8.3 Fluids and Newton's Laws: the buoyant force

Six little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last. After every two stories there is one quick question; if it feels shaky, press Show me another example for one more story. Tap the small round play button next to a caption to hear it read aloud.

1. The bolt and the ship

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A worker at a dock drops a steel bolt into the harbour. It sinks straight to the bottom.

Next to her, a cargo ship made of thousands of tonnes of the same steel floats. A crane lowers containers onto the ship. With each container, the ship sits a little lower in the water. But it still floats.

Steel is about 7.9 times as dense as water. So why does the ship float while the bolt sinks?

The question: what force does water put on an object, how big is it, and when is it big enough to hold the object up?

2. Why water pushes up

Quick check

Why does the water push up on a box held under water?

3. Loading the ship

4. The beach ball

Quick check

Maya holds a beach ball completely under water and then lets go. Right after she lets go, the net force on the ball is:

5. Weighing a stone under water

6. Floating in a salty sea

Quick check

Leo floats in very salty water instead of fresh water. Compared with fresh water, the buoyant force on him is:

7. Check yourself

Think of your answer first, then tap to see it.

a) A ship floats at rest in the harbour. Compare the buoyant force on it with its weight.

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They are equal. The ship is at rest, so the net force is zero (Newton's first law): buoyant force up = weight down.

b) A rock sinks deeper and deeper in a lake. Does the buoyant force on it grow, shrink or stay the same?

Show answer

It stays the same. The buoyant force equals the weight of the water the rock pushes aside, and that volume does not change. The extra push on the bottom face over the top face stays the same at any depth.

c) A stone with a volume of 0.002 m³ is completely under water. What is the buoyant force on it?

Show answer

Fb = ρgV = 1000 × 9.8 × 0.002 = 19.6 N.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead to the simulation. Not sure? No problem: just read on.

1) A block of wood floats at rest on a pond. How does the buoyant force compare with the block's weight?

2) A rock with a volume of 0.003 m³ is completely under water. What is the buoyant force on it?

3) A rock and a block of wood have the same volume. Both are held completely under water. Which has the bigger buoyant force?

The idea in plain words

Picture a beach ball in a swimming pool. Try to push it under the water. It fights you, and the farther you push it in, the harder it fights. Let go, and it shoots back up.

A rock feels lighter when you lift it under water, too. The water is helping you hold it up. Where does that upward help come from?

Water pushes on every surface it touches. The push is always at right angles to the surface. Deeper water pushes harder, because there is more water above it.

Look at a block under water. The water pushes down on the top face and up on the bottom face. The bottom face is deeper, so the upward push is bigger. The pushes on the sides cancel. What is left is a net upward force from the water. We call it the buoyant force, Fb.

water surface block smaller push down bigger push up sidescancel net: F_b up

In symbols, this is the rule from topic 8.2: P = P0 + ρgh. The bottom face is deeper, so h is bigger there. The pressure is bigger, and the push up wins.

How big is it? Picture the same space filled with plain water instead of the block. That water would just sit there, so the water around it must hold it up with a force equal to its weight. The water around the block cannot tell the difference. So:

Fb = weight of the fluid pushed aside = ρfluid Vsub g

Here ρfluid is the density of the fluid (not the object), and Vsub is the volume of the object that is under the fluid. This is Archimedes' principle. The AP equation sheet writes it as Fb = ρVg.

Trap: putting the object's density into Fb = ρVsubg. Instead: use the FLUID's density. A 1.0 × 10⁻³ m³ steel block under water: Fb = 1000 × 1.0 × 10⁻³ × 9.8 = 9.8 N (not 7900 × ...).

Then we use Newton's laws as always. Draw the forces, pick up as positive, and add them:

For a floating object, Fb = Fg gives ρfluid Vsub g = ρobject V g, so the fraction under water is Vsub/V = ρobject/ρfluid.

