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Instrument MI-03-066 · Physics

Buoyancy Experiment Calculator

Two numbers off a scale — weight in air, weight submerged — are all Archimedes' principle needs to hand back an object's volume and density, no calipers required.

Instrument MI-03-066
Sheet 1 OF 1
Rev A
Verified
Type 03 — Fluids SER. 2026-03066

Object density

2,500.000000 kg/m3

F_b = W_air − W_water

2.000000 Buoyant force (N)
0.20394324 Object volume (l)
2.500000 Specific gravity (relative to fluid)
The working Every figure verified twice
  1. buoyantForce = 5 − 3 = 2.000000
  2. volume = 2 ⁄ (1000·9.80665) = 0.00020394
  3. objectDensity = 5 ⁄ 9.80665 ⁄ 0.000204 = 2,500.000000
  4. specificGravity = 2500 ⁄ 1000 = 2.500000
Worksheet log
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How this instrument works

This instrument runs the double-weighing method: hang an object from a scale in air, hang the same object fully submerged in a fluid, and the drop between those two readings is the buoyant force outright — no displacement vessel, no calipers, no assumption about the object's shape. A submerged body is held up partly by the scale and partly by the fluid pushing back against it, so whatever weight the scale stops reporting has been taken over entirely by buoyancy: F_b = W_air − W_water.

Each later result feeds off that first one. Rearranging Archimedes' principle, F_b = ρ_fluid·g·V, for volume turns that buoyant force straight into a size: V = F_b ⁄ (ρ_fluid·g). Divide the object's actual mass — weight in air over standard gravity — by that freshly recovered volume and out comes its density; dividing once more by the fluid's own density cancels every unit and leaves specific gravity, a bare number. Four results out of two scale readings and one known fluid, which is precisely the trick modern labs still run under standards like ASTM B962 when a part's shape is too irregular for a caliper to trust.

The method has real requirements. The submerged object must hang free, touching neither the container's walls nor its bottom, since any contact force gets read by the scale as buoyancy that never happened. Trapped air bubbles clinging to a rough surface do the opposite kind of damage, inflating the apparent lift. And the whole chain depends on Fluid density being right for the temperature in the tank — water's density shifts by roughly half a percent between 4 °C and 30 °C, more than enough to throw off a lab-grade figure.

Fb=WairWwaterF_b = W_{\text{air}} - W_{\text{water}}V=FbρfluidgV = \dfrac{F_b}{\rho_{\text{fluid}}\,g}ρobj=Wair/gV\rho_{\text{obj}} = \dfrac{W_{\text{air}}/g}{V}SG=ρobjρfluidSG = \dfrac{\rho_{\text{obj}}}{\rho_{\text{fluid}}}
F_b — buoyant force, newtons (N) · W_air — weight measured in air, N · W_water — apparent weight fully submerged, N · V — object volume, m³ · ρ_fluid — fluid density, kg/m³ · ρ_obj — object density, kg/m³ · g — standard gravity, 9.80665 m/s² · SG — specific gravity, dimensionless.
  • Enter Weight measured in air — the object's ordinary reading on a scale or force gauge, in newtons, hanging free of anything else.
  • Enter Weight measured fully submerged — the same object on the same gauge, now hanging completely underwater and touching nothing.
  • Set Fluid density to match your liquid: 1000 kg/m³ suits fresh water near room temperature; adjust it for seawater or a different reading.
  • Read Buoyant force and Object volume first — both fall straight out of the two weighings — then Object density and Specific gravity, the figures the experiment was really after.

Worked example — a 5 N rock weighed in air and underwater

Lower a rock on a string from a force gauge and it reads 5 N hanging in air. Submerge it completely in fresh water and the same gauge reads 3 N — the water has taken over 2 N of the holding. That difference is the buoyant force outright: F_b = 5 − 3 = 2.0 N, with nothing yet said about the rock's size or what it is made of.

Rearranging Archimedes' principle turns that force into a volume without a ruler ever touching the rock: V = F_b ⁄ (ρ_fluid·g) = 2.0 ⁄ (1000 × 9.80665) = 0.000203943 m³, close to 203.9 cm³. Divide the rock's actual mass, 5 N ⁄ 9.80665 = 0.50986 kg, by that recovered volume and its density comes out at 2500 kg/m³ — a specific gravity of 2.50, squarely in the range of ordinary dense stone, all from two readings on a scale.

Questions

Why don't I need to measure the object's volume directly?

Because the two weighings already contain it. An object submerged in fluid loses exactly as much apparent weight as the fluid it pushes aside weighs, so Weight measured in air minus Weight measured fully submerged hands you the buoyant force, and rearranging F_b = ρ_fluid·g·V solves for volume without a graduated cylinder, calipers, or any assumption about the object's shape.

What if the submerged weight reads higher than the weight in air?

That means the object floats rather than stays under, and this instrument flags it: the check requires Weight measured fully submerged to be less than Weight measured in air, since the method assumes the object is entirely underwater yet still supported by the scale. A floating sample needs a small sinker of known weight threaded below it so the whole assembly goes under, with the sinker's own contribution subtracted afterward.

Does the string or wire holding the object underwater matter?

Yes, if it is not thin and light. A rigid wire or thick cord displaces a small volume of its own and skews both weighings, which is why hydrostatic-weighing labs favor fine monofilament so its buoyancy stays negligible next to the specimen's. For serious accuracy, weigh the line by itself submerged and subtract that reading from Weight measured fully submerged before entering it.

Who actually uses this double-weighing method?

Materials labs checking sintered or additively manufactured metal parts for hidden porosity, under methods like ASTM B962; gemologists telling a genuine stone from a denser or lighter simulant by its specific gravity; and exercise physiologists, who once ran underwater 'hydrostatic weighing' as a reference method for body-fat percentage before DEXA scanners became common. All three read density from nothing but two scale figures.

Why does the Fluid density figure matter so much?

Every later result is divided or multiplied by it, so an error there carries straight through to volume, density, and specific gravity. Tap water near room temperature is close enough to 1000 kg/m³ for classroom work, but precise labs look up water's density at the measured temperature — it drops to about 998.2 kg/m³ at 20 °C — and enter that instead of the round number.

Air bubbles clung to my object underwater — does that matter?

It matters a lot. A bubble stuck to a rough or porous surface adds its own buoyant lift without adding any of the object's actual mass, so Weight measured fully submerged reads lighter than it should and every result that follows comes out too low. Tap or brush the object gently once it is underwater to dislodge bubbles before reading the scale.

References