in the higher water level till you "haul" the bowl for the winter. It is then returned to the bowl.

Also, if you repeat the two bowls experiment after the displaced water runs down the drain then you will see that even with the smaller amount of water the bowl still floats. In this case there is no displaced water and clearly the water that went down the drain would have no knowledge you where "going around behind it's back and floating things"

 

[Maine Sail]

holy cow..

Wow you guys have set a new records for the sheer number of posts saying essentially the same thing. Winter is certainly upon us!

It's still the Archimedes Principle or to dumb it down displacement..

 

[Bill Roosa]

How water works

See the attachment.
This is how I learned it in college. You can make the same argument with any shape but the math is harder.

I note the following:

The pressure in the water at any depth is equal to the density (of the fluid) times the depth
Pressure(lb/ft^2)=density(lb/ft^3) * depth (ft)

You have to accept that or we can't go further.

For the scuba divers I believe we all memorized that at 34 ft the pressure is 1 atmosphere (2 really since the atmosphere adds one too but the gage pressure at the surface of the earth is set to 0 by decree) or 14.7 psi = 14.7* 12^2=2116.8 lb/ft^2. I prefer to work in units of feet for reasons that will become clear later.

So what does my equation tell us?

Density = 62.4 lb/ft^3 and depth 34 ft
P=62.4*34=2121.6 lb/ft^2 = 2121.6lb/ft^2 or 14.7 psi

I look at is like this:
At any point in the water the weight of a 1'x1' column of water over the point would be the volume (displacement rears its head again) of water times the density (and close on its heels is density) or

W=density * volume

The volume is 1x1xdepth so we can write

W=density*depth assuming we keep our units right.

Now since water cannot sustain any shear (well not much but that is hydrodynamics and this is hydrostatics) you will see that the sides and top of the column can't hold up the column and so all the weight must be supported by the bottom. That bottom is a 1'x1' square area and since we know

Pressure=Force * area

we can say

pressure at a point in water is

P=density * depth * area where I replaced force by weight.

The area is mathematically 1x1=1 so it is not effected by manipulating the other parts of the equation and can be ignored. In calculus we shrink the area to 0 and get pressure at a point but the math is the same.

 

[jibes138]

I completely agree with Bill....nice job I think we can end it here

Another way to think about it is put the boat in the water with a really thin film of saran wrap that will stay in exactly the same shape as the boat when the boat is removed, the saran wrap has zero mass ( just pretend ok). When we extract the boat at the same time we fill the empty hole in the water with water so the saran wrap is still there looking like a hull filled with water. I submit that exactly the same forces are acting on the surface area of the saran wrap as were acting on the hull of the boat when it was there. This time the water pressure is holding up the water inside the saran wrap that weighs the saem as the hull. Now suddenly the saran wrap becomes real saran wrap and we pull it out of the water. The water pressure is still there acting on that exact same shape hoplding up exactly the same amount of weight. So it was water pressure all along that was floating the boat.

 

[DanW9494]

Re: Braaaap!

Roger, you can calculate draft without including atmospheric pressure because pressure is not the buoyant force, the specific weight of water is.

You can and have ignored atmospheric pressure without problems because you design your boats to stay on the surface very well.

The atmospheric pressure at the surface is not zero- that would be outer space. The pressure 1 foot down is not .44. If it was, then you are saying that either there is no atmospheric pressure (outer space) or you have magic sea water that weighs less than 64 lbs. The actual pressure is the addition of the neglible atmospheric and the substantial water pressure.

You are right that the net pressure on the boat is .44 at one foot down, because your boat is floating and the atmospheric pressure cancels out .147 of the water pressure.

Okay, you say this is nitpicking and not practical. I only bring it up to point out that water pressure is not the buoyant force exactly, but it looks very close because the atmosheric pressure is so low.

Buoyant force = water pressure force minus air pressure force.

That way you can do calculations where you define the pressure at the water's surface as zero, which it obviously cannot be under our atmosphere.

 

[Bill Roosa]

Important stuff to know

Gage pressure at the surface of the earth is 0 psi. It is the air pressure that you would read on your propane tank. When my tank is empty it does not read 14.7 it reads 0 because it is designed to read 14.7 psi low all the time.
Gage pressure is the atmospheric pressure that you use when designing boats.

One foot down in the water the pressure is indeed 0.44 pounds per square INCH!!

a column of water 1 foot x 1 foot would weight 62.4 but when you have a column of water 1 inch x 1 inch it is 144 (12x12inches) times smaller in magnitude.

lets see the math

62.4 (lb/ft^2) = 62.4 (lb/ft^2) / 144 (inches^2/1 ft^2) = 0.43333 PSI

62.4 lb/ft^2 is just a unit conversion to 0.433 PSI

As for air pressure:
Imagine a block of wood suitably ballasted so it floats with its upper surface level with the water. Now put any amount of pressure you want over the water (except low pressures that would make the water boil) and you can see that the wood upper surface does indeed get "forced into the water" just like the water surface right next to it does. On a boat it is more complex because of all the surfaces but (just like water on the hull but downward) it presses down on what is effectively the horizontal surface that would run through the boat at its waterline.

I'm not disagreeing with you danw9494. I'm just saying that you typically ignore the contribution of air pressure because it acts downward on everything equally and so has no effect.

 

[Bill Roosa]

jibes138

I hope you are right but alas I fear we still have to convince some.

This stuff is not what I'd call "blindingly obvious to the casual observer" in their defense. That and all the stuff on the definition of "buoyant force" don't help to understand it either.