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## Physics library

### Course: Physics library>Unit 9

Lesson 2: Buoyant Force and Archimedes' Principle

# What is buoyant force?

Why the heck do things float?

## What does buoyant force mean?

Have you ever dropped your swimming goggles in the deepest part of the pool and tried to swim down to get them? It can be frustrating because the water tries to push you back up to the surface as you're swimming downward. The name of this upward force exerted on objects submerged in fluids is the buoyant force.
So why do fluids exert an upward buoyant force on submerged objects? It has to do with differences in pressure between the bottom of the submerged object and the top. Say someone dropped a can of beans in a pool of water.
Because pressure left parenthesis, P, start subscript, g, a, u, g, e, end subscript, equals, rho, g, h, right parenthesis increases as you go deeper in a fluid, the force from pressure exerted downward on the top of the can of beans will be less than the force from pressure exerted upward on the bottom of the can.
Essentially it's that simple. The reason there's a buoyant force is because of the rather unavoidable fact that the bottom (i.e. more submerged part) of an object is always deeper in a fluid than the top of the object. This means the upward force from water has to be greater than the downward force from water.
Knowing conceptually why there should be a buoyant force is good, but we should also be able to figure out how to determine the exact size of the buoyant force as well.
We can start with the fact that the water on the top of the can is pushing down F, start subscript, d, o, w, n, end subscript, and the water on the bottom of the can is pushing up F, start subscript, u, p, end subscript. We can find the total upward force on the can exerted by water pressure (which we call the buoyant force F, start subscript, b, u, o, y, a, n, t, end subscript) by simply taking the difference between the magnitudes of the upward force F, start subscript, u, p, end subscript and downward force F, start subscript, d, o, w, n, end subscript.
F, start subscript, b, u, o, y, a, n, t, end subscript, equals, F, start subscript, u, p, end subscript, minus, F, start subscript, d, o, w, n, end subscript
We can relate these forces to the pressure by using the definition of pressure P, equals, start fraction, F, divided by, A, end fraction which can be solved for force to get F, equals, P, A . So the force exerted upward on the bottom of the can will be F, start subscript, u, p, end subscript, equals, P, start subscript, b, o, t, t, o, m, end subscript, A and the force exerted downward on the top of the can will be F, start subscript, d, o, w, n, end subscript, equals, P, start subscript, t, o, p, end subscript, A. Substituting these expressions in for each F respectively in the previous equation we get,
F, start subscript, b, u, o, y, a, n, t, end subscript, equals, P, start subscript, b, o, t, t, o, m, end subscript, A, minus, P, start subscript, t, o, p, end subscript, A
We can use the formula for hydrostatic gauge pressure P, start subscript, g, a, u, g, e, end subscript, equals, rho, g, h to find expressions for the upward and downward directed pressures. The force from pressure directed upward on the bottom of the can is P, start subscript, b, o, t, t, o, m, end subscript, equals, rho, g, h, start subscript, b, o, t, t, o, m, end subscript and the force from pressure directed downward on the top of the can is P, start subscript, t, o, p, end subscript, equals, rho, g, h, start subscript, t, o, p, end subscript . We can substitute these into the previous equation for each pressure respectively to get,
F, start subscript, b, u, o, y, a, n, t, end subscript, equals, left parenthesis, rho, g, h, start subscript, b, o, t, t, o, m, end subscript, right parenthesis, A, minus, left parenthesis, rho, g, h, start subscript, t, o, p, end subscript, right parenthesis, A
Notice that each term in this equation contains the expression rho, g, A. So we can simplify this formula by pulling out a common factor of rho, g, A to get,
F, start subscript, b, u, o, y, a, n, t, end subscript, equals, rho, g, A, left parenthesis, h, start subscript, b, o, t, t, o, m, end subscript, minus, h, start subscript, t, o, p, end subscript, right parenthesis
Now this term h, start subscript, b, o, t, t, o, m, end subscript, minus, h, start subscript, t, o, p, end subscript is important and something interesting is about to happen because of it. The difference between the depth of the bottom of the can h, start subscript, b, o, t, t, o, m, end subscript and the depth of the top of the can h, start subscript, t, o, p, end subscript is just equal to the height of the can. (see the diagram below)
So we can replace left parenthesis, h, start subscript, b, o, t, t, o, m, end subscript, minus, h, start subscript, t, o, p, end subscript, right parenthesis in the previous formula with the height of the can h, start subscript, c, a, n, end subscript to get,
F, start subscript, b, u, o, y, a, n, t, end subscript, equals, rho, g, A, h, start subscript, c, a, n, end subscript
Here's the interesting part. Since A, times, h is equal to the volume of a cylinder, we can replace the term A, h, start subscript, c, a, n, end subscript with a volume V . The first instinct might be to associate this volume with the volume of the can. But notice that this volume will also be equal to the volume of the water displaced by the can. By displaced water we mean the volume of water that was once in the volume of space now occupied by the can.
So we are definitely going to replace the term A, h with a volume V , but should we write this volume as volume of the can or volume of the displaced fluid? This is important because the two volumes could be different if the object is only partially submerged in the fluid. The short answer is that we need to use the volume of the fluid displaced V, start subscript, f, l, u, i, d, end subscript in the formula because the displaced fluid is the factor that determines the buoyant force.
F, start subscript, b, u, o, y, a, n, t, end subscript, equals, rho, g, V, start subscript, f, end subscript
That pretty much does it. This formula gives the buoyant force on a can of beans (or any other object) submerged wholly or partially in a fluid. Let's take stock of what we have now. Notice how the buoyant force only depends on the density of the fluid rho in which the object is submerged, the acceleration due to gravity g, and the volume of the displaced fluid V, start subscript, f, end subscript.
Surprisingly the buoyant force doesn't depend on the overall depth of the object submerged. In other words, as long as the can of beans is fully submerged, bringing it to a deeper and deeper depth will not change the buoyant force. This might seem strange since the pressure gets larger as you descend to deeper depths. But the key idea is that the pressures at the top and bottom of the can will both increase by the same amount and therefore cancel, leaving the total buoyant force the same.
Something might strike you as being wrong about all this. Some objects definitely sink, but we just proved that there is an upward force on every submerged object. How can an object sink if it has an upward force on it? Well, there is definitely an upward buoyant force on every submerged object, even those that sink. It's just that for sinking objects, their weight is greater than the buoyant force. If their weight was less than their buoyant force they would float. It turns out that it's possible to prove that if the density of a fully submerged object (regardless of its shape) is greater than the density of the fluid it's placed in, the object will sink.

