Archimedes Density Calculation from Apparent Weight

Use the loss of apparent weight

A suspended object usually has a smaller apparent weight when fully immersed in a liquid. The upward buoyant force accounts for that difference. Archimedes' principle relates the force to the weight of displaced liquid: Fᵦ = ρ_liquid gV. With the liquid density known, an immersion experiment can therefore reveal the object's volume and then its average density. This method is different from reading the rise in a measuring cylinder. Both concern displaced fluid, but their recorded quantities and arithmetic are different, so write down which measurements you actually have.

Derive the density ratio

Let W_air be the object's weight in air and W_sub its apparent weight while fully submerged. In the simple model, W_air = ρ_object gV and W_air − W_sub = ρ_liquid gV. Dividing the first equation by the second cancels g and V, giving ρ_object = ρ_liquid W_air/(W_air − W_sub). The denominator must represent a positive loss of apparent weight. If the two readings are equal within instrument resolution, the volume estimate is too uncertain to recover a useful density through this division. A zero difference is not an infinitely precise density measurement.

Calculate from force readings

For a worked example, take a liquid density of 1000 kg/m³, an air reading of 4.905 N and a submerged reading of 2.943 N. The apparent loss is 1.962 N. Density is 1000 × 4.905/1.962 = 2500 kg/m³, or 2.5 g/cm³. In the local Physics Calculator, enter 1000*4.905/(4.905-2.943). That tool evaluates this numerical expression; it has no dedicated immersion form or fluid-property lookup. The 1000 kg/m³ liquid value is an explicit rounded example assumption, not an exact statement about water at every temperature and pressure.

Check by recovering volume and mass

Using g = 9.81 m/s² in that example, displaced volume is 1.962/(1000 × 9.81) = 0.0002 m³. Object mass is 4.905/9.81 = 0.5 kg. Their ratio is 0.5/0.0002 = 2500 kg/m³, matching the direct formula. These two routes provide a useful arithmetic check. They do not supply two independent experiments because they reuse the same observations. Notice also that an apparent weight is a force. Do not enter its numerical value into the Density Calculator's kilogram mass field without first making the appropriate conversion.

Handle a balance that reports mass equivalents

If an instrument consistently reports apparent mass readings rather than newtons, the ratio can use the same mass unit in both readings. For example, air and immersed readings of 200 g and 175 g in a liquid assigned density 800 kg/m³ give 800 × 200/(200 − 175) = 6400 kg/m³. Grams cancel in the ratio; mixing a gram reading with a kilogram reading would not. Correct for the suspension fixture according to the measurement procedure. Do not treat a spring-scale reading in newtons as interchangeable with a balance display merely because both decrease during immersion.

Recognize where the simple method breaks down

The object must be fully submerged without touching the vessel, and the chosen liquid must not react with it or penetrate unaccounted pores. Bubbles increase displaced volume; absorption, evaporation and temperature drift can change the readings. The simple derivation neglects air buoyancy and support corrections, which matter for precision work. For an object that floats, holding it underwater adds another force, so the unsupported two-reading formula needs an appropriate experimental model. Report liquid density, conditions, readings and uncertainty with the result instead of identifying a substance solely from this ratio.

Inspect sensitivity to the submerged reading

With the same 200 g air reading and liquid density 800 kg/m³, changing the immersed reading from 175 g to 174 g changes the density from 6400 to about 6153.85 kg/m³. Changing it to 176 g gives about 6666.67 kg/m³. A one-gram change is amplified because the denominator is a small difference between larger readings. Repeated measurements and an instrument uncertainty estimate are especially important when that apparent loss is small.

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