How this instrument works
Apparent weight is what a scale, spring balance, or lifting hook actually reads while an object sits fully submerged in a fluid — not its true weight, which never changes just because the object got wet. Two forces act on the body at once: gravity pulling down with mg, and the fluid pushing up with a buoyant force equal to the weight of the fluid the object displaces. Archimedes worked this out over two thousand years ago while checking whether a king's crown was pure gold, and the relation still reads the same way today: W_app = mg − ρ_fluid·V·g.
The subtracted term is not a fudge factor — it is a full weight in its own right, ρ_fluid·V·g, the weight of a fluid parcel shaped exactly like the submerged part of the object. Push a beach ball two centimetres under a pool's surface and that thin shell of displaced water already weighs enough to shove the ball back up; push a lead sinker the same depth and the sliver of displaced water is nearly nothing next to the sinker's own mg. The formula treats both cases with the same arithmetic, which is precisely why it works for both.
The volume that belongs in the formula is the displaced volume, not necessarily the object's full geometric volume — a half-submerged log only displaces the half that is underwater, and using its whole volume overstates the buoyant force. The formula also assumes a static, uniform fluid and a rigid object that neither compresses nor traps air pockets; a wetsuit, a hollow pontoon, or a fluid with a density gradient — brackish estuary water layered over salt water — all break that assumption and need a depth-resolved calculation instead of one fixed fluid density.
- Enter Mass — the object's true, dry mass in air; gravity converts it internally to a true weight of mg.
- Enter Fluid density for the medium it sits in — 1000 kg/m³ for fresh water, about 1025 kg/m³ for seawater, or the figure for whatever liquid applies.
- Enter Object volume as the volume actually displaced — the full body volume only if it is completely submerged, less if part breaks the surface.
- Read Apparent (immersed) weight — the downward force a scale, rope, or crane hook feels with the object underwater, always less than its true weight.
Worked example — a 5 kg object submerged in water
Take a 5 kg object with 1 litre, or 0.001 m³, of volume, lowered fully into fresh water at a fluid density of 1000 kg/m³. Its true weight in air is mg = 5 × 9.80665 = 49.03325 N — that number never changes just because the object goes underwater. The water it displaces weighs ρ_fluid·V·g = 1000 × 0.001 × 9.80665 = 9.80665 N, and that buoyant force acts upward against gravity.
Subtracting gives the apparent weight: W_app = 49.03325 − 9.80665 = 39.2266 N. A scale supporting the object underwater reads about a fifth lighter than it would in air — not because the object lost mass, but because a fifth of its true weight is now carried by the displaced water instead of the support. Lift the same object back into air and the reading jumps straight back to 49.03325 N.
This is the same number a crane operator watches during a subsea lift: as a load nears the surface, the hook force climbs smoothly from its fully submerged apparent weight toward the full mg the instant the last surface clears the water — riggers call this breaking the surface, and a crane rated only for the submerged load can be overloaded in the last few centimetres of the lift.
Questions
Why is the immersed object's weight lower without any loss of mass?
Because the scale is no longer supporting the full mg alone — the fluid is doing part of the job. Buoyant force equals the weight of the fluid displaced, Archimedes' principle, and it acts upward, so the support only supplies the difference, mg − ρ_fluid·V·g. The object's actual mass, and its true weight mg, are unchanged; only the force the support feels drops.
What does a negative apparent weight mean?
It means the object floats rather than staying submerged on its own — the buoyant force ρ_fluid·V·g exceeds its true weight mg, so only a downward anchor or held force keeps it under. A block of pine, roughly 500 kg/m³, in water gives a negative result here, which is the calculator's way of saying this object needs holding down, not support from above.
Does the object's shape matter, or only its volume?
Only the volume of fluid it displaces matters, not the shape. Archimedes' principle depends purely on how much fluid is pushed aside, so a 1-litre brass cube and a 1-litre brass sphere immersed in the same water have identical apparent weights. Shape affects drag and stability, not the buoyant force this formula computes.
How do I find Fluid density for something other than water?
Look up the specific liquid: seawater runs about 1025 kg/m³, mercury is 13,534 kg/m³, and light crude oil is roughly 850 kg/m³, all noticeably different from fresh water's 1000 kg/m³. Denser fluids buoy an object up harder, so the same mass and volume give a smaller apparent weight in mercury than in water, and can even turn negative.
Who actually uses an apparent-weight calculation like this?
Archimedes himself used it to catch a fraudulent crown by comparing weight in air against weight in water. The same subtraction now sizes crane loads for subsea lifts, where hook force jumps as a load breaks the surface, sets weight belts for scuba divers aiming for neutral buoyancy, and gives geotechnical engineers the buoyant unit weight of soil below a water table.
Is Object volume the same as the object's total volume?
Only if it is fully submerged. A half-floating log or a partly dipped rod displaces just the volume that is actually underwater, so V in the formula is that displaced volume, not the object's full geometric volume. Using the full volume for a partly submerged object overstates the buoyant force and understates the apparent weight.