MCAT content library / Chemistry & physics / Pressure and buoyancy: distinguish depth, displacement, and interfaces
A submerged block, a hydraulic piston, and a pin at a liquid surface use different force comparisons. Work through the reference pressure and the support before calculating.
About 20 minutes, including the worked examples and checks.
From forces to fluid flow · Lesson 4 of 6 · Continue in this path
Useful foundations
On this page
- Identify which physical comparison supplies the force
- Depth changes pressure relative to a reference
- Worked application: why a submerged object feels lighter
- Pressure transmission can multiply force without multiplying work
- An interface can supply an additional force
- Check your reasoning
- Continue learning
- Sources
Identify which physical comparison supplies the force
A pressure reading, a spring-scale reading, and the observation that a small object stays at a liquid surface do not measure the same thing. Pressure acts over an area; buoyancy comes from the surrounding fluid’s pressure distribution; an interface can add a separate surface-tension force. AAMC’s fluid scope includes density, hydrostatic pressure, pressure transmission, buoyancy, and surface effects.1
Density is mass per volume, ρ = m/V. Its SI unit is kg/m³; 1 g/cm³ = 1000 kg/m³. Relative density compares an object’s density with a stated reference-fluid density and has no units. Pressure is normal force per area, p = F⊥/A, with 1 Pa = 1 N/m². A larger force need not imply a larger pressure if it acts over a larger area.2
Depth changes pressure relative to a reference
For a liquid at rest with effectively constant density, pressure increases downward according to Δp = ρgΔh. If an open surface is exposed to atmospheric pressure, pabsolute = patm + ρgh at depth h. The depth is vertical distance below that surface, not the length of a sloping container wall.2
Gauge pressure subtracts the surrounding reference pressure, commonly atmospheric pressure: pgauge = pabsolute − patm. Zero gauge pressure therefore does not mean no pressure. The expression ρgh is the gauge pressure for this open-surface arrangement; a sealed gas above the liquid can supply a different surface pressure.3
At equal depths within the same connected, stationary liquid, the pressure is the same. Container width does not enter ρgΔh. Pressure forces act normal to each contacting surface, so a net force requires accounting for pressures on all relevant faces, not multiplying one reading by an area and ignoring the opposite face.
Worked application: why a submerged object feels lighter
The following objects, fluid values, and measurements are original illustrative examples. A 0.60 kg rectangular block is suspended at rest by a vertical spring scale, fully submerged without touching the container. Its horizontal cross-sectional area is 25 cm² and its vertical height is 10 cm, giving a volume of 250 cm³ = 2.50 × 10−4 m³. Use water density 1000 kg/m³ and g = 10 m/s². Its top and bottom lie 0.20 m and 0.30 m below the surface.
| Location | Depth | Gauge pressure | Absolute pressure |
|---|---|---|---|
| Top face | 0.20 m | 2.0 kPa | 102.0 kPa |
| Bottom face | 0.30 m | 3.0 kPa | 103.0 kPa |
The pressure difference between the bottom and top is 1.0 kPa. Multiplying by 25 cm² = 2.50 × 10−3 m² gives an upward resultant of 2.5 N. The uniform atmospheric contribution cancels between the opposing faces. The vertical side faces exert horizontal forces, so they add no vertical contribution in this geometry.
Archimedes’ principle expresses that resultant generally as the weight of displaced fluid: Fb = ρfluidgVdisplaced. Here it is (1000)(10)(2.50 × 10−4) = 2.5 N. The block weighs 6.0 N. Its vertical balance is T + Fb − mg = 0, so the scale reads T = 3.5 N.4
Lowering this rigid, fully submerged block farther into the same constant-density liquid raises the pressures on both faces but leaves their difference, displaced volume, and buoyancy unchanged. Its density is 2400 kg/m³, so it would sink if released in water under this model. For an unsupported object floating without a significant interface contribution, the submerged volume adjusts until displaced-fluid weight equals object weight.
