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8.2 - Pressure

Definition of Pressure

Pressure is the magnitude of the perpendicular force per unit area over a surface:
P=FAP = \frac{F}{A}
where PP is pressure in pascals (Pa), FF is the force perpendicular to the surface in Newtons, and AA is the area the force spreads over in m². The SI unit is the pascal, 1 Pa=1Nm21 \text{ Pa} = 1 \frac{\text{N}}{\text{m}^2}.
Pressure tells you how concentrated a force is over an area, the same force spread over a smaller area gives higher pressure, and over a larger area gives lower pressure.
The AP equation sheet writes this as P=F⊥AP = \frac{F_\perp}{A} to remind you that only the perpendicular component of a force creates pressure. If a force is tilted, split it into components: the part perpendicular to the surface pushes into it and counts, while the part parallel to the surface just slides along it and doesn’t count.
Example: you push on a window with a 60 N force at 53° to the surface of the glass, and your palm covers 0.012 m². The perpendicular component is F⊥=(60)sin⁡53∘=(60)(0.80)=48F_\perp = (60)\sin 53^\circ = (60)(0.80) = 48 N, so P=480.012=4.0×103P = \frac{48}{0.012} = 4.0 \times 10^3 Pa. The parallel component, (60)cos⁡53∘=36(60)\cos 53^\circ = 36 N, adds nothing to the pressure.

Pressure Is a Scalar

Pressure comes from a force, but pressure itself has no direction. It’s a scalar, like mass or energy. At any point in a fluid at rest, the pressure pushes equally in every direction, so it doesn’t make sense to say pressure points down or sideways. Direction only shows up when pressure acts on a surface: the force it creates, F⊥=PAF_\perp = PA, always points perpendicular to that surface, pushing into it. That’s how the same water pressure pushes down on the floor of a pool, sideways on its walls, and up on the bottom of a floating raft.

Area Formulas for Common Shapes

For a circle (common for cylinders and pistons), A=πr2A = \pi r^2. For a rectangle, A=l×wA = l \times w. For a square, A=s2A = s^2.

Pascal’s Principle

An incompressible fluid keeps the same volume and density no matter how much pressure is exerted on it. That’s what makes Pascal’s Principle work: when pressure is applied to a confined fluid, that pressure change is transmitted equally to every part of the fluid and to the container’s walls. The fluid can’t squish to absorb the push, so it passes the push along.
Application to hydraulics: in a hydraulic system with two connected pistons of different sizes at the same height, the pressure is the same at both pistons, P1=P2P_1 = P_2, so:
F1A1=F2A2  ⟹  F2=F1⋅A2A1\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies F_2 = F_1 \cdot \frac{A_2}{A_1}
If A2>A1A_2 > A_1, then F2>F1F_2 > F_1, meaning a small input force produces a large output force, exactly the principle behind hydraulic jacks, car lifts, and brake systems.
8.2 - Pressure
Hydraulic systems use incompressible liquids (usually oil, sometimes water), found in car brakes, hydraulic jacks, and construction equipment. Pneumatic systems use compressible gases (usually air), found in air brakes for trucks and trains, pneumatic tools, and air-powered factory equipment.

Where Fluid Pressure Comes From

The pressure a fluid exerts is the combined result of all of its particles interacting with a surface. The particles in a gas or liquid are always moving and bumping into whatever they touch, and each collision gives the surface a tiny push. One collision is nothing, but there are an enormous number of them (around 3×10233 \times 10^{23} collisions every second on each square centimeter of a surface in room air), and together they add up to a steady force spread over the area: the pressure. More particles packed into the same space, or faster-moving particles, means more and harder hits, so more pressure.
In a liquid, the fluid near the bottom also has to support the weight of all the fluid above it, so it pushes harder on everything around it. That’s why pressure increases with depth.

