Buoyant Force vs Weight: The Difference

The distinction, stated once

Two forces act on a submerged object and both are proportional to $g$, which is what makes them so easy to run together. They are not the same quantity and they do not even belong to the same object.

Weight is a fact about the object. It is the gravitational force on that object, $mg$, and its size depends on the object's own mass and nothing else. Move it into water, into mercury, or into air and the number does not change.

The buoyant force is a fact about the fluid. Essential knowledge 8.3.B.1 calls it a net upward force exerted on an object by a fluid, and 8.3.B.3 fixes the size: the magnitude of the buoyant force exerted on an object by a fluid is equivalent to the weight of the fluid displaced by the object. The sheet writes it as

Read those two symbols slowly, because the whole page is in them. The $\rho$ is the density of the fluid, never of the object. The $V$ is the volume of fluid pushed aside, which equals the object's volume only when the object is entirely under the surface.

So write the two forces out side by side and the difference becomes arithmetic:

Every symbol that appears in one is different from the symbol that appears in the other, apart from $g$. That is the reason the outcome of a float-or-sink question is a comparison of densities and not a comparison of weights: divide one expression by the other and $g$ cancels, leaving a ratio of densities and a ratio of volumes.

Buoyant force against weight, row by row

The two rows that decide exam questions are the eighth and ninth, and they are the two nobody checks. Between them they say that the buoyant force and the weight can be changed independently. Change the fluid and only $F_b$ moves. Change the object's mass without changing its outside shape and only $W$ moves. Two independent numbers cannot be the same quantity, whatever they happen to be equal to in a particular case.

A note on the fifth row, because it is a common surprise. The sheet prints $F_b = \rho V g$ as a fluids equation, but it prints no line reading "weight equals $mg$". What it gives you instead is the gravitational field strength in the constants box, $g = 9.8\ \text{N/kg}$, whose units say directly that multiplying by a mass in kilograms returns a force in newtons. The panel's $\Delta U_g = mg\Delta y$ is the only place $mg$ appears as a printed product. So on the sheet, buoyancy has a formula and weight has a unit.

Float, sink or hover, and the density condition for each

Start from Newton's second law with just the two forces, and let $\rho_o$ be the object's average density and $\rho_f$ the fluid's.

For an object fully submerged, the displaced volume is the object's own volume, so $V$ is the same in both expressions and cancels:

The verdict follows immediately from whether that ratio is above or below one.

The algebra behind the first row is worth doing once. A floating object is in equilibrium under gravity and buoyancy alone, so $F_b = W$, which is $\rho_f V_{\text{disp}}\, g = \rho_o V_o\, g$. Cancel $g$ and rearrange:

That fraction has to be at most one for a solution to exist, which is exactly the condition $\rho_o \le \rho_f$. An object denser than the fluid has no floating configuration to find, because even pushing every last cubic meter of itself under the surface does not raise $F_b$ to $W$.

The third row's acceleration comes from the same two expressions. $ma = W - F_b = \rho_o V g - \rho_f V g$, and $m = \rho_o V$, so dividing through gives $a = g(1 - \rho_f/\rho_o)$ downward. Notice that the volume has dropped out entirely: a small dense pebble and a large dense boulder of the same material sink with the same acceleration in the same fluid, ignoring drag.

The case that separates them: the same object, two fluids, opposite verdicts

Take a solid iron block, $\rho_o = 7800\ \text{kg/m}^3$, with a volume of $5.0 \times 10^{-4}\ \text{m}^3$. Its mass is $3.9\ \text{kg}$ and its weight is $38.22\ \text{N}$. That number is now fixed for the rest of this section.

Release it in water, $\rho_f = 1000\ \text{kg/m}^3$. Fully submerged it displaces its own volume, so $F_b = (1000)(5.0 \times 10^{-4})(9.8) = 4.9\ \text{N}$. Against a weight of $38.22\ \text{N}$, the buoyant force is $12.8\ \%$ of the weight, and the block sinks with a net downward force of $33.32\ \text{N}$.

