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8.3 - Fluids and Newton’s Laws

Newton’s Laws Applied to Fluids

Newton’s Laws describe motion within fluids too, and the key force to understand here is the buoyant force.
A fluid is made of particles, and each small chunk of fluid (a fluid parcel) obeys Newton’s laws just like a block or a cart. The forces on a parcel come from two places: the surrounding fluid pushing on it (these pressure forces come from the interactions between the fluid’s particles) and outside forces like gravity, a pump, a piston, or the walls of a pipe. The way a fluid moves as a whole, like water speeding up through a nozzle or sitting still in a glass, comes from adding up these forces.
When does a fluid’s velocity change? Only when the net force on it isn’t zero. If the pressure is the same on both sides of a parcel in a horizontal pipe, the pressure forces cancel and the fluid keeps moving at constant velocity (Newton’s first law). If the pressure behind the parcel is greater than the pressure in front of it, the net force points forward and the fluid speeds up (Newton’s second law). If the pressure ahead is greater, the fluid slows down. The walls of a pipe can push on the fluid too, which is how a bend in a pipe changes the fluid’s direction.
8.3 - Fluids and Newton’s Laws
Fluid at rest (left): a column of still water has zero net force. The push up on its bottom (PbottomAP_{\text{bottom}}A) has to balance the push down on its top (PtopAP_{\text{top}}A) plus its weight (ρAhg\rho A h g). Solving gives Pbottom−Ptop=ρghP_{\text{bottom}} - P_{\text{top}} = \rho g h, which is exactly why pressure increases with depth (8.2).
Fluid speeding up (right): if P1>P2P_1 > P_2, the net force on the parcel is (P1−P2)A(P_1 - P_2)A, pointing forward, so the parcel accelerates forward. This is why fluid speeds up when it flows from a region of high pressure into a region of low pressure, the idea behind Bernoulli’s equation in 8.4.

Worked Example: A Pressure Difference Accelerates Water

Water flows through a horizontal pipe with cross-sectional area 2.0×10−32.0 \times 10^{-3} m². Look at a 0.10 m long parcel of the water. The pressure just behind it is 1.30×1051.30 \times 10^5 Pa, and the pressure just ahead of it is 1.28×1051.28 \times 10^5 Pa. Find the parcel’s acceleration.
Net force: Fnet=(P1−P2)A=(1.30×105−1.28×105)(2.0×10−3)=(2.0×103)(2.0×10−3)=4.0F_{\text{net}} = (P_1 - P_2)A = (1.30 \times 10^5 - 1.28 \times 10^5)(2.0 \times 10^{-3}) = (2.0 \times 10^3)(2.0 \times 10^{-3}) = 4.0 N, forward.
Mass of the parcel: m=ρV=ρAL=(1000)(2.0×10−3)(0.10)=0.20m = \rho V = \rho A L = (1000)(2.0 \times 10^{-3})(0.10) = 0.20 kg.
a=Fnetm=4.00.20=20 m/s2a = \frac{F_{\text{net}}}{m} = \frac{4.0}{0.20} = 20 \text{ m/s}^2
The parcel speeds up at 20 m/s² in the direction of flow. Notice that the size of the pressure (about 1.3×1051.3 \times 10^5 Pa) doesn’t matter, only the difference across the parcel does. The pipe is the same width everywhere, so all the water in it speeds up together, the way it does just after a tap is opened. Once the flow is steady, an ideal fluid needs no pressure difference to keep moving through a level pipe of constant width.