Trap: using the whole volume V for a floating object's buoyant force. Instead: only the part below the surface counts: Vsub. The block of 0.0060 m³ floats with just 0.0036 m³ under, so Fb = 35 N, not 59 N.

The ship floats because it is mostly air inside. Its average density (steel plus air plus cargo, divided by its whole volume) is less than the density of water. The bolt is solid steel, so it sinks.

Newton's laws work inside the fluid too (8.3.A). A bit of fluid speeds up only if the pressure on one side is bigger than on the other side. Water in a hose speeds up toward the nozzle because the pressure behind it is higher than the pressure in front of it.

Worked: a wooden block in a pond

Read the steps as text

A wooden block is 0.30 m long, 0.20 m wide and 0.10 m tall. Its density is 600 kg/m³. It is put in fresh water (1000 kg/m³). Up is positive.

  1. Find the volume: V = 0.30 m × 0.20 m × 0.10 m = 0.0060 m³. Why: we need volume for both the mass and the buoyant force.
  2. Find the weight: m = ρV = 600 kg/m³ × 0.0060 m³ = 3.6 kg, so Fg = mg = 3.6 kg × 9.8 m/s² = 35.3 N (35.28 N). Why: weight is the force the buoyant force must balance.
  3. Find the biggest buoyant force possible (block pushed fully under): Fb,max = 1000 kg/m³ × 0.0060 m³ × 9.8 m/s² = 58.8 N. Why: if even the biggest Fb were less than the weight, the block would sink. 58.8 N > 35.3 N, so it floats.
  4. Floating at rest means net force zero: Fb − Fg = 0, so Fb = 35.28 N. Why: Newton's first law. The block is not accelerating.
  5. Find the submerged volume: Vsub = Fb/(ρwater g) = 35.28 N / (1000 kg/m³ × 9.8 m/s²) = 0.0036 m³. Why: Fb = ρfluidVsubg, solved for Vsub.
  6. Find how deep it sits: depth = Vsub/area of bottom = 0.0036 m³ / (0.30 m × 0.20 m) = 0.060 m. Why: the under-water part is a box with the same bottom area. Check: 0.060 m / 0.10 m = 0.6 = 600/1000, the density ratio.
  7. Force to push it fully under and hold it: Fpush = Fb,max − Fg = 58.8 N − 35.28 N = 23.5 N, down. Why: three forces now (Fb up, Fg down, your push down) and they must add to zero.

Lab 1: float, sink, or hover?

A cube is let go with its bottom face just touching the water. Change the block density, the fluid density and the block volume. Watch the two force arrows (purple): Fb up and Fg down, drawn to the same scale. Up is positive. The height y is the height of the block's bottom face above the water surface.

Real water slows moving objects down. To make the block settle, this sim adds a drag-like force proportional to speed (2.0 s⁻¹ × m × v). Without it, a floating block would bob up and down forever. The final resting place does not depend on the drag.

Buoyant force vs submerged volume

The purple line is Fb = ρfluidVsubg, a straight line through zero. The grey line is the weight. The orange dot is the block right now.

Slope of the line = ρ_fluid × g.

Try this: set the block density to 600 and the fluid to 1000. It settles with 60% under water. Now make the fluid 1200 (salt water). It floats higher: 600/1200 = 50% under. Then set the block to 1000 and the fluid to 1000: it hovers once it is fully under.

Lab 2: a block on a spring scale

A block hangs from a spring scale. The scale is lowered at a steady speed so the block goes into the water. The scale reads the tension in the string. For the block at constant velocity: T + Fb − Fg = 0, so T = Fg − Fb. People call this the "apparent weight".

Scale reading vs depth

Notice: once the block is fully under water, going deeper does not change the reading. The pressure on top and on the bottom both grow, but their difference stays ρfluid g × (block height).

Examples

1. A rock under water basic

A rock with volume 2.0 × 10⁻³ m³ sits fully under fresh water (1000 kg/m³). What is the buoyant force on it?

