AP Physics 1 · Topic 8.3

Topic 8.3: Fluids and Newton's Laws

Unit 8: Fluids10-15% of the multiple-choice section

Topic 8.3 puts fluid forces on a free-body diagram. A fluid pushes harder on the bottom of a submerged object than on its top, and that difference is the buoyant force. Its size is the weight of the fluid displaced, so floating and sinking are ordinary Newton's second law problems.

AP Physics: Unit 8 (topics 8.3 Fluids and Newton's Laws). AP Physics 1 Unit 8, Topic 8.3. Two learning objectives: 8.3.A, describe the conditions under which a fluid's velocity changes, supported by 8.3.A.1 (Newton's laws describe the motion of particles within a fluid) and 8.3.A.2 (macroscopic fluid behavior results from internal interactions between constituent particles and external forces exerted on the fluid); and 8.3.B, describe the buoyant force exerted on an object interacting with a fluid, supported by 8.3.B.1 (the buoyant force is a net upward force exerted on an object by a fluid), 8.3.B.2 (it results from the collective forces exerted on the object by the particles making up the fluid), and 8.3.B.3 (its magnitude is equivalent to the weight of the fluid displaced by the object, with relevant equation F_b = rho V g). Topic 8.3 has no boundary statement. The CED's suggested skills here are 1.A, 2.A, 2.D, and 3.B. Unit 8 carries 10 to 15 percent of the multiple-choice section and a suggested 12 to 17 class periods.

What Topic 8.3 requires

Topic 8.3 carries two learning objectives, and the second one is where almost all of the exam pressure sits.

8.3.A asks you to describe the conditions under which a fluid's velocity changes. Two essential knowledge statements sit under it. 8.3.A.1 states that Newton's laws can be used to describe the motion of particles within a fluid. 8.3.A.2 states that the macroscopic behavior of a fluid results from the internal interactions between the fluid's constituent particles together with the external forces exerted on the fluid.
8.3.B asks you to describe the buoyant force exerted on an object interacting with a fluid. 8.3.B.1 states that the buoyant force is a net upward force exerted on an object by a fluid. 8.3.B.2 states that the buoyant force is a result of the collective forces exerted on the object by the particles making up the fluid. 8.3.B.3 states that 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, and prints the relevant equation $F_b = \rho V g$ .

Read 8.3.B.1 and 8.3.B.3 next to each other, because together they are the whole topic. The first says buoyancy is a net force, which is a promise that it came from adding up something. The third says how big that net force is, in terms of fluid the object is not made of.

Topic 8.3 prints no boundary statement. The only boundary statement in Unit 8 belongs to Topic 8.4, and it is the line that lets you assume ideal fluids. That matters here in one specific way: the CED's ideal fluid has no viscosity (essential knowledge 8.1.A.4), so the buoyancy problems in this topic carry no drag term unless a question builds one in.

The CED lists four suggested skills for Topic 8.3: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 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. Unit 8 is weighted at 10 to 15 percent of the multiple-choice section, with a suggested 12 to 17 class periods for the unit as a whole.

Why a fluid pushes up at all

Essential knowledge 8.3.B.2 says the buoyant force is the collective effect of the fluid particles pushing on the object. You can turn that sentence into the equation in four lines, and doing so is the cleanest answer to "why does anything float," so it is worth carrying in your head.

Start from the pressure relation of Topic 8.2: $P = P_0 + \rho g h$ . Pressure grows with depth. Now submerge a block with horizontal top and bottom faces of area $A$ , the top at depth $h_1$ and the bottom at depth $h_2$ .

The fluid pushes down on the top face with $F_{\text{top}} = P_1 A = (P_0 + \rho g h_1)A$ .
The fluid pushes up on the bottom face with $F_{\text{bottom}} = P_2 A = (P_0 + \rho g h_2)A$ .
The sideways pushes on the vertical faces come in equal and opposite pairs at every depth, so they cancel.
The net upward force is the difference: $F_b = (P_2 - P_1)A = \rho g (h_2 - h_1) A$ .