## What is Archimedes' principle?

The way you will normally see the buoyant force formula written is with the g and the V rearranged like so,
F, start subscript, b, u, o, y, a, n, t, end subscript, equals, rho, V, start subscript, f, end subscript, g
When you rearrange the formula in this way it allows you to notice something amazing. The term rho, V, start subscript, f, end subscript is the density of the displaced fluid multiplied by the volume of the displaced fluid. Since the definition of density rho, equals, start fraction, m, divided by, V, end fraction can be rearranged into m, equals, rho, V, that means the term rho, V, start subscript, f, end subscript corresponds to the mass of the displaced fluid. So, if we wanted to, we could replace the term rho, V, start subscript, f, end subscript with m, start subscript, f, end subscript in the previous equation to get,
F, start subscript, b, u, o, y, a, n, t, end subscript, equals, m, start subscript, f, end subscript, g
But look at that! The mass of the displaced fluid times the magnitude of the acceleration due to gravity is just the weight of the displaced fluid. So remarkably, we can rewrite the formula for the buoyant force as,
F, start subscript, b, u, o, y, a, n, t, end subscript, equals, W, start subscript, f, end subscript
This equation, when stated in words, is called Archimedes' principle. Archimedes' principle is the statement that the buoyant force on an object is equal to the weight of the fluid displaced by the object. The simplicity and power of this idea is striking. If you want to know the buoyant force on an object, you only need to determine the weight of the fluid displaced by the object.
The fact that simple and beautiful (yet not obvious) ideas like this result from a logical progression of basic physics principles is part of why people find physics so useful, powerful, and interesting. And the fact that it was discovered by Archimedes of Syracuse over 2000 years ago, before Newton's laws, is impressive to say the least.

