Exercises — Hydrostatics — pressure = ρgh, derivation

WHAT: we want the extra pressure from water, so use $P_{\text{gauge}}=\rho gh$. WHY that formula: "gauge" means above atmosphere, which is exactly what a fluid column of height $h$ contributes. $P_{\text{gauge}}=\rho gh=1000\times 9.8\times 4=3.92\times 10^4\text{Pa}.$ Answer: $3.92\times 10^4\text{Pa}$ (about $39\text{kPa}$).

WHAT: total pressure = atmosphere + water column. WHY add $P_0$: "absolute" means the total push, and the atmosphere is already sitting on the surface before any water is counted, so its pressure carries all the way down and adds to the water's $\rho gh$. $P_{\text{abs}}=P_0+\rho gh=1.01\times 10^5+3.92\times 10^4=1.402\times 10^5\text{Pa}.$ Answer: $\approx 1.40\times 10^5\text{Pa}$.

WHAT: rewrite the unit, then invert $\rho gh$ to find the matching depth. WHY invert $\rho gh$: the same pressure can be "spoken" as a height of water; solving $\rho gh=P$ for $h$ translates pressure into an equivalent water column. $50\text{kPa}=50000{\text{N/m}}^2$ (since $1\text{Pa}=1{\text{N/m}}^2$ and $\text{k}=1000$). Depth of water giving this gauge pressure: solve $\rho gh=50000$. $h=\frac{50000}{1000\times 9.8}=5.102\dots \text{m}.$ Since the data ($50\text{kPa}$) carries only two significant figures, we round to $h\approx 5.1\text{m}$. Answer: $5.0\times 10^4{\text{N/m}}^2$, equal to about $5.1\text{m}$ of water (2 s.f.).

WHAT first: convert density to SI, because $\rho gh$ needs ${\text{kg/m}}^3$. $1.03{\text{g/cm}}^3=1030{\text{kg/m}}^3.$ WHY convert: mixing ${\text{g/cm}}^3$ with metres gives nonsense units. $P_{\text{gauge}}=1030\times 9.8\times 12=1.2113\times 10^5\text{Pa}.$ Answer: $\approx 1.21\times 10^5\text{Pa}$ (a bit over one atmosphere — makes sense, seawater adds ~1 atm every ~10 m).

WHAT: set $\rho gh=2P_0$ and solve for $h$. $h=\frac{2P_0}{\rho g}=\frac{2\times 1.01\times 10^5}{1000\times 9.8}=20.6\text{m}.$ Answer: $\approx 20.6\text{m}$. (Consistent with the "~10 m per atmosphere" benchmark from the parent.)

WHAT: each layer adds its own $\rho gh$; pressures stack because each column sits on the one below. WHY add them: at the oil–water boundary the pressure is ${\rho }_{o}g{h}_o$; then water adds ${\rho }_{w}g{h}_w$ on top of that. $P={\rho }_{o}g{h}_o+{\rho }_{w}g{h}_w.$ $=800\times 9.8\times 0.20+1000\times 9.8\times 0.30$ $=1568+2940=4508\text{Pa}.$ Answer: $\approx 4.51\times 10^3\text{Pa}$.

Pressure: depends only on depth, not area. $P=\rho gh=1000\times 9.8\times 0.5=4900\text{Pa}\text{for both.}$ Force $=P\times A$ differs: $F_A=4900\times 1.0=4900\text{N},F_B=4900\times 0.001=4.9\text{N}.$ WHY they differ: pressure is force per area; the same pressure over a bigger base gives a bigger total force. The paradox is only about pressure, which is genuinely equal. Answer: same pressure $4900\text{Pa}$; forces $4900\text{N}$ vs $4.9\text{N}$.

WHAT: $h$ in $\rho gh$ is vertical depth, not distance along the slant. WHY: only the vertical stack of fluid weighs down on a point; a tilted length overcounts. Vertical depth: the pipe descends $6\text{m}$ at $30^{\circ }$ below horizontal, so $h=6\sin 30^{\circ }=6\times 0.5=3.0\text{m}.$ $P=\rho gh=1000\times 9.8\times 3.0=2.94\times 10^4\text{Pa}.$ Answer: $\approx 2.94\times 10^4\text{Pa}$ (not the $5.88\times 10^4$ you'd get from the raw $6\text{m}$).

$P_{\text{gauge}}=\rho g(0)=0,P_{\text{abs}}=P_0+0=P_0=1.01\times 10^5\text{Pa}.$ WHY it matters: the formula behaves sensibly at the boundary — the fluid adds nothing at its own surface, and only the atmosphere is left. This is the "anchor" the whole linear law grows from. Answer: gauge $=0$; absolute $=P_0$.

WHAT is new: pressure is not constant on the wall — it grows linearly from $0$ at top to $\rho gH$ at the bottom. So we cannot just do $P\times A$ with one $P$. WHY the average works: because the pressure profile is a straight line (triangle), its average over depth is the mid-value $\frac{1}{2}\rho gH$. $\bar{P}=\frac{1}{2}\rho gH.$ Total force = average pressure $\times$ wetted area $A=wH$: $F=\bar{P}\cdot wH=\frac{1}{2}\rho gH\cdot wH=\frac{1}{2}\rho gw{H}^2.$ $F=\frac{1}{2}\times 1000\times 9.8\times 8\times 5^2=9.8\times 10^5\text{N}.$ Answer: $F=9.8\times 10^5\text{N}$ (about $980\text{kN}$).