And $(h_2 - h_1)A$ is exactly the volume of the block, so

F_b = \rho V g

The surface pressure $P_0$ dropped out on line 4, which is the reason a floating object does not care what the atmosphere is doing above it. Nothing in the derivation referred to what the block is made of. That is skill 2.A, deriving a symbolic expression along a logical pathway, and it is the single most useful thing to be able to reproduce from this topic.

One consequence surprises people: for a fully submerged object in a fluid of uniform density, the buoyant force does not change with depth. Both pressures rise as you go deeper, but the difference between them depends only on the height of the object, not on where it is.

Which density, and which volume

$F_b = \rho V g$ has two symbols in it and both of them belong to the fluid, not to the object. This is the error that costs the most points in Unit 8, so be blunt about it.

$\rho$ is the density of the fluid the object is sitting in. An iron anchor in water uses 1000 kg/m $^3$ , not 7900.
$V$ is the volume of fluid displaced, which is the volume of the object that is actually below the fluid surface. It equals the object's full volume only when the object is fully submerged.
$g$ is the gravitational field strength, 9.8 m/s $^2$ on the AP Physics 1 table of information.

The phrasing in essential knowledge 8.3.B.3 is what keeps this straight: the magnitude of the buoyant force equals the weight of the fluid displaced by the object. Read $\rho V g$ as $(\rho V) g$ , that is, mass of displaced fluid times $g$ . It is a weight, and it is the weight of something that is not there anymore.

Situation	$\rho$ in $F_b = \rho V g$	$V$ in $F_b = \rho V g$
Steel block resting on the bottom of a pool	water, 1000 kg/m $^3$	the block's full volume
Wooden block floating with a third above the surface	water, 1000 kg/m $^3$	two thirds of the block's volume
Same steel block, now in oil of density 800 kg/m $^3$	oil, 800 kg/m $^3$	the block's full volume
Boat hull, mostly air inside	water, 1000 kg/m $^3$	only the part of the hull below the waterline

The object's own density has not disappeared from the physics. It decides the object's weight, $mg = \rho_{\text{obj}} V_{\text{obj}} g$ , which is a different arrow on the diagram. Keep the two products in separate boxes and the algebra stops going wrong.

Buoyancy on a free-body diagram

Suggested skill 1.A for this topic is creating diagrams to represent physical situations, and the CED's own framing for Unit 8 is that students use force and energy representations to describe static and dynamic fluids. So the working method is the one you already own from Topic 2.2: draw the object, draw every force exerted on it, then apply $\sum F = ma$ .

Take up as positive. Every equation and every worked example on this page uses that choice, and you should declare yours in writing before the first line of algebra.

A submerged object typically has three candidate forces: the gravitational force $mg$ down, the buoyant force $F_b$ up, and whatever is holding it, a string tension up or a normal force from the pool floor up. Nothing about the fluid is exotic. Buoyancy is one more arrow.

Object	Newton's second law, up positive	What it tells you
Floating at rest	$F_b - mg = 0$	$F_b = mg$ exactly
Hanging from a string, fully submerged, at rest	$T + F_b - mg = 0$	$T = mg - F_b$
Resting on the bottom, at rest	$F_N + F_b - mg = 0$	$F_N = mg - F_b$
Released from rest, fully submerged, denser than the fluid	$F_b - mg = ma$	$a$ is negative, so the object accelerates downward
Held under and released, less dense than the fluid	$F_b - mg = ma$	$a$ is positive, so it accelerates upward

The last two rows use the same equation as the first three. That is the point of the topic being called Fluids and Newton's Laws. If you can find a net force, you can do fluid statics and dynamics, and the free-body diagram guide is the routine to fall back on when a setup gets crowded.

Floating is a force balance, not a rule about heaviness

One of Unit 8's essential questions in the CED is why some objects float while others sink, and the CED's own note on preparing for the exam warns that students often answer it by calling floating objects "lighter" than water and sinking objects "heavier" than water. That language fails immediately: a steel ship is heavier than a coin and the ship floats. The CED asks students to keep mass, volume, weight, size, and density distinct from one another, and this is the question that punishes anyone who does not.