Show solution
Read the steps as text
  1. The whole rock is under, so Vsub = 2.0 × 10⁻³ m³. Why: Fb uses only the volume under the fluid.
  2. Fb = ρfluidVsubg = 1000 kg/m³ × 2.0 × 10⁻³ m³ × 9.8 m/s² = 19.6 N, upward. Why: Archimedes. We never needed the rock's mass.

2. Aluminium on a spring scale medium

A 2.7 kg aluminium block (density 2700 kg/m³) hangs from a spring scale. It is fully under water and at rest. What does the scale read?

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  1. Volume: V = m/ρ = 2.7 kg / 2700 kg/m³ = 1.0 × 10⁻³ m³. Why: we need the volume for Fb.
  2. Weight: Fg = 2.7 kg × 9.8 m/s² = 26.46 N. Buoyant force: Fb = 1000 × 1.0 × 10⁻³ × 9.8 = 9.8 N. Why: these are the two forces besides the string.
  3. Up positive, at rest: T + Fb − Fg = 0, so T = 26.46 N − 9.8 N = 16.7 N. Why: Newton's first law. The scale reads the tension.

3. The tip of the iceberg medium

Ice has density 917 kg/m³. Sea water has density 1025 kg/m³. What fraction of a floating iceberg is above the water?

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Read the steps as text
  1. Floating: Fb = Fg, so ρseaVsubg = ρiceVg. Why: at rest, net force zero.
  2. Vsub/V = 917/1025 = 0.895. Why: g cancels. The fraction under is the density ratio.
  3. Fraction above = 1 − 0.895 = 0.105, about 10.5%. Why: what is not under is above. "Tip of the iceberg" is literally true.

4. Loading a barge AP

A barge is a flat-bottomed box, 10 m long and 4.0 m wide. Empty, it has a mass of 20 000 kg. It is in fresh water (1000 kg/m³). (a) How deep does the empty barge sit? (b) The bottom may sit at most 1.5 m below the surface. What is the largest cargo mass? (c) The fully loaded barge sails into sea water (1025 kg/m³). Does it rise or sink, and by how much?

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Read the steps as text
  1. (a) Floating: ρwVsubg = Mg, so Vsub = M/ρw = 20 000 kg / 1000 kg/m³ = 20 m³. Depth = 20 m³ / (10 m × 4.0 m) = 0.50 m. Why: g cancels; the under-water part is a box with base 40 m².
  2. (b) At 1.5 m: Vsub = 40 m² × 1.5 m = 60 m³, so total mass = 1000 × 60 = 60 000 kg. Cargo = 60 000 − 20 000 = 40 000 kg. Why: the water pushed aside must weigh as much as barge plus cargo.
  3. (c) Same total mass in denser water: depth = 60 000 kg / (1025 kg/m³ × 40 m²) = 1.463 m. It rises by 1.500 − 1.463 = 0.037 m, about 3.7 cm. Why: denser water gives the same buoyant force with less volume under.

5. A ball let go under water AP

A ball of volume 1.0 × 10⁻³ m³ and density 500 kg/m³ is held at the bottom of a pool and let go. Ignore drag. What is its acceleration just after release?

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Read the steps as text
  1. Mass m = 500 × 1.0 × 10⁻³ = 0.50 kg; Fg = 0.50 × 9.8 = 4.9 N. Why: weight from density and volume.
  2. Fb = 1000 × 1.0 × 10⁻³ × 9.8 = 9.8 N. Why: fully under water.
  3. Up positive: a = (Fb − Fg)/m = (9.8 − 4.9) N / 0.50 kg = 9.8 m/s², upward. Why: Newton's second law. In real water, drag would make it smaller once the ball is moving.

AP-style practice

Use g = 9.8 m/s² and ρwater = 1000 kg/m³ unless told otherwise. Up is positive.

1. A steel cube and a wooden cube have the same volume. Both are put in a tank of water. The steel cube rests on the bottom. The wooden cube floats with part of it above the water. Which cube has the larger buoyant force on it?

