WHAT we need: $F_B={\rho }_f{V}_{\text{disp}}g$. WHY $V_{\text{disp}}=V$: fully submerged means the object displaces its whole volume of water. $F_B=1000\times 2.0\times 10^{-3}\times 9.8=19.6\text{N}$ The force points up, toward the shallower fluid.

WHY we use $850$ not $917$: buoyancy is the fluid pushing, so only the fluid's density enters $F_B$. The object's density decides floating/sinking, not the push size. $F_B=850\times 0.030\times 9.8=249.9\text{N}$

WHAT we do: invert $F_B={\rho }_{f}Vg$ to get $V=F_B/({\rho }_{f}g)$. WHY: we know the push, we want the size. $V=\frac{6.0}{1000\times 9.8}=6.12\times 10^{-4}{\text{m}}^3$

WHY the density ratio: at floating equilibrium the upward push balances the weight, ${\rho }_f(fV)g={\rho }_{o}Vg$. Cancel $Vg$: $f=\frac{{\rho }_o}{{\rho }_f}=\frac{600}{1000}=0.60$ So $60\%$ is underwater, $40\%$ shows above.

WHAT: the "missing" weight is exactly the buoyant force, $F_B=W-W_{\text{app}}=80-52=28\text{N}$. WHY: the scale reads true weight minus the upthrust (Newton's equilibrium: $W_{\text{app}}+F_B=W$, see Newton's laws — equilibrium of forces). $V=\frac{F_B}{{\rho }_{f}g}=\frac{28}{1000\times 9.8}=2.857\times 10^{-3}{\text{m}}^3$

Step 1 — WHAT is the object's density? $V=0.20\times 0.20\times 0.30=0.012{\text{m}}^3,{\rho }_o=\frac{6.0}{0.012}=500{\text{kg/m}}^3$ Since ${\rho }_o=500<1000={\rho }_f$, it floats.

Step 2 — WHY fraction = density ratio: floating equilibrium gives $f={\rho }_o/{\rho }_f=500/1000=0.50$. Half the box is submerged.

Step 3 — WHAT height shows: the vertical side is $0.30\text{m}$; half submerged means $0.15\text{m}$ is underwater, so $h_{\text{above}}=0.30-0.15=0.15\text{m}$ Look at the figure: the blue waterline cuts the box exactly at mid-height.

WHY this is subtle: the water pushes up on the rock with $F_B$; by Newton's third law the rock pushes down on the water with the same $F_B$. That extra downward push lands on the scale. $F_B={\rho }_{f}Vg=1000\times 1.0\times 10^{-4}\times 9.8=0.98\text{N}$ New scale reading: $30+0.98=30.98\text{N}$ The string carries the rock's remaining weight; the scale gains exactly the buoyant force.

Step 1 — WHAT is the maximum upthrust? At the brink of sinking the raft is fully submerged, so it displaces its whole volume: $F_{B,\max }={\rho }_{f}Vg=1000\times 0.50\times 9.8=4900\text{N}$ Step 2 — WHAT weight can that support? The raft's own weight is $W_{\text{raft}}={\rho }_{o}Vg=400\times 0.50\times 9.8=1960\text{N}$ Step 3 — WHY subtract: the extra load's weight must fit in the leftover buoyancy: $W_{\text{load}}=F_{B,\max }-W_{\text{raft}}=4900-1960=2940\text{N}$ $m_{\text{load}}=\frac{2940}{9.8}=300\text{kg}$

In the boat (floating): the ball's weight is displaced as water. Volume of water displaced: $V_1=\frac{m}{{\rho }_f}=\frac{0.50}{1000}=5.0\times 10^{-4}{\text{m}}^3$ Sunk on the bottom: the ball now displaces only its own volume: $V_2=\frac{m}{{\rho }_o}=\frac{0.50}{7870}=6.35\times 10^{-5}{\text{m}}^3$ WHY it falls: since iron is denser than water, $V_2<V_1$. Sunk, it shoves aside less water, so the level drops by the equivalent of $\Delta V=V_1-V_2=5.0\times 10^{-4}-6.35\times 10^{-5}=4.365\times 10^{-4}{\text{m}}^3$

WHY split: buoyancy now comes from two fluids. Let fraction $x$ be in water, so $(1-x)$ is in oil (assume the cube spans the interface, none in air). Equilibrium — total upthrust equals weight: ${\rho }_{\text{oil}}(1-x)Vg+{\rho }_{\text{water}}xVg={\rho }_{o}Vg$ Cancel $Vg$: $800(1-x)+1000x=900\Rightarrow 800+200x=900\Rightarrow x=0.50$ So $50\%$ sits in water, $50\%$ in oil. The figure shows the cube straddling the line, half in each.

Case ${\rho }_f\to 0$: $F_B={\rho }_{f}Vg\to 0$. No fluid, no push — the object simply falls. The "submerged fraction" formula $f={\rho }_o/{\rho }_f\to \infty$, which is nonsense: it signals the object cannot float at all (it needs $f>1$, impossible), i.e. it sinks. WHY the formula breaks: $f\le 1$ is required physically; $f={\rho }_o/{\rho }_f>1$ means "would need to displace more than its own volume," so it sinks fully. This matches the sink condition ${\rho }_o>{\rho }_f$.

Case ${\rho }_f={\rho }_o$: then $f=1$ exactly — the object is neutrally buoyant, floating fully submerged at any depth, in equilibrium wherever you place it. $F_B={\rho }_{f}Vg=W$ exactly.

The scale reads $W_{\text{app}}=W-F_B$, with $F_B={\rho }_f{V}_{\text{disp}}g$. Only $V_{\text{disp}}$ changes.

• (a) Just touching: $V_{\text{disp}}=0\Rightarrow F_B=0$, reading $=12\text{N}$.
• (b) Half in: $V_{\text{disp}}=1.0\times 10^{-3}$, $F_B=1000\times 1.0\times 10^{-3}\times 9.8=9.8\text{N}$, reading $=12-9.8=2.2\text{N}$.
• (c) Fully in: $V_{\text{disp}}=2.0\times 10^{-3}$, $F_B=19.6\text{N}$. But $F_B>W$! The upthrust exceeds the weight, so the spring can't pull down — reading would be $12-19.6=-7.6\text{N}$, i.e. the ball tugs upward on the scale with $7.6\text{N}$. WHY: once submerged the ball is buoyant enough to float; the scale (if it can register tension both ways) reads a negative "weight," meaning the string is now taut upward.