Buoyant Force

The buoyant force is the net upward force a fluid exerts on an object. There are three scenarios: if buoyant force equals weight, the object floats in equilibrium (neutrally buoyant), like a submarine hovering at constant depth. If buoyant force exceeds weight, the object rises to the surface, like a helium balloon in air or wood in water. If buoyant force is less than weight, the object sinks, like a rock in water, or denser fluids settling below less dense ones.
Physically, the buoyant force is the collective result of all the fluid particles pressing on every surface of the object, it’s not one single force, but the sum of countless microscopic forces from the surrounding fluid.
8.3 - Fluids and Newton’s Laws

Where the Buoyant Force Comes From

The fluid pushes on every surface of a submerged object, always perpendicular to that surface. The pushes on the sides cancel, since opposite sides are at the same depth and feel the same pressure. But the bottom of the object is deeper than the top, so the pressure on the bottom is greater. The upward push on the bottom beats the downward push on the top, and the difference is the buoyant force.
8.3 - Fluids and Newton’s Laws
For a block with top and bottom area AA, with its top at depth h1h_1 and its bottom at depth h2h_2, the pressures are P1=P0+ρgh1P_1 = P_0 + \rho g h_1 and P2=P0+ρgh2P_2 = P_0 + \rho g h_2, where ρ\rho is the fluid’s density. So:
Fb=P2A−P1A=ρg(h2−h1)A=ρgVF_b = P_2A - P_1A = \rho g (h_2 - h_1) A = \rho g V
since (h2−h1)A(h_2 - h_1)A is the block’s volume. The surface pressure P0P_0 cancels, and so does the depth: lowering the block deeper raises P1P_1 and P2P_2 by the same amount, so their difference, the buoyant force, doesn’t change. This is Archimedes’ principle (next), built straight from pressure.

Archimedes’ Principle

Archimedes’ Principle states that the magnitude of the buoyant force on an object equals the weight of the fluid it displaces.
Fb=weight of fluid displacedF_b = \text{weight of fluid displaced}
Deriving the buoyant force equation: starting from weight, Fg=mgF_g = mg, and density, ρ=mV  ⟹  m=ρV\rho = \frac{m}{V} \implies m = \rho V, substituting gives Fg=(ρV)gF_g = (\rho V)g. For buoyant force, we use the density and volume of the displaced fluid:
Fb=ρfluidVdisplacedgF_b = \rho_{\text{fluid}} V_{\text{displaced}} g
where ρfluid\rho_{\text{fluid}} is the density of the surrounding fluid (not the object), VdisplacedV_{\text{displaced}} is the volume of fluid displaced, and gg is gravitational acceleration. This distinction is critical, always use the fluid’s density, never the object’s own density.
On the AP equation sheet this appears as Fb=ρVgF_b = \rho V g. The sheet doesn’t say it, but ρ\rho is the fluid’s density and VV is the volume of fluid displaced (the part of the object that’s underwater).

Historical Context: Archimedes

Legend has it that King Hiero II asked Archimedes to figure out whether his crown was pure gold without damaging it. Archimedes realized that submerging the crown in water and comparing the displaced water volume to what pure gold of that mass would displace could reveal if lighter metals had been mixed in. According to the story, he discovered this idea while taking a bath, noticed the water level rise, and ran through the streets shouting “Eureka!” (“I have found it!”). Archimedes is also said to have worked on large ship designs and floating structures, which relied heavily on understanding buoyant forces. Today, his principle underlies ship design, submarine operation, hot air balloons, and measuring the density of irregularly shaped objects.

Free Body Diagrams with Buoyant Force

Always include buoyant force in a free body diagram for objects in fluids.
8.3 - Fluids and Newton’s Laws
In general:
∑Fy=Fb−Fg=may\sum F_y = F_b - F_g = ma_y
For equilibrium (floating or suspended):
Fb=Fg  ⟹  ρfluidVdisplacedg=mobjectg  ⟹  ρfluidVdisplaced=mobjectF_b = F_g \implies \rho_{\text{fluid}} V_{\text{displaced}} g = m_{\text{object}} g \implies \rho_{\text{fluid}} V_{\text{displaced}} = m_{\text{object}}
The buoyant force is just one more force on the diagram. If anything else pushes or pulls on the object (a string, a hand, a spring scale, the bottom of the tank), draw that force too and apply Newton’s second law the same way.