Gauge Pressure vs. Absolute Pressure

Gauge pressure is the pressure from a fluid column above a given point:
Pgauge=ρghP_{\text{gauge}} = \rho gh
where ρ\rho is fluid density, gg is gravitational acceleration, and hh is depth below the surface. This comes directly from the weight of the fluid column above that point. Importantly, the same fluid gives the same pressure at the same depth, regardless of the container’s shape, a swimming pool and a narrow tube of the same water have identical pressure at 2 meters depth, only vertical depth matters.
Where ρgh\rho gh comes from: picture a column of fluid with cross-sectional area AA reaching from the surface down to depth hh. Its volume is AhAh, so its mass is ρAh\rho Ah and its weight is ρAhg\rho Ahg. That weight is spread over the area AA at the bottom of the column:
Pgauge=FgA=ρAhgA=ρghP_{\text{gauge}} = \frac{F_g}{A} = \frac{\rho A h g}{A} = \rho g h
The area cancels, which is exactly why the container’s shape doesn’t matter, only the depth. This is the gauge pressure equation on the AP equation sheet, Pgauge=ρghP_{\text{gauge}} = \rho gh.
8.2 - Pressure
Absolute pressure is the total pressure at a point, including atmospheric pressure:
Pabsolute=Preference+Pgauge=Patm+ρghP_{\text{absolute}} = P_{\text{reference}} + P_{\text{gauge}} = P_{\text{atm}} + \rho gh
Standard atmospheric pressure is Patm=1.0×105P_{\text{atm}} = 1.0 \times 10^5 Pa (sometimes given more precisely as 1.013×1051.013 \times 10^5 Pa, but AP Physics 1 uses 1.0×1051.0 \times 10^5 Pa). Gauge pressure measures relative to atmospheric pressure, while absolute pressure measures relative to a perfect vacuum, and most everyday gauges (like tire pressure gauges) actually read gauge pressure, not absolute.
On the AP equation sheet, absolute pressure is written P=P0+ρghP = P_0 + \rho gh. P0P_0 is the reference pressure, the pressure at the top surface of the fluid, and hh is the depth below that surface. P0P_0 is often atmospheric pressure, but not always: for a sealed tank with pressurized gas above the liquid, P0P_0 is that gas’s pressure.
The same idea compares any two points in one fluid: Pdeep−Pshallow=ρgΔhP_{\text{deep}} - P_{\text{shallow}} = \rho g \Delta h, where Δh\Delta h is the difference in depth. The reference pressure cancels out.
Why don’t we feel the atmosphere? Air presses on every square meter of your body with about 1.0×1051.0 \times 10^5 N, roughly the weight of a 10,000 kg truck. You don’t get crushed because the fluids inside your body push outward with about the same pressure, so the forces nearly cancel. You only notice when the outside pressure changes faster than your body can match, like your ears popping on a plane or at the bottom of a deep pool.

Graphing Pressure vs. Depth

Since P=P0+ρghP = P_0 + \rho gh has the same form as y=b+mxy = b + mx, a graph of absolute pressure vs. depth is a straight line. The vertical intercept is the pressure at the surface, P0P_0, and the slope is ρg\rho g. A gauge pressure graph has the same slope but starts at zero.
8.2 - Pressure
For fresh water (using g=10g = 10 m/s², the value the AP exam uses for numerical answers), the slope is ρg=(1000)(10)=1.0×104\rho g = (1000)(10) = 1.0 \times 10^4 Pa/m. Every meter of depth adds 1.0×1041.0 \times 10^4 Pa, so 10 m of water adds a whole atmosphere, and the absolute pressure there is 2.0×1052.0 \times 10^5 Pa, double the pressure at the surface. A denser liquid gives a steeper line. Going the other way, if you’re given a pressure vs. depth graph, the fluid’s density is the slope divided by gg.
Example: a pressure sensor is lowered into a tank of an unknown liquid. It reads 1.00×1051.00 \times 10^5 Pa at the surface and 1.36×1051.36 \times 10^5 Pa at a depth of 3.0 m. The slope is 1.36×105−1.00×1053.0=1.2×104\frac{1.36 \times 10^5 - 1.00 \times 10^5}{3.0} = 1.2 \times 10^4 Pa/m, so ρ=slopeg=1.2×10410=1.2×103\rho = \frac{\text{slope}}{g} = \frac{1.2 \times 10^4}{10} = 1.2 \times 10^3 kg/m³. The liquid is denser than water.

Worked Example: Pressure Under Chair Legs

A 65 kg person sits on a 5.0 kg chair with 4 circular legs, each 1.8 cm in diameter. Find the pressure each leg exerts on the floor.
Total weight: Ftotal=(mp+mc)g=(65+5.0)(10)=70×10=700F_{\text{total}} = (m_p + m_c)g = (65+5.0)(10) = 70 \times 10 = 700 N.
Force per leg: Fleg=7004=175F_{\text{leg}} = \frac{700}{4} = 175 N.
Area of each leg: d=0.018d = 0.018 m, so r=0.009r = 0.009 m, giving A=πr2=π(0.009)2≈2.545×10−4A = \pi r^2 = \pi(0.009)^2 \approx 2.545 \times 10^{-4} m².
Pressure: P=1752.545×10−4≈6.88×105P = \frac{175}{2.545 \times 10^{-4}} \approx 6.88 \times 10^5 Pa.
Each leg exerts roughly 6.9×1056.9 \times 10^5 Pa, about 6.9 times atmospheric pressure, since that force concentrates onto such a small area.

Real-World Applications of Pressure

High heels vs. flat shoes: heels have a very small contact area, so the same body weight creates much higher pressure, potentially damaging soft floors or causing foot and back pain. Flats have a larger contact area, spreading that same weight into lower pressure. For example, a 600 N person on heels with 4 cm² total area gets P=6000.0004=1.5×106P = \frac{600}{0.0004} = 1.5 \times 10^6 Pa, versus flats with 200 cm² giving P=6000.02=3.0×104P = \frac{600}{0.02} = 3.0 \times 10^4 Pa, the heels create 50 times more pressure.
Bed of nails: a person’s weight spreads across hundreds of nails, keeping the pressure at each contact point low enough not to break skin. With a 700 N person on 1000 nails, each nail bears just 0.7 N, and if each tip has area 10−610^{-6} m², the pressure per nail is 0.710−6=7×105\frac{0.7}{10^{-6}} = 7 \times 10^5 Pa, tolerable when spread out, but standing on just one nail would give 70010−6=7×108\frac{700}{10^{-6}} = 7 \times 10^8 Pa, easily piercing skin.
Snowshoes: a large surface area keeps you from sinking into snow, since spreading the same weight over more area lowers the pressure, letting you walk on soft snow that couldn’t support a concentrated point load.