Release the same block in mercury, $\rho_f = 13{,}600\ \text{kg/m}^3$. Its weight is still $38.22\ \text{N}$, unchanged, because nothing about the iron changed. But fully submerged it would now displace $F_b = (13{,}600)(5.0 \times 10^{-4})(9.8) = 66.64\ \text{N}$ of mercury, which exceeds its weight. It cannot stay under. It rises and floats, settling with $\rho_o/\rho_f = 7800/13{,}600 = 57.4\ \%$ of its volume below the surface.

One object, one weight, two fluids, and the answer reversed. If the buoyant force were a property of the object, or if floating were about being heavy or light, that could not happen.

Run the comparison the other way for the second half of the point. Take a wooden block and an iron block that happen to have the same weight, and drop both in water. The wood floats and the iron sinks, even though gravity is pulling on each of them with exactly the same force. Equal weights, opposite outcomes. What differs is the volume each one needs to occupy in order to have that weight, which is another way of saying what differs is the density.

Why a steel ship floats and a steel bolt sinks

This is the question the comparison exists to answer, and once the two forces are separated it takes three lines.

Take $78\ \text{kg}$ of iron. As a solid lump it occupies $78/7800 = 0.010\ \text{m}^3$. Its weight is $764.4\ \text{N}$, and the largest buoyant force water can ever supply to it is the weight of $0.010\ \text{m}^3$ of water, which is $(1000)(0.010)(9.8) = 98\ \text{N}$. That is not close. It sinks, and it would sink no matter how the lump were shaped, so long as it stayed solid.

Now beat the same $78\ \text{kg}$ of iron into an open box measuring $1.0\ \text{m}$ by $1.0\ \text{m}$ by $0.40\ \text{m}$ deep. Nothing has been added and nothing removed, so the weight is still exactly $764.4\ \text{N}$. But the volume the hull can displace is now $0.40\ \text{m}^3$, and the largest buoyant force available is $(1000)(0.40)(9.8) = 3920\ \text{N}$, forty times what it was. The box only needs $764.4\ \text{N}$ of it, so it floats, displacing $764.4/9800 = 0.078\ \text{m}^3$ of water and settling $7.8\ \text{cm}$ deep in a hull $40\ \text{cm}$ tall.

The density bookkeeping says the same thing more compactly. The box's average density, counting the air it encloses, is $78/0.40 = 195\ \text{kg/m}^3$, well under water's $1000$, and $195/1000 = 19.5\ \%$ is the fraction submerged, which matches the $7.8\ \text{cm}$ out of $40\ \text{cm}$ exactly.

So shipbuilding does not fight the weight. It leaves the weight alone and raises the maximum buoyant force by increasing the volume of water the object can push aside. Any argument phrased as "heavy things sink" has to explain why the identical $78\ \text{kg}$ does both, and it cannot.

When it costs a mark

• Putting the object's density into $F_b = \rho V g$. The sheet's symbol list defines $\rho$ as density without a subscript, so nothing on the page stops you. Write $\rho_{\text{fluid}}$ next to the symbol before you substitute.
• Using the object's full volume for a floating object. $V$ in $F_b = \rho V g$ is the displaced volume. For anything floating, that is less than the object's volume, and the two are equal only for a fully submerged object.
• Assuming $F_b = W$ always. It holds for an object floating or hovering in equilibrium. It fails the moment the object is accelerating, held by a string, or resting on the bottom. Reaching for it automatically turns a dynamics question into a wrong statement.
• Deciding float or sink by weight. A $1\ \text{g}$ steel pin sinks and a $50{,}000\ \text{tonne}$ ship floats. Compare $\rho_o$ with $\rho_f$, never $W$ with anything.
• Thinking the buoyant force grows with depth. The pressure on both faces grows with depth, but their difference does not, so a fully submerged object in a uniform fluid feels the same $F_b$ at $2\ \text{m}$ and at $200\ \text{m}$. Gauge vs absolute pressure shows the cancellation term by term.
• Forgetting the normal force on the bottom. A sunken object at rest has three forces on it, not two: $N = W - F_b$. Leaving $N$ off the free-body diagram makes the object accelerate on paper while it sits still in the tank.
• Reporting apparent weight as a new weight. The weight has not changed. A scale supporting a submerged object reads $W - F_b$ because part of the support is being provided by the fluid, and the object's own gravitational force is untouched.

