# Hydrostatic force on the bottom of a container

• I

## Main Question or Discussion Point

Is the force from a liquid against the horizontal bottom of a container a function of depth and cross sectional area at the bottom only, i.e., computed by the formula Force=unit weight of water x depth x cross sectional area? If so the force could be greater than the weight of water and that defies logic.

Related Other Physics Topics News on Phys.org
Dale
Mentor
If so the force could be greater than the weight of water and that defies logic.
If the fluid is in equilibrium then $F=mg=\rho h A g$

Homework Helper
Gold Member
Is the force from a liquid against the horizontal bottom of a container a function of depth and cross sectional area at the bottom only, i.e., computed by the formula Force=unit weight of water x depth x cross sectional area? If so the force could be greater than the weight of water and that defies logic.
An interesting question. What you are basically asking is if we take a funnel shaped container that is wider at the bottom than at the top, can we have more force on the bottom face than the total weight of water? And the answer is yes. The total forces on the water must be zero though. The bottom pushes up with more force than the weight (downward gravitational force), but the sides of the funnel container press partly downward on the water as the water presses (from water pressure) against this surface. Thereby the downward forces from the slanted walls plus gravity will equal (in magnitude) the upward force from the bottom face on the water so that the water is in equilibrium with zero net forces. The force that the bottom face presses upward is the same for a funnel shape as it is for a cylinder shape of the same height that is filled with water. The pressure at the bottom (force per unit area) simply depends on the depth of the water, plus any atmospheric pressure...editing... a follow-on: I believe a hydraulic brake works by a similar arrangement. You apply force (and pressure) at the narrow funnel opening. The pressure balances everywhere, but the bottom face has much larger area, thereby more total force on the brake pad than the force you apply with your foot. Instead of a funnel shape, you could have a somewhat narrow cylinder containing the water, with a wider bottom section, and it requires very little (weight of the) water in the cylinder to cause considerable force on the bottom face.

Last edited:
An interesting question. What you are basically asking is if we take a funnel shaped container that is wider at the bottom than at the top, can we have more force on the bottom face than the total weight of water? And the answer is yes. The total forces on the water must be zero though. The bottom pushes up with more force than the weight (downward gravitational force), but the sides of the funnel container press partly downward on the water as the water presses (from water pressure) against this surface. Thereby the downward forces from the slanted walls plus gravity will equal (in magnitude) the upward force from the bottom face on the water so that the water is in equilibrium with zero net forces. The force that the bottom face presses upward is the same for a funnel shape as it is for a cylinder shape of the same height that is filled with water. The pressure at the bottom (force per unit area) simply depends on the depth of the water, plus any atmospheric pressure...editing... a follow-on: I believe a hydraulic brake works by a similar arrangement. You apply force (and pressure) at the narrow funnel opening. The pressure balances everywhere, but the bottom face has much larger area, thereby more total force on the brake pad than the force you apply with your foot. Instead of a funnel shape, you could have a somewhat narrow cylinder containing the water, with a wider bottom section, and it requires very little (weight of the) water in the cylinder to cause considerable force on the bottom face.
Thanks very much Charles.
An interesting question. What you are basically asking is if we take a funnel shaped container that is wider at the bottom than at the top, can we have more force on the bottom face than the total weight of water? And the answer is yes. The total forces on the water must be zero though. The bottom pushes up with more force than the weight (downward gravitational force), but the sides of the funnel container press partly downward on the water as the water presses (from water pressure) against this surface. Thereby the downward forces from the slanted walls plus gravity will equal (in magnitude) the upward force from the bottom face on the water so that the water is in equilibrium with zero net forces. The force that the bottom face presses upward is the same for a funnel shape as it is for a cylinder shape of the same height that is filled with water. The pressure at the bottom (force per unit area) simply depends on the depth of the water, plus any atmospheric pressure...editing... a follow-on: I believe a hydraulic brake works by a similar arrangement. You apply force (and pressure) at the narrow funnel opening. The pressure balances everywhere, but the bottom face has much larger area, thereby more total force on the brake pad than the force you apply with your foot. Instead of a funnel shape, you could have a somewhat narrow cylinder containing the water, with a wider bottom section, and it requires very little (weight of the) water in the cylinder to cause considerable force on the bottom face.
.

• billy_joule
• russ_watters
Dale
Mentor
a funnel shaped container
Oh, I misunderstood the question. I thought he was talking about a cylinder.