## What's confusing about the buoyant force and Archimedes' principle?

Sometimes people forget that the density rho in the formula for buoyant force F, start subscript, b, end subscript, equals, rho, V, start subscript, f, end subscript, g is referring to the density of the displaced fluid, not the density of the submerged object.
People often forget that the volume in the buoyancy formula refers to the volume of the displaced fluid (or submerged volume of the object), and not necessarily the entire volume of the object.
Sometimes people think the buoyant force increases as an object is brought to deeper and deeper depths in a fluid. But the buoyant force does not depend on depth. It only depends on volume of the displaced fluid V, start subscript, f, end subscript, density of the fluid rho, and the acceleration due to gravity g.
Many people, when asked to state Archimedes' principle, usually give a look of confused exasperation before launching into a wandering discussion about people jumping naked out of bathtubs. So, make sure you understand Archimedes' principle well enough to state it clearly: "Every object is buoyed upwards by a force equal to the weight of the fluid the object displaces."

## What do solved examples involving buoyant force look like?

### Example 1: (an easy one)

A 0, point, 650, start text, space, k, g, end text garden gnome went snorkeling a little too low and found himself at the bottom of a fresh water lake of depth 35, point, 0, start text, space, m, end text . The garden gnome is solid (with no holes) and takes up a total volume of 1, point, 44, times, 10, start superscript, minus, 3, end superscript, start text, space, m, end text, cubed . The density of fresh water in the lake is 1000, start fraction, start text, k, g, end text, divided by, start text, m, end text, cubed, end fraction .
What is the buoyant force on the gnome?
F, start subscript, b, end subscript, equals, rho, V, g, start text, left parenthesis, U, s, e, space, b, u, o, y, a, n, t, space, f, o, r, c, e, space, e, q, u, a, t, i, o, n, comma, space, w, h, i, c, h, space, i, s, space, j, u, s, t, space, A, r, c, h, i, m, e, d, e, s, apostrophe, space, p, r, i, n, c, i, p, l, e, space, i, n, space, m, a, t, h, space, f, o, r, m, right parenthesis, end text
F, start subscript, b, end subscript, equals, left parenthesis, 1000, start fraction, start text, k, g, end text, divided by, start text, m, end text, cubed, end fraction, right parenthesis, left parenthesis, 1, point, 44, times, 10, start superscript, minus, 3, end superscript, start text, space, m, end text, cubed, right parenthesis, left parenthesis, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, start text, left parenthesis, P, l, u, g, space, i, n, space, n, u, m, e, r, i, c, a, l, space, v, a, l, u, e, s, right parenthesis, end text
F, start subscript, b, end subscript, equals, 14, point, 1, start text, space, N, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, comma, space, a, n, d, space, c, e, l, e, b, r, a, t, e, right parenthesis, end text

