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.
Bean pollution is a crime. If you see someone throwing beans into a pool or ocean call the Society for Bean Free Waterways immediately.
Because pressure (Pgauge=ρgh)(P_{gauge}=\rho gh) 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.
OK, so it doesn't completely follow. After all, what if we considered an object where the area of the bottom was smaller than the area of the top (like a cone). Since F=PAF=PA, could the greater area on the top of the cone compensate for the smaller pressure on the top? Would this make the object experience a net downward buoyant force? The answer is no. It turns out that no matter what shape you make your object, the net force from water pressure will always point upward. For the example of the cone, the tapered sides make it so that a component of the pressure on the sides also points up which makes it so the net buoyant force again points upward.
It's fun to try and think of more examples of other shapes, and then try to figure out why they won't make the buoyant force point downward.
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 FdownF_{down}, and the water on the bottom of the can is pushing up FupF_{up}. We can find the total upward force on the can exerted by water pressure (which we call the buoyant force FbuoyantF_{buoyant}) by simply taking the difference between the magnitudes of the upward force FupF_{up} and downward force FdownF_{down}.
Fbuoyant=FupFdownF_{buoyant} =F_{up} - F_{down}
We can relate these forces to the pressure by using the definition of pressure P=FAP=\dfrac{F}{A} which can be solved for force to get F=PAF=PA . So the force exerted upward on the bottom of the can will be Fup=PbottomAF_{up}=P_{bottom}A and the force exerted downward on the top of the can will be Fdown=PtopAF_{down}=P_{top}A. Substituting these expressions in for each FF respectively in the previous equation we get,
Fbuoyant=PbottomAPtopA F_{buoyant} =P_{bottom} A - P_{top}A
We can use the formula for hydrostatic gauge pressure Pgauge=ρghP_{gauge}=\rho gh to find expressions for the upward and downward directed pressures. The force from pressure directed upward on the bottom of the can is Pbottom=ρghbottomP_{bottom}=\rho gh_{bottom} and the force from pressure directed downward on the top of the can is Ptop=ρghtopP_{top}=\rho gh_{top} . We can substitute these into the previous equation for each pressure respectively to get,
Fbuoyant=(ρghbottom)A(ρghtop)AF_{buoyant} =(\rho gh_{bottom}) A - (\rho gh_{top})A
Notice that each term in this equation contains the expression ρgA\rho g A. So we can simplify this formula by pulling out a common factor of ρgA\rho g A to get,
Fbuoyant=ρgA(hbottomhtop)F_{buoyant} =\rho gA(h_{bottom} -h_{top})
Now this term hbottomhtoph_{bottom} -h_{top} is important and something interesting is about to happen because of it. The difference between the depth of the bottom of the can hbottomh_{bottom} and the depth of the top of the can htoph_{top} is just equal to the height of the can. (see the diagram below)
So we can replace (hbottomhtop)(h_{bottom} -h_{top}) in the previous formula with the height of the can hcanh_{can} to get,
Fbuoyant=ρgAhcanF_{buoyant} =\rho gAh_{can}
Here's the interesting part. Since A×hA \times h is equal to the volume of a cylinder, we can replace the term AhcanAh_{can} with a volume VV . 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.
Since there is no water left in the region of space where the can is now, all that water went somewhere else in the fluid.
So we are definitely going to replace the term AhAh with a volume VV , 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 VfluidV_{fluid} in the formula because the displaced fluid is the factor that determines the buoyant force.
Well, imagine the can was floating with half of its volume submerged beneath the surface of the fluid.
There would no longer be any downward force from the water pressure on the top of the can. And the depth hbottomh_{bottom} of the bottom of the can would now only be a fraction of the can's height. So if we solved for the buoyant force like we did before we would get,
Fbuoyant=ρgA(hbottom0)\Large F_{buoyant} =\rho gA(h_{bottom} -0)
But the term AhbottomAh_{bottom} is not equal to the entire volume of the can. It's only equal to the volume of the can submerged, or in other words the volume of the displaced fluid VfV_f.
Fbuoyant=ρgVfluid\Large F_{buoyant} =\rho gV_{fluid}
You could of course choose to write the formula in terms of the volume of the can VcanV_{can} as long as you knew that the only part of the volume that counts is the volume submerged.
Fbuoyant=ρgVcan\Large F_{buoyant} =\rho gV_{can}
Fbuoyant=ρgVf\Large F_{buoyant} =\rho gV_{f}
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 ρ\rho in which the object is submerged, the acceleration due to gravity gg, and the volume of the displaced fluid VfV_f.
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.
The net vertical force (including gravity now) on a submerged object will be the buoyant force on the object minus the magnitude of the weight of the object.
Fnet=FbWF_{net} = F_b-W
We can use the formula we derived for buoyant force to rewrite FbF_b as ρfVfg\rho_f V_f g where ρf\rho_f is the density of the fluid,
Fnet=ρfVfgmgF_{net} = \rho_fV_fg - mg
We can make that second term in the formula look a whole lot more like the first term if we use the rearranged definition of density to write the mass of the object mm in terms of the density of the object ρo\rho_o and the volume of the object submerged VoV_o. Subbing in the formula m=ρoVom=\rho_oV_o for the mass in the previous formula we get,
Fnet=ρfVfgρoVogF_{net} = \rho_fV_fg - \rho_oV_og
If the object is fully submerged the two volumes VV are the same and we can pull out a common factor of VgVg to get,
Fnet=Vg(ρfρo)F_{net} = Vg(\rho_f- \rho_o)
So there it is! If the density of the object is greater than the density of the fluid the net force will be negative which means the object will sink if released in the fluid.

