Mistake Master
Mistake Master
Drop a wooden block in water and it floats. Drop a steel block in water and it sinks. Drop a steel block in mercury and it floats again. The fluid pushes up on whatever sits in it with a buoyant force $F_b = \rho_\text{fluid} V_\text{disp} g$, and Newton's second law settles the rest. The trap: that formula uses the fluid's density, not the object's.
The buoyant force is the net upward push a fluid exerts on an object sitting in it. The microscopic story: fluid pressure rises with depth (that's the $P = \rho g h$ result from Topic 8.2), so the bottom of a submerged object sits in higher-pressure fluid than the top. The fluid pushes inward on every surface, but the larger push from below beats the smaller push from above, and the difference is a net upward force.
Archimedes pinned the magnitude:
$$F_b \;=\; \rho_\text{fluid} \, V_\text{disp} \, g$$The buoyant force on an object equals the weight of the fluid the object displaces. Two things to notice. The density is the fluid's density, never the object's. The volume is the displaced volume: what the object pushes out of the way. For a fully submerged object, $V_\text{disp} = V_\text{obj}$. For a floating object with part of its volume above the surface, $V_\text{disp}$ is the part below the waterline.
The formula is dead specific about whose properties matter. The object's color, mass, and density do not appear. Only the fluid's density and the displaced volume do. A steel ball and a wood ball with the same shape, both fully submerged in the same lake, feel the same buoyant force. They behave differently because their weights differ, not because the fluid pushes on them differently.
Apply Newton's second law to a submerged object. Only two forces matter: the buoyant force up and the weight down. The net is
$$F_\text{net} \;=\; F_b - W$$Three outcomes follow.
If $F_b > W$, the object rises. The net force is upward and the object accelerates that way. It does not rise forever; once part of it breaks the surface, the displaced volume drops, $F_b$ shrinks, and the object settles into a floating equilibrium with $F_b = W$.
If $F_b = W$, the object is in equilibrium. Either it floats with the right fraction above water, or, if fully submerged with matching density, it hovers.
If $F_b < W$, the object sinks. The net force is downward and the object accelerates that way until it hits the bottom or some external force catches it.
For a fully submerged object, substitute $F_b = \rho_\text{fluid} V g$ and $W = \rho_\text{obj} V g$. The $V$ and $g$ cancel, leaving a clean comparison:
$$\rho_\text{obj} < \rho_\text{fluid} \;\Longrightarrow\; \text{floats}; \quad \rho_\text{obj} > \rho_\text{fluid} \;\Longrightarrow\; \text{sinks}$$Density-vs-density, not weight-vs-weight. A grain of sand sinks in water; a ten-thousand-tonne tanker floats. Density decides.
A wooden block of volume $V = 0.010 \text{ m}^3$ and density $\rho_\text{wood} = 600 \text{ kg/m}^3$ floats in freshwater ($\rho_\text{water} = 1000 \text{ kg/m}^3$). Use $g = 10 \text{ N/kg}$. What fraction of the block is below the waterline, and what is the buoyant force on it?
Weight. $W = \rho_\text{wood} \, V \, g = (600)(0.010)(10) = 60 \text{ N}$.
Floating equilibrium. $F_b = W$, so $F_b = 60 \text{ N}$.
Submerged fraction. From $F_b = \rho_\text{water} V_\text{disp} g$, solve for the displaced volume: $V_\text{disp} = F_b / (\rho_\text{water} g) = 60 / (1000 \cdot 10) = 0.006 \text{ m}^3$. The fraction is $V_\text{disp} / V = 0.006 / 0.010 = 0.60$.
Read it. Equivalently, $V_\text{disp}/V = \rho_\text{wood}/\rho_\text{water} = 600/1000 = 0.60$. Both routes hit $60\%$ submerged.
For a floating object in equilibrium, the submerged fraction equals the density ratio: $V_\text{disp}/V_\text{obj} = \rho_\text{obj}/\rho_\text{fluid}$. Less-dense object sits higher; more-dense object sinks the moment that ratio crosses one.
Iron has density $7870 \text{ kg/m}^3$, far above water's $1000$. A solid iron block dropped into a lake sinks. Yet a $5000$-tonne iron ship sits happily on the ocean. How?
The trick is average density. A ship's hull encloses a huge volume of air, and the density that matters for the density-comparison rule is the total mass divided by the total enclosed volume, hull plus air. For most ships that number is well below $1000 \text{ kg/m}^3$, and the ship floats.
Punch a hole in the hull. Water replaces the air inside, raising the average density of the hull-plus-contents. Once that average density exceeds water's $1000 \text{ kg/m}^3$, the ship sinks.
The same logic explains why a solid iron block sinks in water but floats in mercury ($\rho = 13600 \text{ kg/m}^3$). Iron's $7870$ is above $1000$ (sinks in water) but below $13600$ (floats in mercury). The fluid sets the bar.
"Average density of the object" is what to compare to the fluid. For a solid uniform object that's just its material density. For a hollow or compound object, count the empty interior in the volume.
It's tempting to think the fluid pushes up harder on heavier objects, the way the ground does. The ground does that. The fluid does not. The buoyant force depends on the fluid's density and the displaced volume, not on the object's own properties. A $1 \text{ kg}$ foam cube displaces about $10 \text{ L}$ of water and gets $100 \text{ N}$ of lift. A $1 \text{ kg}$ iron cube displaces about $0.13 \text{ L}$ and gets only $1.3 \text{ N}$. Same weight, very different $F_b$.
Fix. $F_b = \rho_\text{fluid} V_\text{disp} g$. The object's own density does not appear. To change $F_b$, change the fluid or change how much of the object is submerged.
A grain of sand weighing a fraction of a gram sinks in water; a five-thousand-tonne cargo ship floats. Weight alone cannot tell you which way an object will go in a fluid. Density-vs-density does.
Fix. Compare $\rho_\text{obj}$ to $\rho_\text{fluid}$. Less dense than the fluid: floats. More dense: sinks. For a hollow object like a ship, use the average density (mass divided by total enclosed volume, hull plus air inside).
That equality is the floating-equilibrium condition, not a universal rule. A sinking block has $F_b$ less than $W$; that's the whole reason there is a net downward force on it. A balloon rising through air has $F_b$ greater than $W$. A buoy held underwater by a tether is the same story: $F_b > W$, and the tether pulls down to balance.
Fix. $F_b = W$ only when the object is floating in equilibrium (or hovering, neutrally buoyant). For any other case, compute $F_b$ and $W$ separately and apply Newton's second law.
Ten scenarios mixing focus-and-choice, symbolic, and numeric prompts. Each one targets a specific way buoyancy goes sideways. Pick the right option and you'll see the trap baked into the others. Your progress saves in the browser as you go.