Show answer

(A). The steel cube is fully under, so it pushes aside its whole volume of water. The wooden cube pushes aside only part of its volume. Fb = ρfluidVsubg depends on Vsub, not the total volume and not the mass.

2. A block floats in water with three quarters of its volume under the surface. What is the density of the block?

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(B). Floating: Vsub/V = ρblock/ρwater, so ρblock = 0.75 × 1000 = 750 kg/m³.

3. A sealed metal can is fully under water and is pulled from 1 m deep to 5 m deep. Assume the can and the water do not compress. What happens to the buoyant force on the can?

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(C). The pressure on both the top and the bottom grows by the same amount, so their difference does not change. Equivalently, Fb = ρVg and none of ρ, V, g change.

4. A toy boat floats in a bathtub. A 0.20 kg stone is placed in the boat, and the boat still floats. By how much does the buoyant force on the boat increase?

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(C). A floating object has Fb = total weight. The weight grows by 0.20 kg × 9.8 m/s² = 1.96 N, so Fb grows by 1.96 N. The boat sinks a little lower to push aside 0.20 kg more water.

5. A rock of weight Fg rests on the bottom of a tank of water. The buoyant force on it is Fb. What is the normal force from the tank bottom on the rock?

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(B). Three forces, at rest, up positive: FN + Fb − Fg = 0, so FN = Fg − Fb.

6. Water flows to the right in a level pipe and speeds up as it goes. Which statement about the pressure in the water is correct?

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(A). A bit of water speeds up to the right, so the net force on it points right (Newton's second law). The only horizontal forces are from pressure, so the push from the left side must be bigger: higher pressure on the left.

7. (Short free response, calculation) A spring scale reads 30.0 N when a metal block hangs in air, and 20.0 N when the block is fully under water. (a) Find the volume of the block. (b) Find its density.

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  1. Fb = 30.0 N − 20.0 N = 10.0 N. Why: T = Fg − Fb, so the drop in reading is Fb.
  2. (a) V = Fb/(ρwg) = 10.0 N / (1000 × 9.8) = 1.02 × 10⁻³ m³. Why: fully under, so Vsub = V.
  3. (b) m = 30.0 N / 9.8 m/s² = 3.06 kg, so ρ = 3.06 kg / 1.02 × 10⁻³ m³ = 3000 kg/m³. Shortcut: ρ/ρw = Fg/Fb = 30/10 = 3.

8. (Qualitative/quantitative translation) A rectangular block of height h and top area A is fully under a liquid of density ρ, with its top face at depth d. (a) In words, explain why the liquid exerts a net upward force on the block. (b) Using P = P0 + ρgh, derive an expression for the net force from the liquid in terms of ρ, g, A and h. (c) Explain how your expression shows that the buoyant force does not depend on d.

Show answer
  1. (a) Liquid pressure grows with depth. The bottom face is deeper than the top face, so the liquid pushes up on the bottom harder than it pushes down on the top. Side pushes cancel. The net force is up.
  2. (b) Top: Ptop = P0 + ρgd, force down = (P0 + ρgd)A. Bottom: Pbot = P0 + ρg(d + h), force up = (P0 + ρg(d + h))A. Net up: Fb = ρghA = ρVg, where V = Ah.
  3. (c) The d terms (and P0) cancel when we subtract. Only h, the block's own height, is left, so Fb is the same at any depth.

9. (Experimental design) You have a spring scale, a set of metal cylinders of known volumes, a beaker of an unknown liquid, and string. Describe a procedure to find the density of the liquid using a graph.

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  1. For each cylinder, hang it from the scale in air and record Fg. Then lower it until it is fully under the liquid (not touching the bottom or sides) and record the reading T.
  2. Compute Fb = Fg − T for each cylinder. Repeat each reading three times and average.
  3. Plot Fb (vertical) against volume V (horizontal). Draw a best-fit straight line.
  4. The theory is Fb = (ρg)V, so the slope is ρg. The density is ρ = slope / 9.8 m/s². The line should pass close to the origin; if not, check for a zero error in the scale.