Worked Example: Block Held Underwater by a String

A 0.60 kg block of wood (density 600 kg/m³) is tied to the bottom of a tank by a string, so it’s held completely underwater. Find the buoyant force, the tension in the string, and the block’s acceleration right after the string is cut. Use g=10g = 10 m/s².
8.3 - Fluids and Newton’s Laws
Volume: V=mρwood=0.60600=1.0×10−3V = \frac{m}{\rho_{\text{wood}}} = \frac{0.60}{600} = 1.0 \times 10^{-3} m³. The block is fully underwater, so Vdisplaced=VV_{\text{displaced}} = V.
Buoyant force: Fb=ρwaterVg=(1000)(1.0×10−3)(10)=10F_b = \rho_{\text{water}} V g = (1000)(1.0 \times 10^{-3})(10) = 10 N, upward.
Tension: the block is at rest, so ∑Fy=0\sum F_y = 0. Up: FbF_b. Down: Fg=mg=(0.60)(10)=6.0F_g = mg = (0.60)(10) = 6.0 N, and the tension TT (the string pulls the block down toward the bottom of the tank).
Fb−Fg−T=0  ⟹  T=Fb−Fg=10−6.0=4.0 NF_b - F_g - T = 0 \implies T = F_b - F_g = 10 - 6.0 = 4.0 \text{ N}
Right after the string is cut: the tension disappears, but FbF_b and FgF_g don’t change, so Fnet=10−6.0=4.0F_{\text{net}} = 10 - 6.0 = 4.0 N upward and a=4.00.60≈6.7a = \frac{4.0}{0.60} \approx 6.7 m/s² upward (ignoring water resistance). The block rises until it floats at the surface with Fb=FgF_b = F_g, with 6001000=60%\frac{600}{1000} = 60\% of its volume underwater.

Worked Example: Buoyant Force on a Weather Balloon

A weather balloon has volume V=3.5V = 3.5 L =3.5×10−3= 3.5 \times 10^{-3} m³, filled with hydrogen. Air density is ρair=1.29\rho_{\text{air}} = 1.29 kg/m³, hydrogen density is ρH2=0.0899\rho_{H_2} = 0.0899 kg/m³, and the balloon material has mass 2.0×10−32.0 \times 10^{-3} kg. Use g=10g = 10 m/s². Find the buoyant force in air, and the net vertical force when released.
Buoyant force: the balloon displaces air, so we use air’s density (not the density of what’s inside):
Fb=ρairVg=(1.29)(3.5×10−3)(10)≈0.0452 NF_b = \rho_{\text{air}} V g = (1.29)(3.5 \times 10^{-3})(10) \approx 0.0452 \text{ N}
Net vertical force: three forces act here: buoyant force (up, 0.0452 N), the weight of the balloon material (down), and the weight of the hydrogen inside (down).
Mass of hydrogen: mH2=ρH2V=(0.0899)(3.5×10−3)≈3.15×10−4m_{H_2} = \rho_{H_2} V = (0.0899)(3.5 \times 10^{-3}) \approx 3.15 \times 10^{-4} kg, giving weight Fg,H2=(3.15×10−4)(10)=3.15×10−3F_{g,H_2} = (3.15 \times 10^{-4})(10) = 3.15 \times 10^{-3} N.
Weight of the balloon material: Fg,balloon=(2.0×10−3)(10)=0.0200F_{g,\text{balloon}} = (2.0 \times 10^{-3})(10) = 0.0200 N.
Net force:
Fnet=Fb−Fg,balloon−Fg,H2=0.0452−0.0200−0.00315≈0.0220 NF_{\text{net}} = F_b - F_{g,\text{balloon}} - F_{g,H_2} = 0.0452 - 0.0200 - 0.00315 \approx 0.0220 \text{ N}
The net upward force is about 0.022 N, so the balloon rises.
Finding acceleration (an extension): total mass is mtotal=2.0×10−3+3.15×10−4≈2.315×10−3m_{\text{total}} = 2.0 \times 10^{-3} + 3.15 \times 10^{-4} \approx 2.315 \times 10^{-3} kg, giving:
a=Fnetmtotal=0.02202.315×10−3≈9.50 m/s2a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{0.0220}{2.315 \times 10^{-3}} \approx 9.50 \text{ m/s}^2
The balloon accelerates upward at about 9.5 m/s².