Pressure transmission can multiply force without multiplying work
Pascal’s principle concerns a pressure change applied to an enclosed fluid: in the static idealization, that change is transmitted through the fluid. For two frictionless pistons at the same fluid height, an applied pressure increment gives ΔF1/A1 = ΔF2/A2. Different piston heights would require adding the hydrostatic pressure difference.5
Consider input and output piston areas of 4.0 cm² and 40 cm². Neglect piston weights and friction; both outside faces experience the same atmospheric pressure. A 12 N input force corresponds to 12/(4.0 × 10−4) = 30,000 Pa of additional fluid pressure. Acting on the larger area, this supports an additional 120 N output load.
For small quasistatic displacements in the same-height approximation, incompressibility requires A1d1 = A2d2. An input travel of 2.0 cm produces an output travel of 0.20 cm. The ideal input and output work are both 0.24 J: (12 N)(0.020 m) = (120 N)(0.0020 m). The greater output force is paired with a smaller displacement.5
| Quantity | Value |
|---|---|
| Input piston area / force | 4.0 cm² / 12 N |
| Output piston area / force | 40 cm² / 120 N |
| Transmitted pressure increment | 30 kPa |
| Block scale force upward | 3.5 N |
| Block buoyancy upward | 2.5 N |
| Block weight downward | 6.0 N |
An interface can supply an additional force
A small dense needle may rest at a liquid–air interface even though its material is denser than the liquid. Surface tension has units of force per length and acts tangentially to the interface. The vertical components of the surface-tension forces at the needle’s contact line can supplement the pressure forces. Their magnitude depends on the contact length and interface geometry; density and immersed volume alone do not determine the full support.6
Once the needle is fully submerged, there is no liquid–air contact line on it to supply that contribution. If its weight exceeds displaced-liquid weight and no other support acts, it sinks. A change in wetting or interface shape can therefore change whether the needle remains at the surface without changing the density of its material.
Choose the force model from the situation: a depth difference gives ρgΔh; a pressure difference over an area gives a force; displaced-fluid weight gives bulk buoyancy; and a contact line may require an additional interface term. Combine those forces in a balance only after identifying which are actually present.
ACTIVE PRACTICE · FLUID STATICS
Lower the Block Without Changing Its Buoyancy
The original rigid block is 0.60 kg, with 25 cm² top area and 10 cm height, fully submerged without contacting the container. Its volume is 250 cm³. Start in water at density 1,000 kg/m³, using g = 10 m/s² and atmospheric pressure 100 kPa.
Top-face gauge pressure: kPa. Bottom-face gauge pressure: kPa.
Pressure difference: kPa. Buoyant force: N. Spring-scale tension: N.
Force balance T + Fb = mg applies to the suspended block at rest. Depth affects each face’s hydrostatic pressure; the difference depends on fixed block height. This ideal model assumes the entire block stays submerged and fluid density is uniform.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
A sealed vessel contains a liquid of density 800 kg/m³. The gas above it has absolute pressure 120 kPa. A sensor sits 0.50 m vertically below the liquid surface; external atmospheric pressure is 100 kPa. With g = 10 m/s², what are the sensor’s absolute pressure and gauge pressure relative to that atmosphere?
A. Absolute 4.0 kPa; gauge 4.0 kPa.
Review this choice. The 4.0 kPa value is only the liquid-depth increment, ρgh. The gas already supplies 120 kPa at the surface, and gauge pressure also needs its stated atmospheric reference.
B. Absolute 24 kPa; gauge 124 kPa.
Review this choice. These values are interchanged. Absolute pressure adds the 4.0 kPa depth increment to 120 kPa; gauge pressure then subtracts 100 kPa.
C. Absolute 120 kPa; gauge 20 kPa.
Review this choice. This gives the pressures at the liquid surface and omits the additional 4.0 kPa from the liquid column above the sensor.
D. Absolute 124 kPa; gauge 24 kPa.
Correct. The depth increment is (800)(10)(0.50) = 4000 Pa = 4.0 kPa. Absolute pressure is 120 + 4 = 124 kPa, and gauge pressure is 124 − 100 = 24 kPa.