The defensible version starts from the force balance. A floating object is in equilibrium, so $F_b = mg$ . Write both sides out:

\rho_{\text{fluid}} V_{\text{sub}} \, g = \rho_{\text{obj}} V_{\text{obj}} \, g

The $g$ cancels, and the submerged fraction is

\frac{V_{\text{sub}}}{V_{\text{obj}}} = \frac{\rho_{\text{obj}}}{\rho_{\text{fluid}}}

So the comparison that decides floating is between two densities, never between two weights. If the object's average density is less than the fluid's, the ratio is less than 1 and some of the object stays above the surface. If it is greater, no amount of the object is enough and the object sinks. If the two are equal, the object is in equilibrium at any depth.

The word average is doing real work. A ship's average density counts the air in the hull, which is why a hollow steel object floats and a solid one of the same steel does not. The same reasoning explains why a submarine controls depth by flooding and emptying ballast tanks: it is changing its own average density while its volume stays fixed. The fluids guide walks the standard float-or-sink routine end to end if you want the procedure rather than the framing.

Apparent weight, scales, and Newton's third law

Lower an object hanging from a spring scale into water and the reading drops. Nothing about the object changed, and its weight $mg$ is exactly what it was. What changed is the tension the scale reads, because the fluid is now carrying part of the load: $T = mg - F_b$ . That reading is what people mean by apparent weight.

This is a favorite experimental setup precisely because two scale readings, one in air and one submerged, are enough to get the density of an irregular object with no ruler involved. The CED suggests exactly this as a Unit 8 activity for Topic 8.3: hand students an irregularly shaped metal object, a spring scale, and a sink, and have them use buoyancy to find its volume and density.

The algebra behind that experiment is short. In air the scale reads $T_1 = mg$ . Submerged it reads $T_2 = mg - \rho_{\text{fluid}} V g$ , so the difference $T_1 - T_2 = \rho_{\text{fluid}} V g$ gives the volume directly, and then $\rho_{\text{obj}} = m/V$ .

One more piece of bookkeeping the exam likes. By Newton's third law, if the fluid exerts an upward buoyant force on the object, the object exerts an equal downward force on the fluid. Put the whole beaker on a balance and lower a hanging object into it without letting it touch the bottom, and the balance reading increases by exactly $F_b$ .

Learning objective 8.3.A: Newton's laws inside the fluid

Buoyancy takes most of the attention, but 8.3.A is the objective about the fluid itself, and it is phrased as a question about velocity change. A fluid's velocity changes when a net force is exerted on it, and nothing else. That is Newton's second law applied to a parcel of fluid rather than to a block.

Essential knowledge 8.3.A.2 says the macroscopic behavior you can see comes from two sources: the internal interactions among the fluid's own particles, and the external forces exerted on the fluid. Both halves show up in familiar situations.

Water accelerates out of a nozzle because the pressure behind it is higher than the pressure in front of it. A pressure difference across a parcel of fluid is a net force on that parcel.
A column of liquid sitting in a container is not accelerating, so the net force on every parcel is zero. The upward push from the fluid below has to balance the weight of everything above, and that requirement is what produces $P = P_0 + \rho g h$ in the first place.
Buoyancy is the same idea run one level up: replace a parcel of fluid with a solid object of the same shape, and the surrounding fluid keeps pushing exactly as it did, which is why the force is the weight of the fluid that used to be there.

That last bullet is worth rereading. It explains why $F_b$ equals a weight of fluid rather than anything about the object, and it is the bridge to Topic 8.4, where essential knowledge 8.4.A.1 states outright that a difference in pressure between two locations causes a fluid to flow. Topic 8.3 gives you the force reasoning; Topic 8.4 turns it into conservation laws.

Predicting factors of change, and reading the graphs

Suggested skill 2.D for this topic is predicting new values or factors of change using functional dependence between variables, and the CED's own sample multiple-choice question for learning objective 8.3.B is built on it. A rock sits fully submerged at the bottom of a cup of water and the upward force from the water on it is $F_0$ . The water is poured out and replaced by an oil that is three quarters as dense, and the rock is again completely covered. The correct answer is $\tfrac{3}{4} F_0$ , and one of the wrong choices claims the question cannot be answered without knowing the rock's volume.