Worked Example: Determining if an Object Floats

A solid plastic sphere has mass 0.45 kg and radius 0.080 m. Water density is 1000 kg/m³. Does it float?
Comparing densities: Vsphere=43π(0.080)3≈2.14×10−3V_{\text{sphere}} = \frac{4}{3}\pi(0.080)^3 \approx 2.14 \times 10^{-3} m³, giving ρsphere=0.452.14×10−3≈210\rho_{\text{sphere}} = \frac{0.45}{2.14 \times 10^{-3}} \approx 210 kg/m³. Since 210 kg/m³ is less than water’s 1000 kg/m³, the sphere floats.
Comparing forces (checking): the buoyant force if fully submerged would be Fb=ρwaterVsphereg=(1000)(2.14×10−3)(10)=21.4F_b = \rho_{\text{water}} V_{\text{sphere}} g = (1000)(2.14 \times 10^{-3})(10) = 21.4 N, while the weight is Fg=(0.45)(10)=4.5F_g = (0.45)(10) = 4.5 N. Since FbF_b exceeds FgF_g, it floats, confirming our density comparison.
How much is submerged: the volume submerged is just enough to satisfy ρwaterVsubmergedg=mg\rho_{\text{water}} V_{\text{submerged}} g = mg:
Vsubmerged=mρwater=0.451000=4.5×10−4 m3V_{\text{submerged}} = \frac{m}{\rho_{\text{water}}} = \frac{0.45}{1000} = 4.5 \times 10^{-4} \text{ m}^3
Fraction submerged: 4.5×10−42.14×10−3≈0.21\frac{4.5 \times 10^{-4}}{2.14 \times 10^{-3}} \approx 0.21, so about 21% of the sphere sits underwater, with 79% above the surface.

Worked Example: Object Suspended in Water

A metal block hangs from a spring scale, reading 45.0 N in air and 39.2 N fully submerged in water. Water density is 1000 kg/m³. Find the buoyant force, the block’s volume, and its density.
Buoyant force: the difference in readings equals the buoyant force:
Fb=Fair−Fwater=45.0−39.2=5.8 NF_b = F_{\text{air}} - F_{\text{water}} = 45.0 - 39.2 = 5.8 \text{ N}
When submerged, buoyancy pushes up, reducing the tension the scale reads.
Volume: from Fb=ρwaterVgF_b = \rho_{\text{water}} V g:
V=Fbρwaterg=5.8(1000)(10)=5.8×10−4 m3=580 cm3V = \frac{F_b}{\rho_{\text{water}} g} = \frac{5.8}{(1000)(10)} = 5.8 \times 10^{-4} \text{ m}^3 = 580 \text{ cm}^3
Density: the block’s mass from its weight in air is m=Fgg=45.010=4.50m = \frac{F_g}{g} = \frac{45.0}{10} = 4.50 kg. So:
ρmetal=mV=4.505.8×10−4≈7.8×103 kg/m3\rho_{\text{metal}} = \frac{m}{V} = \frac{4.50}{5.8 \times 10^{-4}} \approx 7.8 \times 10^3 \text{ kg/m}^3
That’s close to iron or steel (roughly 7,800 kg/m³).