Free response (Translation between representations). A 0.50 kg metal block, 0.050 m tall with volume 2.0 × 10⁻⁴ m³, hangs from a spring scale. It is slowly lowered into water (1000 kg/m³) until it is fully under.

(a) Describe the free-body diagram of the block when it is fully under water: name each force and its direction.

(b) Calculate the scale reading when the block is fully under water.

(c) Describe the graph of scale reading against how far the block's bottom is below the surface, from 0 to 0.10 m.

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(a) Three forces: weight Fg = mg straight down; tension T from the scale straight up; buoyant force Fb straight up. Up arrows together are as long as the down arrow (at rest).

(b) Fg = 0.50 × 9.8 = 4.9 N. Fb = 1000 × 2.0 × 10⁻⁴ × 9.8 = 1.96 N. T = 4.9 − 1.96 ≈ 2.9 N.

(c) It starts at 4.9 N and falls in a straight line while the block goes in (Vsub grows steadily), reaching 2.9 N at 0.050 m. From 0.050 m to 0.10 m it is flat at 2.9 N: once fully under, Fb does not depend on depth.

Point guide (4 points): 1 for all three forces with directions; 1 for Fb = 1.96 N; 1 for 2.9 N; 1 for the falling then flat graph with reason.

Free response (Mathematical routines) A 0.40 kg block of wood with volume 6.0 × 10⁻⁴ m³ is held fully under water by a string tied to the bottom of a tank. Water has ρ = 1000 kg/m³.

(a) Calculate the buoyant force on the block.

(b) Calculate the tension in the string.

(c) The string is cut. Calculate the block's acceleration just after the cut.

(d) When the block floats at rest, what fraction of its volume is under water?

Show answer

(a) FB = ρVg = 1000 × 6.0 × 10⁻⁴ × 9.8 = 5.88 N.

(b) Weight = 0.40 × 9.8 = 3.92 N. FB = W + T, so T = 5.88 − 3.92 = 1.96 N.

(c) Fnet = 1.96 N up, so a = 1.96 / 0.40 = 4.9 m/s² upward.

(d) Floating: ρVsubg = mg, so Vsub = 0.40 / 1000 = 4.0 × 10⁻⁴ m³, which is 2/3 of the block.

Point guide (4 points): 1 for FB = 5.88 N; 1 for T = 1.96 N; 1 for 4.9 m/s² up; 1 for 2/3 submerged.

Common mistakes

The mistake: using the object's density in Fb = ρVg.

Why it is wrong: the buoyant force is the weight of the fluid pushed aside. It does not know what the object is made of.

How to spot it: if two objects of the same submerged volume get different buoyant forces in your answer, check which ρ you used.

The mistake: using the whole volume of a floating object.

Why it is wrong: only the part under the surface pushes fluid aside. Fb uses Vsub.

How to spot it: if your Fb for a floating object is bigger than its weight, you used the total volume. Floating at rest means Fb = Fg exactly.

The mistake: "deeper means more buoyant force".

Why it is wrong: deeper means more pressure on top and on the bottom. The difference, which is what matters, stays the same for an incompressible object and fluid.

How to spot it: once an object is fully under, the depth d should drop out of your answer.

The mistake: "the buoyant force always equals the weight".

Why it is wrong: that is only true for an object floating at rest with no other forces. A rock on the bottom or a block on a string has Fb less than its weight; a ball rising has Fb more than its weight.

How to spot it: always draw the free body diagram first. If there is a third force (string, normal force, your hand), Fb ≠ Fg.

The mistake: "heavy things sink and light things float".

Why it is wrong: what matters is average density compared with the fluid. A 50 000-tonne ship floats; a 5 g steel bolt sinks.

How to spot it: if your reasoning uses mass alone, rewrite it with density: ρobject < ρfluid floats.

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