• An interesting question. What you are basically asking is if we take a funnel shaped container that is wider at the bottom than at the top, can we have more force on the bottom face than the total weight of water? And the answer is yes. The total forces on the water must be zero though. The bottom pushes up with more force than the weight (downward gravitational force), but the sides of the funnel container press partly downward on the water as the water presses (from water pressure) against this surface. Thereby the downward forces from the slanted walls plus gravity will equal (in magnitude) the upward force from the bottom face on the water so that the water is in equilibrium with zero net forces. The force that the bottom face presses upward is the same for a funnel shape as it is for a cylinder shape of the same height that is filled with water. The pressure at the bottom (force per unit area) simply depends on the depth of the water, plus any atmospheric pressure...editing... a follow-on: I believe a hydraulic brake works by a similar arrangement. You apply force (and pressure) at the narrow funnel opening. The pressure balances everywhere, but the bottom face has much larger area, thereby more total force on the brake pad than the force you apply with your foot. Instead of a funnel shape, you could have a somewhat narrow cylinder containing the water, with a wider bottom section, and it requires very little (weight of the) water in the cylinder to cause considerable force on the bottom face.
There's a hole in your logic (although it is textbook). The hole is the explanation that the liquid exerts a force against the that exceeds its weight because the funnel-shaped sides of the container press partly downward on the water as the water presses (from water pressure) against this surface, thereby adding the "additional'" force on the bottom that results in the liquid exerting a force exceeding its weigh. This explanation is flawed because it states the sides of the container exert an additional force, which is is impossible since the force exerted by the walls is a reaction, thus a passive force, and thus equal to the force exerted by the liquid against the walls, and thus being "equal" cannot produce "additional" force. As for the hydraulic brake explanation, although textbook, it is flawed because there is no vehicle to allow the force exerted by the brake pad to exceed that applied. The only "vehicles" which would permit this for a liquid a constant temperature would be an increase in density or expansion (as a gas), neither of which is possible for a liquid.

thank you-that is very interesting
Actually Pascal's Barrel is an anecdote-based, qualitative rather than quantitative, experiment/demonstration, as are all demonstrations "confirming" the principle that the force exerted by a liquid against the walls of the container is a function of the hydrostatic pressure rather than being based on mass.

Chestermiller
Mentor
There's a hole in your logic (although it is textbook). The hole is the explanation that the liquid exerts a force against the that exceeds its weight because the funnel-shaped sides of the container press partly downward on the water as the water presses (from water pressure) against this surface, thereby adding the "additional'" force on the bottom that results in the liquid exerting a force exceeding its weigh. This explanation is flawed because it states the sides of the container exert an additional force, which is is impossible since the force exerted by the walls is a reaction, thus a passive force, and thus equal to the force exerted by the liquid against the walls, and thus being "equal" cannot produce "additional" force.
Actually the flaw in logic is yours and not @Charles Link's. When you have an action-reaction pair between two bodies A and B, if you are doing a force balance on A, you only include the force exerted by B on A (not the force that A exerts on B), and, if you are doing a force balance on B, you only include the force exerted by A on B (not the force that B exerts on A). Didn't they teach you that in your course. If they diden't, shame on them.

• The total force on the bottom for any liquid is (density X gravity X height X area of the bottom). The total force on the bottom can be much greater than the weight of the liquid. This may be counter intuitive but that's the way it works. It's all hydraulics. There are good articles about how things like hydraulic presses and jacks work that should clear up any confusion.

Chestermiller
Mentor
The total force on the bottom for any liquid is (density X gravity X height X area of the bottom). The total force on the bottom can be much greater than the weight of the liquid. This may be counter intuitive but that's the way it works. It's all hydraulics. There are good articles about how things like hydraulic presses and jacks work that should clear up any confusion.
@Charles Link, in post #3, provided a perfect explanation of all this, without reference to hydraulic presses and jacks.

• Actually the flaw in logic is yours and not @Charles Link's. When you have an action-reaction pair between two bodies A and B, if you are doing a force balance on A, you only include the force exerted by B on A (not the force that A exerts on B), and, if you are doing a force balance on B, you only include the force exerted by A on B (not the force that B exerts on A). Didn't they teach you that in your course. If they diden't, shame on them.
It appears my comment went over your head. Look a little closer. Try evaluating the logic of my statement.

Homework Helper
Gold Member
It appears my comment went over your head. Look a little closer. Try evaluating the logic of my statement.
It appears you don't seem to understand our inputs on this. The thread was a year old, and it took an effort to read through the comments that we made a year ago, (Edit: Upon seeing @russ_watters post, I looked more closely at the date=yes, it is two years ago), but upon re-reading them, I think they give a very satisfactory answer to the problem.

Last edited:
• russ_watters
Chestermiller
Mentor
It appears my comment went over your head. Look a little closer. Try evaluating the logic of my statement.
There are three forces acting on the water in the container:
1. The downward gravitational force
2. The upward pressure force exerted by the bottom of the container
3. The the downward pressure force exerted by the slanted side(s) of the container.

Do you disagree with any of this? If so, which part.