Worked Example: Hydraulic Car Jack

An input piston has radius r1=2.0r_1 = 2.0 cm, and an output piston has radius r2=15r_2 = 15 cm, lifting a 12,000 N car. Find the required input force.
Areas: A1=π(0.02)2≈1.26×10−3A_1 = \pi(0.02)^2 \approx 1.26 \times 10^{-3} m², A2=π(0.15)2≈7.07×10−2A_2 = \pi(0.15)^2 \approx 7.07 \times 10^{-2} m².
Applying Pascal’s Principle:
F1=F2⋅A1A2=12,000×1.26×10−37.07×10−2=12,000×0.0178≈214 NF_1 = F_2 \cdot \frac{A_1}{A_2} = 12{,}000 \times \frac{1.26 \times 10^{-3}}{7.07 \times 10^{-2}} = 12{,}000 \times 0.0178 \approx 214 \text{ N}
Just 214 N (about 48 pounds) can lift a 12,000 N car! The mechanical advantage is F2F1=12,000214≈56\frac{F_2}{F_1} = \frac{12{,}000}{214} \approx 56.

Worked Example: Pressure at Ocean Depth

A submarine sits at 350 m depth, with seawater density ρ=1025\rho = 1025 kg/m³, atmospheric pressure 1.0×1051.0 \times 10^5 Pa, and g=10g = 10 m/s². Find the gauge and absolute pressure.
Gauge pressure:
Pgauge=ρgh=(1025)(10)(350)≈3.59×106 PaP_{\text{gauge}} = \rho gh = (1025)(10)(350) \approx 3.59 \times 10^6 \text{ Pa}
Absolute pressure:
Pabsolute=Patm+Pgauge=1.0×105+3.59×106≈3.69×106 PaP_{\text{absolute}} = P_{\text{atm}} + P_{\text{gauge}} = 1.0 \times 10^5 + 3.59 \times 10^6 \approx 3.69 \times 10^6 \text{ Pa}
That’s about 37 times atmospheric pressure, which is exactly why submarine hulls need such specialized design.

Common Mistakes

Confusing force and pressure: force is measured in Newtons, pressure in Pascals (N/m²), pressure is force divided by area, not just force alone.
Forgetting to divide among support points: if an object has 4 legs, each leg only supports a quarter of the total weight, don’t use the full weight when finding pressure under just one leg.
Using diameter instead of radius: area formulas use radius, if given diameter, divide by 2 first, using diameter directly makes the area 4 times too large.
Unit conversion errors: convert cm to m (divide by 100), mm to m (divide by 1000), and remember 1 cm² equals 10−410^{-4} m², not 10−210^{-2}.
Confusing gauge and absolute pressure: gauge pressure is above atmospheric, absolute pressure includes atmospheric, Pabsolute=Patm+PgaugeP_{\text{absolute}} = P_{\text{atm}} + P_{\text{gauge}}, and most everyday gauges read gauge pressure.
Using the wrong fluid pressure equation: it’s Pgauge=ρghP_{\text{gauge}} = \rho gh, not mghmgh (that’s potential energy), and there’s no need for mghV\frac{mgh}{V} (for a fluid column it equals ρgh\rho gh anyway, since m=ρVm = \rho V).
Thinking container shape affects pressure at depth: pressure at depth hh only depends on hh, not on the container’s width or shape, a wide pool and a narrow tube have the same pressure at the same depth.
Mixing up areas and forces in hydraulic systems: Pascal’s principle equates pressures, F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}, not the forces or areas themselves, the larger area gets a proportionally larger force.
Forgetting to square the radius: A=πr2A = \pi r^2 requires squaring, if r=0.02r = 0.02 m, A=π(0.02)2≈1.26×10−3A = \pi(0.02)^2 \approx 1.26 \times 10^{-3} m², not π×0.02\pi \times 0.02.
Not accounting for angled forces: pressure needs the force component perpendicular to the surface. If the force is angled, use F⊥=Fcos⁡θF_\perp = F\cos\theta when θ\theta is measured from the perpendicular (the normal), or F⊥=Fsin⁡θF_\perp = F\sin\theta when θ\theta is measured from the surface itself.
Getting the atmospheric pressure value wrong: AP Physics 1 uses Patm=1.0×105P_{\text{atm}} = 1.0 \times 10^5 Pa, though you might see the more precise 1.013×1051.013 \times 10^5 Pa elsewhere, check your formula sheet.
Assuming pressure equals weight: a heavy object over a large area can have lower pressure than a light object over a tiny area, pressure depends on both force and area together.

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