The principle to carry forward
Pressure differences from depth are added to a surface pressure. Gauge pressure then subtracts a separately stated reference.
Check 2
The worked 0.60 kg block, volume 250 cm³, is now fully submerged at rest in a liquid of density 1500 kg/m³. A vertical scale supports it from above; it touches neither walls nor bottom. Use g = 10 m/s². What does the scale read?
A. 2.25 N
Correct. Displaced-fluid weight is (1500)(10)(250 × 10⁻⁶) = 3.75 N. The block’s 6.0 N weight is balanced by buoyancy plus tension, giving T = 6.0 − 3.75 = 2.25 N.
B. 3.50 N
Review this choice. This retains the water-case reading. The denser replacement fluid supplies greater buoyancy, so less upward scale force is needed.
C. 3.75 N
Review this choice. This is the buoyant force, not the scale reading. The scale supplies the remaining support after buoyancy is included.
D. 6.00 N
Review this choice. This is the block’s weight. It would be the required support if buoyancy were negligible, but the displaced liquid now supplies 3.75 N.
The principle to carry forward
The scale measures one support force, not the unchanged gravitational force. Recalculate displaced-fluid weight before solving the force balance.
Check 3
Two identical pins have material density greater than water. One rests at an undisturbed water–air interface with a curved contact line. The other is released fully submerged, with no trapped air or external support. Which statement best explains how the first can remain supported while the second sinks?
A. Bulk buoyancy necessarily grows when a smaller fraction of the pin is immersed.
Review this choice. A smaller immersed volume does not by itself provide greater displaced-fluid weight. This claim leaves out the interface force that distinguishes the two conditions.
B. The first pin’s gravitational force decreases because atmospheric pressure acts above it.
Review this choice. The pins have the same mass and gravitational field, so their weights are the same. Pressure and interface forces contribute additional forces; they do not reduce mg.
C. The contact line can supply an upward surface-tension component in addition to pressure forces.
Correct. The first pin interacts with a liquid–air interface whose geometry can provide upward support. The fully submerged pin has no such contact line; with weight greater than buoyancy and no other support, its net force is downward.
D. The water surrounding the first pin must become denser than the pin’s material.
Review this choice. The interface can provide a distinct support mechanism without reversing the materials’ density ordering. No such local density change is supplied by the experiment.
The principle to carry forward
Identify the interface before deciding whether a bulk displaced-volume calculation accounts for every force.
Continue learning
From forces to fluid flow
Lesson 4 of 6.
- Previous in From forces to fluid flow: Work, energy, and power: choose the useful conservation model
- Next in From forces to fluid flow: Ideal fluid flow: use continuity before Bernoulli
- Return to this path’s overview
See every step in From forces to fluid flow
From forces to fluid flow
Select the physical model that explains pressure, speed and flow resistance.
- Motion: translate between vectors, units, and graphs
- Net force and friction: build the right free-body diagram
- Work, energy, and power: choose the useful conservation model
- Pressure, buoyancy, and the role of fluid interfaces · You are here
- Ideal fluid flow: use continuity before Bernoulli
- Viscous flow: explain pressure loss and radius dependence
Related lessons
- Net force and friction: build the right free-body diagram
- Work, energy, and power: choose the useful conservation model
- Ideal fluid flow: use continuity before Bernoulli
- Viscous flow: explain pressure loss and radius dependence
Browse chemistry & physics · MCAT content library
MCAT preparation and roadmap · Aim High, Score Higher
Sources
1. AAMC: C/P Content Category 4B
2. OpenStax University Physics 1: Fluids, Density, and Pressure
3. OpenStax University Physics 1: Measuring Pressure
4. OpenStax University Physics 1: Archimedes’ Principle and Buoyancy
5. OpenStax University Physics 1: Pascal’s Principle and Hydraulics
6. OpenStax College Physics 2e: Surface Tension and Capillary Action