When they coincide, and why that lulls you

For a floating object at rest, the two are numerically equal. Only gravity and buoyancy act, the object is not accelerating, so $F_b = W$ exactly. Since floating objects are the ones you see every day, the equality is what intuition records, and the distinction never gets tested.

Watch how strange that equality really is. Move the floating oak block from fresh water into seawater and the buoyant force on it does not change, because it is still holding up the same weight. What changes is the displaced volume, which shrinks from $0.0216\ \text{m}^3$ to $0.0211\ \text{m}^3$, so the block rides a little higher. For a floater, the buoyant force is pinned to the weight and the geometry does the adjusting. For a fully submerged object, the reverse is true: the volume is pinned and the buoyant force does the moving.

That is the sharpest way to hold the two apart. Ask which quantity is free to vary.

• Floating: $F_b$ is locked to $W$; $V_{\text{disp}}$ adjusts.
• Fully submerged: $V_{\text{disp}}$ is locked to $V_o$; $F_b$ is whatever the fluid density makes it, and the difference from $W$ shows up as acceleration, tension or a normal force.

A second coincidence hides in the neutrally buoyant case, $\rho_o = \rho_f$. Here $F_b = W$ and the displaced volume equals the object's volume, so both of the relationships above hold at once and the object hovers wherever you leave it. It looks like the simplest case and it is actually the only case where the two very different quantities agree for a reason other than equilibrium: they agree because the densities do.

One last lull. Both forces scale with $g$, so both would be smaller on the Moon in the same proportion, and every float-or-sink verdict on this page would be unchanged there. That makes $g$ feel unimportant, and for the verdict it is. It is not unimportant for the numbers: an apparent weight, a string tension, or an acceleration all carry it.

What the CED asks of Topic 8.3

Buoyancy is Topic 8.3 of AP Physics 1, Fluids and Newton's Laws, inside Unit 8, which is weighted at $10$ to $15\ \%$ of the multiple-choice section across a suggested $12$ to $17$ class periods. The topic has two learning objectives.

8.3.A, describe the conditions under which a fluid's velocity changes. 8.3.A.1 says Newton's laws can be used to describe the motion of particles within a fluid, and 8.3.A.2 says the macroscopic behavior of a fluid results from the internal interactions between the fluid's constituent particles and external forces exerted on the fluid.

8.3.B, describe the buoyant force exerted on an object interacting with a fluid.

• 8.3.B.1: the buoyant force is a net upward force exerted on an object by a fluid.
• 8.3.B.2: the buoyant force exerted on an object by a fluid is a result of the collective forces exerted on the object by the particles making up the fluid.
• 8.3.B.3: the magnitude of the buoyant force exerted on an object by a fluid is equivalent to the weight of the fluid displaced by the object, with the relevant equation $F_b = \rho V g$.

Two details in that wording repay attention. First, 8.3.B.1 says net upward force. The fluid pushes on every face of the object; buoyancy is already the sum, which is why you never add a separate downward fluid force alongside it on a free-body diagram. Second, 8.3.B.3 says the weight of the fluid displaced by the object, not the weight of the object. The CED puts the trap's answer in the sentence that states the principle.

Topic 8.3 prints no boundary statement. Unit 8's single boundary statement sits under Topic 8.4 and assumes all fluids are ideal unless a question says otherwise, which for buoyancy problems means constant fluid density with depth and no viscous drag while an object rises or sinks. Ideal vs real fluid sets out what that assumption covers.

The suggested skills against Topic 8.3 are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Skill 2.D is the one this page is built for: a question that doubles the fluid density and asks what happens to the submerged fraction is asking whether you know which of the two forces just changed. The unit's own essential questions include why some objects float while others sink, and what implications there would be for our lives if nothing floated. Topic 8.3 carries the full framing, and the fluids guide runs the standard procedures.