What is Archimedes' principle?

The way you will normally see the buoyant force formula written is with the gg and the VV rearranged like so,
Fbuoyant=ρVfgF_{buoyant} =\rho V_{f}g
When you rearrange the formula in this way it allows you to notice something amazing. The term ρVf\rho V_f is the density of the displaced fluid multiplied by the volume of the displaced fluid. Since the definition of density ρ=mV\rho=\dfrac{m}{V} can be rearranged into m=ρVm=\rho V, that means the term ρVf\rho V_f corresponds to the mass of the displaced fluid. So, if we wanted to, we could replace the term ρVf\rho V_f with mfm_f in the previous equation to get,
Fbuoyant=mfgF_{buoyant} =m_{f}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,
Fbuoyant=WfF_{buoyant} =W_f
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.
Since there is no water left in the region of space where the can is now, all the water that was in that volume must have been displaced elsewhere in the fluid.
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 ρ\rho in the formula for buoyant force Fb=ρVfgF_b=\rho V_{f}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 VfV_f, density of the fluid ρ\rho, and the acceleration due to gravity gg.
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 0.650 kg0.650 \text{ kg} garden gnome went snorkeling a little too low and found himself at the bottom of a fresh water lake of depth 35.0 m35.0 \text{ m} . The garden gnome is solid (with no holes) and takes up a total volume of 1.44×103 m31.44 \times 10^{-3}\text{ m}^3 . The density of fresh water in the lake is 1000kgm31000 \dfrac{\text{kg}}{\text{m}^3} .
What is the buoyant force on the gnome?
Fb=ρVg(Use buoyant force equation, which is just Archimedes’ principle in math form)F_b=\rho Vg \qquad \text{(Use buoyant force equation, which is just Archimedes' principle in math form)}
Fb=(1000kgm3)(1.44×103 m3)(9.8ms2)(Plug in numerical values)F_b=(1000 \dfrac{\text{kg}}{\text{m}^3})(1.44 \times 10^{-3}\text{ m}^3)(9.8\dfrac{\text{m}}{\text{s}^2}) \qquad \text{(Plug in numerical values)}
Fb=14.1 N(Calculate, and celebrate)F_b= 14.1 \text{ N}\qquad \text{(Calculate, and celebrate)}

Example 2: (a slightly harder one)

A cube, whom you have developed a strong companionship with, has a total mass of 2.33kg2.33 \text{kg} .
What must be the minimum side length of the cube so that it floats in sea water of density 1025kgm31025 \dfrac{\text{kg}}{\text{m}^3}?
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,
Wcube=Fb(Weight of cube equals magnitude of buoyant force)W_{cube}=F_b \qquad \text{(Weight of cube equals magnitude of buoyant force)}
mg=ρVg(Plug in expressions for the weight of the cube and buoyant force)mg=\rho Vg \qquad \text{(Plug in expressions for the weight of the cube and buoyant force)}
mg=ρL3g(Insert the formula for the volume of a cube L3)mg=\rho L^3g \qquad \text{(Insert the formula for the volume of a cube } L^3)
L3=mgρg(Solve symbolically for L3)L^3 = \dfrac{mg}{\rho g} \qquad \text {(Solve symbolically for } L^3)
L=(mρ)1/3(Cancel the factor of g and take cubed root of both sides)L = \left(\dfrac{m}{\rho }\right)^{1/3} \qquad \text{(Cancel the factor of g and take cubed root of both sides)}
L=(2.33 kg1025kgm3)1/3(Plug in numbers)L = \left(\dfrac{2.33\text{ kg}}{1025 \dfrac{\text{kg}}{\text{m}^3}}\right)^{1/3} \qquad \text{(Plug in numbers)}
L=0.131m(Calculate, and celebrate)L = 0.131 \text{m} \qquad \text{(Calculate, and celebrate)}

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 9.20 kg9.20 \text{ kg} . The diameter of the balloon is 3.50 m3.50\text{ m} . The density of the air is 1.23kgm31.23 \dfrac{\text{kg}}{\text{m}^3} .
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.
Fb=W+FT(Upward and downward forces are equal/balanced)F_b = W +F_{T} \qquad \text{(Upward and downward forces are equal/balanced)}
ρVg=mg+FT(Insert the formulas for buoyant force and weight of balloon respectively)\rho Vg = mg + F_T \qquad \text{(Insert the formulas for buoyant force and weight of balloon respectively)}
FT=ρVgmg(Solve symbolically for the tension and isolate it on one side of the equation)F_T = \rho Vg -mg \qquad \text{(Solve symbolically for the tension and isolate it on one side of the equation)}
FT=ρ(43πr3)gmg(Plug in the formula for the volume of a sphere)F_T = \rho (\dfrac{4}{3}\pi r^3)g -mg \qquad \text{(Plug in the formula for the volume of a sphere)}
FT=(1.23kgm3)[43π(3.50 m2)3]g(9.20 kg)g(Plug in numbers. Convert diameter to radius!)F_T = (1.23 \dfrac{\text{kg}}{\text{m}^3}) [\dfrac{4}{3}\pi (\dfrac{3.50 \text{ m}}{2})^3]g -(9.20 \text{ kg})g \qquad \text{(Plug in numbers. Convert diameter to radius!)}
FT=180 N(Calculate, and celebrate)F_T = 180 \text{ N} \qquad \text{(Calculate, and celebrate)}
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