### Example 2: (a slightly harder one)

A cube, whom you have developed a strong companionship with, has a total mass of 2, point, 33, start text, k, g, end text .
What must be the minimum side length of the cube so that it floats in sea water of density 1025, start fraction, start text, k, g, end text, divided by, start text, m, end text, cubed, end fraction?
We know that in order to float the buoyant force when the object is submerged must be equal to the magnitude of the weight of the cube. So we put this in equation form as,
W, start subscript, c, u, b, e, end subscript, equals, F, start subscript, b, end subscript, start text, left parenthesis, W, e, i, g, h, t, space, o, f, space, c, u, b, e, space, e, q, u, a, l, s, space, m, a, g, n, i, t, u, d, e, space, o, f, space, b, u, o, y, a, n, t, space, f, o, r, c, e, right parenthesis, end text
m, g, equals, rho, V, g, start text, left parenthesis, P, l, u, g, space, i, n, space, e, x, p, r, e, s, s, i, o, n, s, space, f, o, r, space, t, h, e, space, w, e, i, g, h, t, space, o, f, space, t, h, e, space, c, u, b, e, space, a, n, d, space, b, u, o, y, a, n, t, space, f, o, r, c, e, right parenthesis, end text
m, g, equals, rho, L, cubed, g, start text, left parenthesis, I, n, s, e, r, t, space, t, h, e, space, f, o, r, m, u, l, a, space, f, o, r, space, t, h, e, space, v, o, l, u, m, e, space, o, f, space, a, space, c, u, b, e, space, end text, L, cubed, right parenthesis
L, cubed, equals, start fraction, m, g, divided by, rho, g, end fraction, start text, left parenthesis, S, o, l, v, e, space, s, y, m, b, o, l, i, c, a, l, l, y, space, f, o, r, space, end text, L, cubed, right parenthesis
L, equals, left parenthesis, start fraction, m, divided by, rho, end fraction, right parenthesis, start superscript, 1, slash, 3, end superscript, start text, left parenthesis, C, a, n, c, e, l, space, t, h, e, space, f, a, c, t, o, r, space, o, f, space, g, space, a, n, d, space, t, a, k, e, space, c, u, b, e, d, space, r, o, o, t, space, o, f, space, b, o, t, h, space, s, i, d, e, s, right parenthesis, end text
L, equals, left parenthesis, start fraction, 2, point, 33, start text, space, k, g, end text, divided by, 1025, start fraction, start text, k, g, end text, divided by, start text, m, end text, cubed, end fraction, end fraction, right parenthesis, start superscript, 1, slash, 3, end superscript, start text, left parenthesis, P, l, u, g, space, i, n, space, n, u, m, b, e, r, s, right parenthesis, end text
L, equals, 0, point, 131, start text, m, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, comma, space, a, n, d, space, c, e, l, e, b, r, a, t, e, right parenthesis, end text

