Question

(a) Using principles of buoyancy, estimate the density of the human body, using reasonable assumptions
from your experience of swimming in fresh water, and from reports of how you “float” in the Dead Sea.
(b) Estimate the buoyancy force you experience in air, while walking around on the UNH campus.
(c) Estimate the buoyancy force you experience while hiking the last few steps to the peak after driving
up the Mount Evans Scenic Byway outside of Denver, CO. What is the difference in buoyancy force
compared to (b)?

Answer 1

a)

Suppose a 60.0-kg woman floats in freshwater with 97 of her volume submerged when her lungs are full of air. We can find the woman’s density by solving the equation

for the density of the object. This yields

We know both the fraction submerged and the density of water, and so we can calculate the woman’s density. Entering the known values into the expression for her density, we obtain

Her density is less than the fluid density. We expect this because she floats. Body density is one indicator of a person’s per cent body fat.

b)

Let’s say my weight is 120 pounds = 534 N. I estimate my weight-density to be a bit less than that of water, say 9500 N/m³. (water’s mass-density is 1000 kg/m³ which corresponds to a weight-density of 9800 N/m^3). So, since weight-density = weight/volume, this means

my volume = weight/weight-density = 534 N/(9500 N/m³) = 0.056 m³

The buoyant force that air exerts on me equals the weight-density of air multiplied by my volume, and since the density of air is about 1.25 kg/m³, so weight-density of air is 9.8 times this, i.e. 12.25 N/m³, we have,

buoyant force = my volume x weight-density-of-air = 0.056 m³ x 12.25 N/m³ = 0.69 N

i.e. the buoyant force on me is about 0.7 N.

This is much smaller than my weight!

c)

If you attempt to compare your weight at sea level with its value up a mountain there are two important effects that need to be taken into account if the comparison is to be meaningful - gravity and air density.

Weight is the gravitational force acting on a body (literally in this case!) and the value of the gravitational force changes with latitude (the Earth is not spherical) but also with height above sea level - or more accurately with distance from the centre of mass of the Earth. It changes by roughly 0.3 parts per million per metre so a change in height of, say, 3 000 m would reduce the gravitational force, and hence your weight, by about 900 parts per million (~0.1% in more round-figures). Thus if you weigh around 70 kg your weight would be about 0.07 kg less at 3 000 m.

But this is a weight loss rather than a mass loss and if you try to measure it using a properly set-up balance you will not detect it. If the balance is of the two-pan variety, or a single-pan balance used as a comparator, the weight-loss - through reduced gravitational acceleration - will apply equally to both you and to the reference mass against which you were being compared and would thus cancel-out. If a single-pan balance was used in direct reading mode, its scale should effectively have been compensated for the different value of g, through the setting-up and calibration process - involving the use of known masses (and mass does not change with g). A weight loss would thus only be apparent when weighing at altitude using a single-pan balance, in direct reading mode, that had not been properly set-up or calibrated since moving it from ground level.

The second effect is due to the density of the air; at greater heights the air is less dense, or thinner, due to its reduced pressure (air density depends on air pressure, temperature and humidity) and whilst this does not affect your weight - the gravitational force acting on you - it will affect your buoyancy which in turn affects the weighing process.

The density of air at sea level is nominally 1.2 kg/m³; it reduces very roughly by 1 part in 10 000 per metre increase in altitude so at 3 000 m its density will be something over about 0.8 kg/m³. To know what effect this has on the weighing (not weight) of a person we need to know their volume. Assuming a corporal density of 1 000 kg/m³ (humans are more-or-less neutrally buoyant in water of this density) and knowing that volume is mass/density, the volume of our 70 kg 'standard' (!) person would be around 0.07 m³. Therefore the reduction in buoyancy at 3 000 m would be about [0.07 × (1.2 - 0.8)] kg - that is around 0.03 kg.

So, within the assumptions and approximations stated, at an altitude of 3000 m

• the weight of a 70 kg person would reduce by about 0.07 kg
• this reduction is in weight, not mass
• the weight reduction would not be detected by a properly set-up and calibrated balance but your leg muscles would have marginally less work to do in moving you around
• reduced buoyancy would be detected by the balance (making you appear to be heavier than you would otherwise be by about 0.03 kg) but, by definition, this would not be a weight increase and the buoyancy correction should be added algebraically [see note 2 below] to your measured weight
• the reduced buoyancy would cause your leg muscles to have to work slightly harder.

In summary

• you would have the same mass but slightly less weight
• the effect would not be detected by a properly set up balance
• about half the 'gain' in reduced muscle effort would be cancelled out because of reduced air buoyancy

Of course, if you climb the mountain under your own steam rather than using a cable car you may genuinely lose mass (and hence 'really' lose weight) and this would be detectable with a balance; provided you did not eat too much on the descent it would not reappear afterwards!