The total force on the bottom for any liquid is (density X gravity X height X area of the bottom). The total force on the bottom can be much greater than the weight of the liquid. This may be counter intuitive but that's the way it works. It's all hydraulics. There are good articles about how things like hydraulic presses and jacks work that should clear up any confusion.
I am familiar with all you state - I initially acknowledged that which I am debating as textbook. As for the hydraulic jack, its principle is based on the principle that the applied force is transferred as a pressure, not force, a principle that has yet to be verified quantitatively.

There are three forces acting on the water in the container:
1. The downward gravitational force
2. The upward pressure force exerted by the bottom of the container
3. The the downward pressure force exerted by the slanted side(s) of the container.

Do you disagree with any of this? If so, which part.
I would add that the forces said to be exerted by the container differ from the forces exerted by the water in that they are passive forces. There is also the lateral force exerted by the water resisted by the container.

jbriggs444
Homework Helper
2019 Award
I would add that the forces said to be exerted by the container differ from the forces exerted by the water in that they are passive forces. There is also the lateral force exerted by the water resisted by the container.
There is no such thing as a passive force. There is no such thing as an active force. There is no such thing as an action force. There is no such thing as a reaction force. There are only forces. They come in pairs.

Chestermiller
Mentor
I would add that the forces said to be exerted by the container differ from the forces exerted by the water in that they are passive forces.
I was asking only about the forces acting on the water. Was that not clear?
There is also the lateral force exerted by the water resisted by the container.
Like I said, I am asking only about the forces acting on the water. There are indeed lateral forces acting on the water by the walls of the container, but, if the system is in equilibrium, these cancel out. And, in our discussion here, I wish to focus only on the components of the forces in the vertical direction (since we are interested in the vertical force from the bottom of the container). Is this OK with you?

russ_watters
Mentor
It appears my comment went over your head. Look a little closer. Try evaluating the logic of my statement.
No need to get testy, but there is a need to pay better attention to what you are being told.

Or: it's been two years. At this point, why don't you just build yourself a test rig for this? Here's how:

Take a clear, flexible, rubber or plastic tube and attach it to a yardstick. Mount the cap of a 3 liter soda bottle to a board, with legs. Drill a hole in the cap and attach the tube. Cut the bottom off the soda bottle and screw the rest into the cap. fill completely with water. Note the height of the water in the tube. squeeze the bottle from the "bottom", making a flat wedge shape. Note that the water height doesn't change.

It should take you less than an hour! (if you have the parts)

To jbriggs444: "There is no such thing as a passive force. There is no such thing as an active force. There is no such thing as an action force. There is no such thing as a reaction force. There are only forces. They come in pairs."

I will explain what I mean: When a force is applied to a stationary object as opposed to the application of two opposing forces, the force being applied is active. The resisting force is passive. And while they are equal in one sense (magnitude) the are not in another since one is being applied and the other a reaction.

I was asking only about the forces acting on the water. Was that not clear?

Like I said, I am asking only about the forces acting on the water. There are indeed lateral forces acting on the water by the walls of the container, but, if the system is in equilibrium, these cancel out. And, in our discussion here, I wish to focus only on the components of the forces in the vertical direction (since we are interested in the vertical force from the bottom of the container). Is this OK with you?
As for you being clear but not completely correct. I am fine with focusing on vertical direction.

jbriggs444
Homework Helper
2019 Award
TWhen a force is applied to a stationary object as opposed to the application of two opposing forces, the force being applied is active. The resisting force is passive. And while they are equal in one sense (magnitude) the are not in another since one is being applied and the other a reaction.
This distinction is physically unjustified. It is a distinction without a difference.

No need to get testy, but there is a need to pay better attention to what you are being told.

Or: it's been two years. At this point, why don't you just build yourself a test rig for this? Here's how:

Take a clear, flexible, rubber or plastic tube and attach it to a yardstick. Mount the cap of a 3 liter soda bottle to a board, with legs. Drill a hole in the cap and attach the tube. Cut the bottom off the soda bottle and screw the rest into the cap. fill completely with water. Note the height of the water in the tube. squeeze the bottle from the "bottom", making a flat wedge shape. Note that the water height doesn't change.

It should take you less than an hour! (if you have the parts)
My attention is on correcting the misinformation presented. I have performed experiments sufficient to prove my points.

This should be interesting. I am always open to improving what I know about a subject. I don't expect to see any experiment here that greatly changes what I know about hydrostatic forces, but maybe the OP can prove that wrong. $\\$ A couple of things make the hydrostatic forces relatively straightforward: 1) Nothing is moving 2) The fluid (water) is basically incompressible 3) The forces are pressure forces that are always perpendicular to the surfaces. In order to have equilibrium, the force per unit volume from the minus of the pressure gradient, i.e. $\vec{f}_V=-\nabla p$, must be equal and opposite the gravitational force per unit volume $\vec{f}_g=-\rho g \hat{z}$, where $\rho$ is the density. This along with Archimedes principle is usually sufficient to solve most of these problems.