Work that as a ratio and the reason is immediate. The rock is fully covered in both cases, so $V$ is the same, and $g$ is the same, so $F_b \propto \rho_{\text{fluid}}$ and the ratio is just the density ratio. You never needed a number for $V$ . Building that reflex, cancel everything that repeats and read off the proportionality, is most of skill 2.D.

The proportionalities worth having ready:

Fully submerged, $F_b \propto \rho_{\text{fluid}}$ and $F_b \propto V_{\text{obj}}$ . Double the object's volume and the buoyant force doubles.
Fully submerged, $F_b$ is independent of depth and independent of the object's density and mass.
Floating, $F_b$ is pinned to $mg$ and cannot change at all. Move the same floating block from fresh water to denser salt water and $F_b$ stays equal to $mg$ ; what changes is $V_{\text{sub}}$ , which shrinks.

The graph version appears in another Unit 8 activity the CED suggests: raise an object with a rope from the bottom of a deep pool and graph the rope's tension against the height of the object's bottom above the pool floor. While the object is fully submerged the tension is constant, because $F_b$ does not depend on depth. As the object breaks the surface, the displaced volume falls, $F_b$ falls, and the tension climbs until it reaches the full weight in air. The shape of that climbing section encodes the object's shape, since a cube, a sphere, and a cone lose displaced volume at different rates as they emerge.

How Topic 8.3 is tested, and where it leads

The patterns worth rehearsing, mapped to the skills the CED suggests here:

Draw and label a free-body diagram for an object floating, hanging submerged, or resting on the bottom, then write $\sum F = ma$ from it (skill 1.A).
Derive a symbolic result such as the submerged fraction $\rho_{\text{obj}}/\rho_{\text{fluid}}$ , or the acceleration of a released object, starting from a force balance rather than from a memorized shortcut (skill 2.A).
Predict a factor of change when the fluid, the depth, or the object's volume changes, without numbers (skill 2.D).
Justify a claim about whether an object floats using density rather than weight, in words a reader can grade (skill 3.B).
Design or analyze a measurement of an unknown object's volume or density from two scale readings.

Four of those five ask for reasoning rather than a calculator, and the free-response section has a question type for each habit: the four questions are Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation.

From here, Topic 8.4 takes the same fluid and asks what is conserved when it moves. Behind you, Topic 8.1 supplies density and Topic 8.2 supplies the pressure-with-depth relation this page differentiated. Every fluid equation named here is printed on the AP Physics 1 equation sheet, and the Unit 8 overview shows how the four topics stack.

Getting the buoyant force twice: from pressures, then from the formula

A solid cube of side 0.20 m is held fully submerged in fresh water ( $\rho_w = 1000$ kg/m $^3$ ) with its top and bottom faces horizontal. The top face is 0.50 m below the surface, and atmospheric pressure at the surface is $1.0 \times 10^5$ Pa. Find the force the water exerts on the top face, the force it exerts on the bottom face, and the net upward force. Then check the result against $F_b = \rho V g$ .

Set up the geometry. Each face has area $A = (0.20)^2 = 0.040$ m $^2$ , the cube's volume is $V = (0.20)^3 = 8.0 \times 10^{-3}$ m $^3$ , and the bottom face sits at depth $0.50 + 0.20 = 0.70$ m. Take up as positive.
Pressure at the top face: $P_1 = P_0 + \rho_w g h_1 = 1.0 \times 10^5 + (1000)(9.8)(0.50) = 1.0 \times 10^5 + 4900 = 1.049 \times 10^5$ Pa.
Pressure at the bottom face: $P_2 = P_0 + \rho_w g h_2 = 1.0 \times 10^5 + (1000)(9.8)(0.70) = 1.0 \times 10^5 + 6860 = 1.0686 \times 10^5$ Pa.
Forces on the horizontal faces: $F_{\text{top}} = P_1 A = (1.049 \times 10^5)(0.040) = 4196$ N downward, and $F_{\text{bottom}} = P_2 A = (1.0686 \times 10^5)(0.040) = 4274.4$ N upward. The four vertical faces contribute nothing, because at each depth the push from one side is matched by an equal push from the opposite side.
Net upward force: $F_{\text{bottom}} - F_{\text{top}} = 4274.4 - 4196 = 78.4$ N.
Check with the printed equation, using the water's density and the displaced volume: $F_b = \rho_w V g = (1000)(8.0 \times 10^{-3})(9.8) = 78.4$ N.