WHAT: the gas pushes one column down and the other up; the height difference $\Delta h$ measures the pressure gap. WHY $\rho g\Delta h$: at the lower common level, pressures from both arms must match; the difference is exactly one water column of height $\Delta h$. $P_{\text{gas, gauge}}=\rho g\Delta h=1000\times 9.8\times 0.15=1470\text{Pa}.$ Answer: $\approx 1.47\times 10^3\text{Pa}$. (See Manometers for the general rule.)

WHAT balances what: pick the level of the mercury–oil interface on the left arm. Pressure there equals pressure at the same horizontal level in the right arm.

• Left arm at interface: oil column presses down, $P={\rho }_{o}g{H}_o$ (gauge above the interface).
• Right arm at that same level: a mercury column of height $\Delta h=0.012\text{m}$ sits above it, $P={\rho }_{Hg}g\Delta h$.

WHY equal: connected fluid at rest has equal pressure at equal heights (same fluid path). ${\rho }_{o}g{H}_o={\rho }_{Hg}g\Delta h.$ WHY $g$ can be removed: it multiplies both sides identically, so dividing both sides by $g$ leaves the equation unchanged — gravity affects both columns equally, so its strength never enters the density ratio. ${\rho }_o={\rho }_{Hg}\frac{\Delta h}{H_o}=13600\times \frac{0.012}{0.20}=816{\text{kg/m}}^3.$ Answer: ${\rho }_o\approx 816{\text{kg/m}}^3$ (a typical light oil). See Density and Specific Gravity.

WHAT: in an ideal barometer, atmosphere = $\rho gh$ of mercury (vacuum on top). Here the trapped gas adds pressure on top, so it doesn't need to push mercury as high. Balance at the dish surface: $P_{\text{atm}}=P_{\text{gas}}+{\rho }_{Hg}g{h}_{\text{read}}.$ Convert heights to metres: true $760\text{mm}=0.760\text{m}$, read $735\text{mm}=0.735\text{m}$. $P_{\text{atm}}={\rho }_{Hg}g(0.760)=13600\times 9.8\times 0.760=1.0129\times 10^5\text{Pa}.$ $P_{\text{gas}}=P_{\text{atm}}-{\rho }_{Hg}g(0.735)={\rho }_{Hg}g(0.760-0.735).$ WHY ${\rho }_{Hg}g$ factors out: both $P_{\text{atm}}$ and the read column share the same fluid (${\rho }_{Hg}$) and the same gravity $g$, so their difference is just ${\rho }_{Hg}g$ times the height gap $(0.760-0.735)$ — the common factor is pulled out cleanly. $=13600\times 9.8\times 0.025=3332\text{Pa}.$ Answer: $P_{\text{gas}}\approx 3.33\times 10^3\text{Pa}$ (about $25\text{mm Hg}$ worth). See Atmospheric Pressure & Barometer.

WHAT is the mechanism: the bottom face is deeper than the top by $a$, so it feels more pressure. That difference, times area, is the buoyant force. Step 1 — pressure on each horizontal face (gauge; horizontal forces on the four side faces cancel by symmetry, so we ignore them):

• Top face depth $d$: $P_{\text{top}}=\rho gd=1000\times 9.8\times 0.30=2940\text{Pa}$.
• Bottom face depth $d+a=0.40\text{m}$: $P_{\text{bot}}=\rho g(d+a)=1000\times 9.8\times 0.40=3920\text{Pa}$.

Step 2 — turn pressures into forces on the face area $A=a^2=(0.10{)}^2=0.01{\text{m}}^2$:

• Up push on bottom: $F_{\text{bot}}=P_{\text{bot}}A=3920\times 0.01=39.2\text{N}$.
• Down push on top: $F_{\text{top}}=P_{\text{top}}A=2940\times 0.01=29.4\text{N}$.

Step 3 — net upward force = up − down: $F=F_{\text{bot}}-F_{\text{top}}=39.2-29.4=9.8\text{N}.$ WHY $d$ dropped out: algebraically, $F=(P_{\text{bot}}-P_{\text{top}})a^2=\rho g[(d+a)-d]a^2=\rho ga\cdot a^2=\rho g{a}^3.$ Only the difference in depth ($a$) survives; the shared depth $d$ cancels. This is Buoyancy & Archimedes' Principle falling straight out of $P=\rho gh$. Answer: $F=\rho g{a}^3=1000\times 9.8\times (0.10{)}^3=9.8\text{N}$, exactly the weight of the displaced water.

Does the container's shape change the pressure at a given depth? ::: No — pressure depends only on vertical depth $h$ (hydrostatic paradox). To get force on a straight dam wall, which pressure do you multiply by area? ::: The average pressure, $\frac{1}{2}\rho gH$, not the bottom value. In a manometer, what does the height difference measure? ::: The pressure difference, $\rho g\Delta h$. Why is buoyant force $\rho g{a}^3$ and not depth-dependent? ::: Only the top-vs-bottom depth difference survives; common depth cancels.