AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/8/8-3)
Unit 8 · Topic 8.3
8.3 Fluids and Newton’s Laws
Fluids obey Newton's laws: a bit of fluid speeds up, slows down or turns only when the forces on it are unbalanced. Because pressure grows with depth, a fluid pushes up harder on the bottom of a submerged object than down on its top, giving an upward buoyant force equal to the weight of the fluid the object displaces, F_b = ρVg.
Key terms
- buoyant force
- Archimedes' principle
- displaced fluid
- fluid density
- free-body diagram
Newton's laws inside a fluid
Think of a small parcel of fluid as an object. The fluid around it pushes on every side, and gravity pulls it down. If those forces balance, the parcel stays at rest or keeps moving at constant velocity. If they don't, it accelerates in the direction of the net force.
In a still fluid, the pressure below each parcel is a little higher than the pressure above it, by just enough to support its weight. That balance is why pressure increases with depth (8.2).
In a moving fluid, a parcel speeds up when it moves from a region of higher pressure toward lower pressure, because the push from behind is bigger than the push from ahead. That idea returns in 8.4.
Where buoyancy comes from
Hold a block of height h and top area A completely underwater. The water pushes down on the top and up on the bottom. The bottom is deeper, so the pressure there is higher by ρgh. The sideways pushes cancel.
The net upward force is the buoyant force: F_b = (ΔP)A = ρghA = ρVg, where V = Ah is the block's volume and ρ is the density of the fluid, not the block.
ρV is the mass of fluid that would fill the space the object takes up, so ρVg is the weight of that displaced fluid. This is Archimedes' principle: the buoyant force equals the weight of the fluid displaced. For a partly submerged object, V is only the volume below the surface.
What the buoyant force depends on
- The fluid's density: the same object feels a bigger buoyant force in seawater than in fresh water.
- The submerged volume: push more of a floating object under and the buoyant force grows.
- Not the object's own mass or density: a steel ball and a plastic ball of the same volume, both fully underwater, feel the same buoyant force.
- Not the depth, once the object is fully submerged: the top and bottom pressures both rise by the same amount, so their difference stays the same.
Sink, float or hover
Draw a free-body diagram with the weight mg down and the buoyant force up, plus any string tension or normal force. For an object fully underwater, compare densities: if ρ_object > ρ_fluid, its weight beats the buoyant force and it sinks; if it's less, it rises; if they're equal, it hovers.
A floating object sinks until the buoyant force equals its weight. Setting ρ_fluid V_sub g = ρ_object V g gives V_sub/V = ρ_object/ρ_fluid: the fraction underwater equals the density ratio.
An object hanging from a scale underwater seems lighter. The scale reads the tension, T = mg − F_b, sometimes called the apparent weight.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Weighing a block underwater
A 0.50 kg aluminum block (ρ = 2700 kg/m³) hangs from a spring scale, fully submerged in water (ρ = 1000 kg/m³). Use g = 9.8 m/s². Find the buoyant force and the scale reading.
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- Step 1: Volume of the block: V = m/ρ = 0.50/2700 ≈ 1.85 × 10⁻⁴ m³. It's fully submerged, so it displaces that much water.
- Step 2: F_b = ρ_water V g = (1000)(1.85 × 10⁻⁴)(9.8) ≈ 1.81 N upward.
- Step 3: Forces on the block: weight mg = 4.9 N down, buoyant force 1.81 N up, tension T up. It's at rest, so T + F_b = mg.
- Step 4: T = 4.9 − 1.81 ≈ 3.1 N. The scale reads about 3.1 N instead of 4.9 N.
Answer: F_b ≈ 1.8 N upward; the scale reads about 3.1 N
- Example 2Calculator allowed
How much of an iceberg is underwater?
Ice has a density of 917 kg/m³. What fraction of a floating iceberg is below the surface in seawater (1025 kg/m³)? In fresh water (1000 kg/m³)?
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- Step 1: Floating means the buoyant force equals the weight: ρ_fluid V_sub g = ρ_ice V g.
- Step 2: So V_sub/V = ρ_ice/ρ_fluid.
- Step 3: Seawater: 917/1025 ≈ 0.89, so about 89% is underwater.
- Step 4: Fresh water: 917/1000 ≈ 0.92, so about 92%. Less dense water gives less buoyant force per cubic meter, so more of the ice must sit below the surface.
Answer: About 89% in seawater and 92% in fresh water
- Example 3Calculator allowed
Releasing a block held underwater (classic trap)
A 2.0 × 10⁻³ m³ block of wood (ρ = 600 kg/m³) is held completely underwater and then released. Use g = 9.8 m/s². Find its acceleration just after release.
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- Step 1: The trap is to say the buoyant force equals the weight. That's true only for an object floating at rest, not for one held under.
- Step 2: Mass: m = ρV = (600)(2.0 × 10⁻³) = 1.2 kg, so its weight is (1.2)(9.8) = 11.76 N down.
- Step 3: Fully submerged, it displaces 2.0 × 10⁻³ m³ of water: F_b = (1000)(2.0 × 10⁻³)(9.8) = 19.6 N up.
- Step 4: Net force = 19.6 − 11.76 = 7.84 N up, so a = 7.84/1.2 ≈ 6.5 m/s² upward (ignoring drag from the water).
Answer: About 6.5 m/s² upward
Common mistakes
- Using the object's density in F_b = ρVg. Use the density of the fluid.
- Assuming the buoyant force always equals the object's weight. That's only for an object floating in equilibrium.
- Thinking a fully submerged object feels more buoyant force the deeper it goes. For an incompressible fluid, it's the same at any depth.
- Using the whole volume of a floating object instead of only the submerged part.
On the exam
- Free-body diagrams are central here. Draw the buoyant force as a single upward arrow from the fluid, separate from any tension or normal force, and label every force.
- Expect questions that compare buoyant forces on objects of the same volume but different densities, or the same object in different fluids, along with an explanation in terms of the pressure difference between top and bottom.
Connected topics
Videos
Check yourself
4 questions on 8.3 Fluids and Newton’s Laws. Pick an answer to see if you got it, and why.
A block with a volume of 2.0 × 10⁻³ m³ is held completely under water (ρ = 1000 kg/m³). What is the magnitude of the buoyant force on it? Use g = 9.8 m/s².
A block of aluminum and a block of lead have the same volume. Both are held completely under water. How do the buoyant forces on them compare?
A steel ball is dropped into a deep lake and sinks. Treat the water as an ideal fluid. How does the buoyant force on the ball change as it sinks from 2 m deep to 20 m deep?
A cube with edges of length s is held completely under a liquid of density ρ, with its top face horizontal at depth h. Which expression gives the magnitude of the net upward force the liquid exerts on the cube?
0 of 4 answered