The water pushes down on the top face with 4196 N and up on the bottom face with 4274.4 N, and the difference is a net upward force of 78.4 N, matching $\rho_w V g = 78.4$ N exactly. Two forces of over 4000 N differ by less than 80 N, which is worth noticing: buoyancy is a small residue of two enormous opposing pushes, and it is why the atmospheric term cancels and why the answer does not depend on how deep the cube is.

A floating block: how deep it sits, and what it takes to push it under

A rectangular block measuring 0.30 m by 0.20 m by 0.10 m has a density of 650 kg/m $^3$ . It is placed in fresh water ( $\rho_w = 1000$ kg/m $^3$ ) and floats with its 0.30 m by 0.20 m face horizontal. (a) Find the buoyant force on it and the depth of the submerged part. (b) Find the extra downward force needed to hold it exactly fully submerged.

Volume and weight of the block: $V = (0.30)(0.20)(0.10) = 6.0 \times 10^{-3}$ m $^3$ , so $m = \rho V = (650)(6.0 \times 10^{-3}) = 3.9$ kg and $mg = (3.9)(9.8) = 38.22$ N.
(a) Floating means equilibrium. With up positive, $F_b - mg = 0$ , so $F_b = 38.22$ N. The buoyant force is fixed by the weight, not by anything about the water.
Now invert $F_b = \rho_w V_{\text{sub}} g$ to find the displaced volume: $V_{\text{sub}} = F_b / (\rho_w g) = 38.22 / [(1000)(9.8)] = 3.9 \times 10^{-3}$ m $^3$ .
The submerged part is a box on the same horizontal footprint, area $(0.30)(0.20) = 0.060$ m $^2$ , so its depth is $d = V_{\text{sub}} / A = (3.9 \times 10^{-3}) / 0.060 = 0.065$ m.
Cross-check with the density ratio: $V_{\text{sub}}/V = (3.9 \times 10^{-3})/(6.0 \times 10^{-3}) = 0.65$ , and $\rho_{\text{obj}}/\rho_w = 650/1000 = 0.65$ . The block sits 6.5 cm into the water out of its 10 cm height.
(b) Fully submerged, the displaced volume becomes the whole block: $F_b' = \rho_w V g = (1000)(6.0 \times 10^{-3})(9.8) = 58.8$ N.
Equilibrium while held under, up positive: $F_b' - mg - F_{\text{push}} = 0$ , so $F_{\text{push}} = 58.8 - 38.22 = 20.58$ N downward.

(a) $F_b = 38.2$ N and the block floats 0.065 m deep, that is 65 percent of its height, matching the density ratio 650/1000. (b) Holding it fully under takes an extra 20.6 N pushing down. Note which quantity changed between the two parts: the buoyant force went from 38.2 N to 58.8 N because $V_{\text{sub}}$ grew, while $\rho_w$ , $g$ , and the block's weight all stayed put.

Released from rest under water: the acceleration does not need the volume

A stone of density 2500 kg/m $^3$ and volume $4.0 \times 10^{-4}$ m $^3$ is released from rest while fully submerged in fresh water ( $\rho_w = 1000$ kg/m $^3$ ). Find its initial acceleration, then show the answer does not depend on the volume you were given. Treat the water as ideal, so there is no drag.