### Example 3: (an even harder one)

A huge spherical helium filled balloon painted to look like a cow is prevented from floating upward by a rope tying it to the ground. The balloon plastic structure plus all the helium gas inside of the balloon has a total mass of 9, point, 20, start text, space, k, g, end text . The diameter of the balloon is 3, point, 50, start text, space, m, end text . The density of the air is 1, point, 23, start fraction, start text, k, g, end text, divided by, start text, m, end text, cubed, end fraction .
What is the tension in the rope?
This one is a little harder so we should first draw a free body diagram (i.e. force diagram) for the balloon. There are lots of numbers here too so we could include our known variables in our diagram so that we can see them visually. (Note that in this case, the fluid being displaced is the air.)
Since the spherical cow balloon is not accelerating, the forces must be balanced (i.e. no net force). So we can start with a statement that the magnitudes of the total upward and downward forces are equal.
F, start subscript, b, end subscript, equals, W, plus, F, start subscript, T, end subscript, start text, left parenthesis, U, p, w, a, r, d, space, a, n, d, space, d, o, w, n, w, a, r, d, space, f, o, r, c, e, s, space, a, r, e, space, e, q, u, a, l, slash, b, a, l, a, n, c, e, d, right parenthesis, end text
rho, V, g, equals, m, g, plus, F, start subscript, T, end subscript, start text, left parenthesis, I, n, s, e, r, t, space, t, h, e, space, f, o, r, m, u, l, a, s, space, f, o, r, space, b, u, o, y, a, n, t, space, f, o, r, c, e, space, a, n, d, space, w, e, i, g, h, t, space, o, f, space, b, a, l, l, o, o, n, space, r, e, s, p, e, c, t, i, v, e, l, y, right parenthesis, end text
F, start subscript, T, end subscript, equals, rho, V, g, minus, m, g, start text, left parenthesis, S, o, l, v, e, space, s, y, m, b, o, l, i, c, a, l, l, y, space, f, o, r, space, t, h, e, space, t, e, n, s, i, o, n, space, a, n, d, space, i, s, o, l, a, t, e, space, i, t, space, o, n, space, o, n, e, space, s, i, d, e, space, o, f, space, t, h, e, space, e, q, u, a, t, i, o, n, right parenthesis, end text
F, start subscript, T, end subscript, equals, rho, left parenthesis, start fraction, 4, divided by, 3, end fraction, pi, r, cubed, right parenthesis, g, minus, m, g, start text, left parenthesis, P, l, u, g, space, i, n, space, t, h, e, space, f, o, r, m, u, l, a, space, f, o, r, space, t, h, e, space, v, o, l, u, m, e, space, o, f, space, a, space, s, p, h, e, r, e, right parenthesis, end text
F, start subscript, T, end subscript, equals, left parenthesis, 1, point, 23, start fraction, start text, k, g, end text, divided by, start text, m, end text, cubed, end fraction, right parenthesis, open bracket, start fraction, 4, divided by, 3, end fraction, pi, left parenthesis, start fraction, 3, point, 50, start text, space, m, end text, divided by, 2, end fraction, right parenthesis, cubed, close bracket, g, minus, left parenthesis, 9, point, 20, start text, space, k, g, end text, right parenthesis, g, start text, left parenthesis, P, l, u, g, space, i, n, space, n, u, m, b, e, r, s, point, space, C, o, n, v, e, r, t, space, d, i, a, m, e, t, e, r, space, t, o, space, r, a, d, i, u, s, !, right parenthesis, end text
F, start subscript, T, end subscript, equals, 180, start text, space, N, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, comma, space, a, n, d, space, c, e, l, e, b, r, a, t, e, right parenthesis, end text

## Want to join the conversation?

• if upward force is more than downward force then in zero gravity, will a object in water will go upward?
(20 votes)
• You're correct if there was gravity acting on the water but not the object. However, in zero gravity there would be no water pressure at all and therefore the water wouldn't push you upwards. Buoyancy is a result of gravity acting on a liquid.
(7 votes)
• why do some objects float when partially submerged, but sink when fully submerged?
(3 votes)
• Nice answer Charles. (I was thinking bucket :)

There is also the situation in which 'surface tension' can play a part.

So, for a small object such a a needle, it might float on the surface but this is not due to displacement. When you push it under water, then surface tension can no longer keep it up so it sinks.

ok??
(12 votes)
• Can we calculate the pressure acting on a submerged object in a liquid, on it's side?
(3 votes)
• Yes you can! It's more complicated though, because you need to break it down with some of your calculus and geometry tools, see why? The classic types of problem are "calculate the pressure on the sides of a pool, dam, tank.... etc." You should find problems like these in the section of integral calculus applications of many calculus books :)
If you'd want me to recommend a book and a page with the answers to the problems, i'll be gald to do so.

Cheers,
(10 votes)
• if we put a cone inside a fluid such that it is in the lowest most point and there is no water beneath it but there is water around it. then the buoyant force should act downward . but if the density of the fluid is greater than the cone then who will provide the force to float??
(6 votes)
• If there is no chance for the fluid to get beneath the cone, the cone might stay in place. But I've seen a video on which a screwdriver floats in mid air just because the air pressure is decreased at the right spot above the screwdriver. Of course things are a bit different with gasses and fluids.

