A body falling at its terminal velocity is not accelerating. By Newton's 2nd Law, it therefore has no net force acting upon it, i.e., the sum of the forces acting upon it must be zero. This does not imply that there are no forces
acting upon it. Any falling body in the earth's atmosphere will experience at least two* forces, its
weight, W, which acts downward and a force of air resistance, FA which acts upward (in the absence
of crosswinds).The gravitational force acting on the body (its weight) is proportional to its mass which
is proportional to its volume. The air resistance force acting on the body is proportional to its surface
area (upon which the air acts) and also proportional to its speed with respect to the air squared, so
as a falling body speeds up, the air resistance force acting upon it increases from zero (when it is
at rest) until it matches its weight. Thereafter, since the net force has become zero, there is no
further acceleration and the body continues to fall at a constant speed (its terminal velocity).
To illustrate further, consider the simple case of a falling cube which has reached its
terminal velocity, vT. The size of the cube pictured at the left is 2 units ´ 2 units ´ 2 units.
Hence its volume is 8 units3 and its surface area is 6 ´ 4 units2 = 24 units2. The weight of
the objectis therefore proportional to 8 units3 and its air resistance force is proportional to
vT 2 ´ 24 units2.
Now suppose the cube is cut along the dotted lines into eight equal pieces of size
1 unit ´ 1 unit ´ 1 unit. The volume of each resulting piece is 1 unit3 and its surface area
is 6 ´ 1 units2 = 6 units2. Note then that the volume and therefore the mass and weight is
only ⅛ that of the larger cube, but the surface area has only dropped to ¼ of its value on
the larger cube. It follows that vT 2 of the smaller cube must only be ½ of the value of vT 2
for the larger cube in order for the two forces to cancel so that the net force is zero.
Therefore vT 2(small) = vT 2(large)/2 or vT (small) » 0.7 vT (large), i.e., the smaller the body, the smaller the associated terminal velocity. The shape of the body is immaterial. The same result would hold for bodies of other shapes as well, including nearly spherical water droplets. (Actually the decrease in terminal velocity is more sensitive to particle size than is suggested by this oversimplified example. See the table in the next image for quantitative, experimentally verified results.)
*Actually
there is a third force, the buoyant force, which affects all bodies immersed
in any fluid, including the earth's atmosphere. The buoyant force, which
acts upward, is
due to the
air pressure at the base of a body being greater than the pressure at the
top of a body. Hence air pressure exerts a net force. Since
an air
parcel at rest
experiences no air resistance, the sum of the two remaining forces acting on
such a parcel, its weight and the buoyant force, must cancel
each other and
therefore must be equal in magnitude and opposite in direction. From
this fact we can infer Archimedes Principle, namely that the buoyant force
on any body,
immersed in any fluid, must be equal to the weight of the fluid it
displaces. (Hence the buoyant force on a helium balloon = the weight
of the air
it displaces, is greater
than the weight of the helium-filled balloon, and the net force is upward.)
If an immersed body is considerably denser than air, e.g., a
water
droplet, then
the buoyant force is so much smaller than its weight that it can, for many
purposes, be
ignored.