Worked Example: Same Boat, Fresh Water vs. Seawater

A small boat floats in a freshwater lake (ρ=1000\rho = 1000 kg/m³), displacing 2.00 m³ of water. It then sails out into the ocean (ρ=1025\rho = 1025 kg/m³). Does it float higher or lower, and how much seawater does it displace?
Key idea: a floating object always has Fb=FgF_b = F_g. The boat’s weight doesn’t change, so the buoyant force doesn’t change either. Since Fb=ρVgF_b = \rho V g stays the same, ρV\rho V has to stay the same: if the fluid’s density goes up, the displaced volume goes down by the same factor.
Vsea=Vfresh⋅ρfreshρsea=(2.00)(10001025)≈1.95 m3V_{\text{sea}} = V_{\text{fresh}} \cdot \frac{\rho_{\text{fresh}}}{\rho_{\text{sea}}} = (2.00)\left(\frac{1000}{1025}\right) \approx 1.95 \text{ m}^3
The boat floats slightly higher in seawater, displacing about 2.4% less volume. (Its mass is ρfreshVfresh=(1000)(2.00)=2.00×103\rho_{\text{fresh}} V_{\text{fresh}} = (1000)(2.00) = 2.00 \times 10^3 kg in both cases.) This is why ships have load lines painted on their hulls for different kinds of water.

Common Mistakes

Using the object’s density instead of the fluid’s: the buoyant force equation is Fb=ρfluidVgF_b = \rho_{\text{fluid}} V g, always use the surrounding fluid’s density (water, air, oil), never the object’s own density.
Forgetting multiple downward forces: for balloons or hollow objects, remember to include both the weight of the container material and the weight of whatever’s inside, both act downward.
Using total volume instead of displaced volume: for a floating object, only the submerged portion displaces fluid, use VdisplacedV_{\text{displaced}}, not the total object volume (though for a fully submerged object, they’re the same).
Forgetting unit conversions with volume: 1 liter equals 10−310^{-3} m³ (0.001 m³), always convert to m³ before calculating buoyancy.
Thinking buoyant force depends on depth: it depends on volume displaced, not depth, Fb=ρfluidVgF_b = \rho_{\text{fluid}} V g has no depth term. An object fully submerged at 1 m has the same buoyant force as at 10 m (pressure varies with depth, but buoyant force doesn’t for incompressible objects).
Getting the direction wrong: buoyant force always points upward, opposite gravity, even for sinking objects, it’s just smaller than the weight in that case.
Forgetting gg in the buoyant force equation: it’s Fb=ρfluidVgF_b = \rho_{\text{fluid}} V g, don’t drop the gg, buoyancy is fundamentally tied to the weight of displaced fluid.
Comparing the wrong quantities to determine floating: compare ρobject\rho_{\text{object}} to ρfluid\rho_{\text{fluid}}, or FbF_b (if fully submerged) to FgF_g. If ρobject<ρfluid\rho_{\text{object}} < \rho_{\text{fluid}}, it floats; if greater, it sinks.
Saying something floats because it’s “lighter” than water: a steel cargo ship with a mass of thousands of tonnes floats, while a small pebble sinks. What matters is whether the buoyant force can match the weight, which for a solid object comes down to comparing densities. A ship floats because its hull holds a lot of air, so its average density (total mass divided by the total volume it encloses) is less than water’s.
Using the wrong density for gas-filled objects: for a helium balloon in air, use air’s density for buoyancy, for an air bubble underwater, use water’s density, always use the density of whatever fluid the object is in, not what it contains.
Incomplete free body diagrams: always include both FbF_b (up) and FgF_g (down), plus any additional weights (like gas inside a balloon).
Thinking the scale reading is the true weight: when submerged, a scale reads the apparent weight, which equals true weight minus buoyant force, the true weight is what the scale reads in air.
Assuming floating objects are fully submerged: floating objects sit partly above the surface, only calculate buoyant force for the submerged portion, and use ρfluidVsubmergedg=mobjectg\rho_{\text{fluid}} V_{\text{submerged}} g = m_{\text{object}} g for equilibrium.

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