Keep looking and asking around and I'm sure you'll find out soon enough how things really are!
(6 votes)
• Does the laws stated above also apply to an object moving to the surface of the water??
(6 votes)
• Yes they do. It doesn't matter whether it is stationary or moving, the follow the same laws.
(2 votes)
• After reaching its terminal velocity, woud a fully submerged object sink at a constant speed or at a gradually slower speed as it sinks deeper?
(4 votes)
• 1. equation
weight_object = buoyant force_by_fluid
m_obj * g = m_fld * g
#obj stands for object | fld for fluid
#g cancel
density_obj * volume_obj = density_fld * volume_fld
#vol_obj = ovl_fld, thus cancel
density_obj = density_fld

2. cases
at the end of the day the net force, acceleration, and thus velocity only depend on the density difference of the two (object and fluid)
1) d_obj > d_fld
it sinks with net force of (d_obj-d_fld)*v_obj*g to the bottom
2) d_obj = d_fld
it stays sill wherever you put it inside the fluid as there's no net force
3) d_obj < d_fld
it floats with net force of (d_fld-d_obj)*v_obj*g toward the water surface

3. interpretations
case 1) and 3)
: net force is constant (positive or negative) cause their densities stay same, thus giving a constant acceleration. so the velocity is constantly increasing
case 2)
: 0 velocity, as there's no motion
(1 vote)
• As stated above the buoyant force can be calculated by subtracting the downward force from the upward force exerted by the water on the object.But the upward force itself is buoyant force by water ,then how do we arrive at the above equation?please make it a little clear I'm too confused.
(3 votes)
• F (upwards) stands for the force applied by water in an upward direction while F(downwards) stands for the force applied by the water on the can in the downward direction. Buoyant force is the net upward force on the can by the water.
(3 votes)
• what is the difference between buoyant force and archimedes principle?
(2 votes)
• buoyant force is the upward force a fluid exerts on an object.
Archimedes' Principle is the fact that buoyant force is equal to the weight of the displaced fluid.
(3 votes)
• If I place an object in water. It will exert a force equal to its weight on the water below. Now, the water below will exert a reaction force upwards on the object. Isn't that buoyant force?
(3 votes)
• sorry but it's not. buoyant force is different from reaction force

1. condition of reaction force
reaction force (normal force in the case of gravity) only works when two objects meet on a surface

2. analogy of reaction force
in the case of a ball in water, we don't treat water or whatever fluid around it as another object. it is environment. as much as we don't consider a freefalling ball in air as having any reaction force from any surfaces, even though it keeps contacting with air pressure (this is a buoyant force too)

3. definition & example of buoyant force
1) definition
then what pushes the object upward (and any-wards)? the whole water itself and its weight by gravity.
2) example
let's say a ball is inside a sea. then the sea itself has a mass thus weight by gravity. so all the water molecules in the sea are pushed down. and there's something less dense than them. what will happen? they want to occupy that very volume of that less densy invader! in all possible directions. that's what happens inside the mercury barometer we checked previous chapters here. no matter how small the density differene is (vacuum to a tiny tiny mismatch), water molecules don't miss that chance to exploit it and then steal that volume. and the left is what we've learned. h of a lower part of the ball is a bit larger than that of upper part, thus the net force pushes upward it. that's buoyant force. and that very force is not from the ball itself. or its reaction force from water "surface". but from water molecules around it (insdie the entire sea, to be exact)

plus, it's after the ball lands on the bottom of the sea when the reaction force works. the bottom ground functions as a surface. it keeps the ball from drilling down underground by forcing the same magnitude but opposite direction of reaction force to the ball's weight. thus the ball stays still (happily ever after, i hope)
(2 votes)
• Wood and Iron blocks have equal volumes. The wood floats while the iron
sinks in water.
how is a greater buoyant force acting on iron?
AND..
Two identical inflated balloons; one with helium while the other with CO2. The helium balloon rises while the CO2 balloon falls to the ground.
how is the buoyant force is equal on both balloons?
(1 vote)
• "Wood and Iron blocks have equal volumes. The wood floats while the iron
sinks in water.
how is a greater buoyant force acting on iron?"

What makes you think that buoyant force is bigger on iron when the volumes are same.
For your next question as well the answer depends on the word "identical" which tells you something about the volume. What is it?
(4 votes)