Mass and weight of the stone: $m = \rho_s V = (2500)(4.0 \times 10^{-4}) = 1.0$ kg, so $mg = (1.0)(9.8) = 9.8$ N.
Buoyant force, using the water's density and the stone's full volume since it is fully submerged: $F_b = \rho_w V g = (1000)(4.0 \times 10^{-4})(9.8) = 3.92$ N.
Newton's second law with up positive: $F_b - mg = ma$ , so $(3.92) - (9.8) = (1.0)a$ and $a = -5.88$ m/s $^2$ . The minus sign is the direction: 5.88 m/s $^2$ downward.
Now do it symbolically. $\rho_f V g - \rho_s V g = \rho_s V a$ , and every term carries a factor of $V$ , so $V$ cancels: $a = g\left(\dfrac{\rho_f}{\rho_s} - 1\right)$ .
Substitute: $a = 9.8\left(\dfrac{1000}{2500} - 1\right) = 9.8(0.40 - 1) = 9.8(-0.60) = -5.88$ m/s $^2$ , the same value.
Sanity-check the limits of that expression. If $\rho_f = \rho_s$ then $a = 0$ and the object hovers. If $\rho_f = 0$ , no fluid at all, then $a = -g$ and you have free fall. If $\rho_f > \rho_s$ then $a$ is positive and the object accelerates upward, which is what a released cork does.

$a = 5.88$ m/s $^2$ downward, and symbolically $a = g(\rho_f/\rho_s - 1)$ , so the volume never mattered. A stone two and a half times as dense as water sinks with 60 percent of the acceleration it would have in free fall. This is a skill 2.A and 2.D question in disguise: the useful output is the formula, because it answers every later version of the question without a calculator.

Frequently asked questions

What is the buoyant force in AP Physics 1?

The buoyant force is the net upward force a fluid exerts on an object placed in it, from essential knowledge 8.3.B.1 of AP Physics 1 Topic 8.3. Its magnitude equals the weight of the fluid the object displaces, written Fb = rho V g, where rho is the fluid's density and V is the displaced volume. It exists because pressure grows with depth, so the fluid pushes up on the object's bottom harder than it pushes down on its top.

Which density do you use in Fb = rho V g?

The density of the fluid, never the density of the object. An iron block submerged in water uses 1000 kg/m^3 in the buoyant force equation, not iron's 7900 kg/m^3. The object's own density belongs in its weight, mg, which is a separate force on the diagram. Essential knowledge 8.3.B.3 states the rule as the weight of the fluid displaced by the object, and reading rho V g as the mass of displaced fluid times g keeps it straight.

Is V in the buoyant force equation the object's volume?

Only when the object is completely submerged. V is the volume of fluid displaced, which is the volume of the object below the fluid's surface. A fully submerged rock displaces its own volume. A floating block displaces less than its own volume, and a boat displaces far less than the volume of its hull, because the air inside the hull is above the waterline and displaces nothing.

Why do some objects float and others sink?

Compare average densities, not weights. A floating object is in equilibrium, so the buoyant force equals its weight, and writing both sides out gives the submerged fraction as the object's density divided by the fluid's density. If the object's average density is less than the fluid's, that fraction is under 1 and part of the object stays above the surface; if it is greater, the object sinks; if the two are equal, it stays put at any depth. A steel ship floats because its average density counts the air in the hull.

Does the buoyant force change with depth?

No, not for an object that is already fully submerged in a fluid of uniform density. The pressure on the top face and the pressure on the bottom face both increase as the object goes deeper, but the difference between them depends only on the object's height, so the net upward force stays the same. Buoyancy changes only when the displaced volume or the fluid's density changes, which is what happens while an object is entering or leaving the surface.

How do you draw a free-body diagram for a floating object?

Draw the object, then two arrows: the gravitational force mg pointing down and the buoyant force Fb pointing up. For an object floating at rest they are equal in length, because the object is in equilibrium. Add a tension arrow if a string holds it, or a normal-force arrow if it rests on the bottom. Declare which direction is positive before writing Newton's second law, then solve the resulting equation exactly as you would for a block on a table.

Why does an object weigh less underwater?

Its weight is unchanged. What drops is the reading on the scale, because the scale reads the tension it supplies, and the water is now carrying part of the load: T = mg minus Fb. That difference is called apparent weight. It also gives a laboratory method for finding an irregular object's volume, since the change in the two scale readings equals the density of the fluid times the displaced volume times g.