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us how hard it is to change its motion. Its weight measures the strength of
the gravitational attraction between the apple and the planet earth. The
apple’s weight is less on the moon, but its mass is the same. Astronauts
assembling the International Space Station in zero gravity cannot just pitch
massive modules back and forth with their bare hands; the modules are
weightless, but not massless.
We have already seen the experimental evidence that when weight (the
force of the earth’s gravity) is the only force acting on an object, its accelera-
tion equals the constant g, and g depends on where you are on the surface of
the earth, but not on the mass of the object. Applying Newton’s second law
then allows us to calculate the magnitude of the gravitational force on any
object in terms of its mass:
|F
W
| = mg .
(The equation only gives the magnitude, i.e. the absolute value, of F
W
,
because we’re defining g as a positive number, so it equals the absolute value
of a falling object’s acceleration.)
Example: calculating terminal velocity
Question: Experiments show that the force of air friction on a
falling object such as a skydiver or a feather can be
approximated fairly well with the equation |F
air
|=c
.
Av
2
, where c is
a constant,
.
is the density of the air, A is the cross-sectional
area of the object as seen from below, and v is the object’s
velocity. Predict the object’s terminal velocity, i.e. the final
velocity it reaches after a long time.
Solution: As the object accelerates, its greater v causes the
upward force of the air to increase until finally the gravitational
force and the force of air friction cancel out, after which the
object continues at constant velocity. We choose a coordinate
system in which positive is up, so that the gravitational force is
negative and the force of air friction is positive. We want to find
the velocity at which
F
air
+ F
W
=0 , i.e.
c
.
Av
2
– mg=0 .
Solving for v gives
v
terminal
=
mg
c
.
A
Self-Check
It is important to get into the habit of interpreting equations. These two self-
check questions may be difficult for you, but eventually you will get used to this
kind of reasoning.
(a) Interpret the equation v
terminal
=
mg/c
.
A
in the case of
.
=0.
(b) How would the terminal velocity of a 4-cm steel ball compare to that of a 1-
cm ball.
A simple double-pan balance works by
comparing the weight forces exerted
by the earth on the contents of the two
pans. Since the two pans are at almost
the same location on the earth’s
surface, the value of g is essentially
the same for each one, and equality
of weight therefore also implies
equality of mass.
(a) The case of
.
=0 represents an object falling in a vacuum, i.e. there is no density of air. The terminal velocity
would be infinite. Physically, we know that an object falling in a vacuum would never stop speeding up, since there
would be no force of air friction to cancel the force of gravity. (b) The 4-cm ball would have a mass that was greater
by a factor of 4x4x4, but its cross-sectional area would be greater by a factor of 4x4. Its terminal velocity would be
greater by a factor of
4
3
/4
2
=2.
Section 4.3Newton’s Second Law