Derive buoyancy and distinguish support from free motion

1. Frame the question

A submerged object receives pressure forces on all wetted surfaces. In a uniform static fluid, pressure on a horizontal lower face exceeds pressure on an equal upper face because the lower face is deeper. Horizontal forces on opposite vertical sides cancel for a simple symmetric block. The remaining upward resultant is buoyancy, not an extra unrelated force to add on top of the same pressure resultant.

2. Build the mechanism

For a block of face areaA and heightH, B=(Pbottom−Ptop)A=ρgHA=ρgVdisplaced. The argument extends through a displaced-fluid equilibrium construction to other shapes: buoyancy equals the weight of the displaced fluid, not the weight or total volume of the object unless the relevant conditions make those equal. A partially immersed object displaces only its immersed external volume.

An object can have buoyancy while sinking. Its free-motion force balance is B−mg=ma upward positive. If it is held motionless by an external restraint, the required signed holding force is mg−B. A negative value means downward restraint is needed; it is not a negative tension that an overhead string can supply. For an overhead spring scale and B<mg, the apparent weight is the positive tensionmg−B.

For a rigid fully immersed object in uniform liquid, buoyancy does not increase merely by moving it deeper: both opposing absolute forces rise by the same amount. Compressible objects, changing fluid density, contact with the bottom and incomplete wetting may invalidate this conclusion. The displayed release acceleration is a basic particle force model that omits drag and added-fluid inertia, not a precise prediction of real underwater motion.

3. Work an example

A block has area0.01 m² and height0.20 m, so volume0.002 m³. In water withρ=1000 andg=10, buoyancy is20 N. If top depth is0.20 m and atmospheric pressure100000 Pa, top and bottom force magnitudes are1020 and1040 N. For mass3 kg, weight30 N requires10 N upward support; the simple released acceleration is−10/3 m/s².

4. Predict and investigate

Does increasing immersion depth change buoyancy when volume and fluid density remain fixed?

Hold mass3 kg and compare top depths0.1 and0.5 m, then compare masses1 and5 kg at fixed depth. Distinguish opposing pressure forces, their resultant, signed holding force and the ideal released acceleration. Explain which cases require downward rather than overhead restraint.

5. Transfer to a new case

Try a new practice case

A fully immersed object has volume0.003 m³ and mass4 kg in fluidρ800 kg/m³. With g=10, find the upward holding force needed for rest.

Your numerical result (N) Check number

Reveal worked check (practice, not mastery)

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Explain the boundary

Why is setting buoyancy equal to weight invalid for every submerged object?

These bounded virtual recipes and supplied observations are not independent measurements, observed laboratory work or equipment safety calculations. No lifting, pressurizing, deep-water activity or physical jet construction is requested. Practical evidence requires an appropriate supervised setting.

Sources

Original explanations, cases and numerical models. Framework checked 30 October 2026: AP Physics 1 CED, topics 8.3. Mechanism references: OpenStax University Physics 1, fluids overview, buoyancy, and Bernoulli energy balance. No official assessment items reproduced.