2
v = v
+ v
BG
BGx
BGy
v =
11
(
.
)
m/s
299
2 + 10
(
.
)
m/s
82
2
BG
v = 15. m/s
64
BG
and the definition of the tangent to determine the direction of v
:
BG
v
tan
BG y
θ = vBGx
v
1
θ = tan− BGy
vBGx
−
10.
m/s
82
1
θ = tan 11. m/s
299
o
θ = 43 8
.
Hence, v
= 15.6 m/s at 43.8° North of East.
BG
68
Chapter 12 Gravitational Force Near the Surface of the Earth, First Brush with Newton’s 2nd Law
12 Gravitational Force Near the Surface of the Earth, First
Brush with Newton’s 2nd Law
Some folks think that every object near the surface of the earth has an
acceleration of 9.8 m/s2 downward relative to the surface of the earth. That just
isn’t so. In fact, as I look around the room in which I write this sentence, all the
objects I see have zero acceleration relative to the surface of the earth. Only
when it is in freefall, that is, only when nothing is touching or pushing or pulling
on the object except for the gravitational field of the earth, will an object
experience an acceleration of 9.8 m/s2 downward relative to the surface of the
earth.
Gravitational Force near the Surface of the Earth
We all live in the invisible gravitational field of the earth. Mass is always accompanied by a
surrounding gravitational field. Any object that has mass, including the earth, is surrounded by a
gravitational field. The greater the mass of the object, the stronger the field is. The earth has a
huge mass; hence, it creates a strong gravitational field in the region of space around it. The
gravitational field is a force-per-mass at each and every point in the region around the object,
always ready and able to exert a force on any particle that finds itself in the gravitational field.
The earth’s gravitational field exists everywhere around the earth, not only everywhere in the air,
but out beyond the atmosphere in outer space, and inside the earth as well. The effect of the
gravitational field is to exert a force on any particle, any “victim,” that finds itself in the field.
The force on the victim depends on both a property of the victim itself, namely its mass, and on a
property of the point in space at which the particle finds itself, the force-per-mass of the
gravitational field at that point. The force exerted on the victim by the gravitational field is just
the mass of the victim times the force-per-mass value of the gravitational field at the location of
the victim.
Hold a rock in the palm of your hand. You can feel that something is pulling the rock
downward. It causes the rock to make a temporary indentation in the palm of your hand and you
can tell that you have to press upward on the bottom of the rock to hold it up against that
downward pull. The “something” is the field that we have been talking about. It is called the
gravitational field of the earth. It has both magnitude and direction so we use a vector variable,
the symbol g to represent it. In general, the magnitude and the direction of a gravitational field
both vary from point to point in the region of space where the gravitational field exists. The
gravitational field of the earth, near the surface of the earth is however, to a very good
approximation, much simpler than that. To a very good approximation, the gravitational field of
the earth has the same value at all points near the surface of the earth, and it always points
toward the center of the earth, a direction that we normally think of as downward. To a very
good approximation,
N
=
g 9 80
.
downward
(12-1)
kg
at all points near the surface of the earth. The fact that the gravitational field is a force-per-mass
at every point in space means that it must have units of force-per-mass. Indeed, the N (newton)
69
Chapter 12 Gravitational Force Near the Surface of the Earth, First Brush with Newton’s 2nd Law
N
appearing in the value 9.80
is the SI unit of force (how strong the push or pull on the object
kg
is) and the kg (kilogram) is the SI unit of mass, so the N/kg is indeed a unit of force-per-mass.
The gravitational force exerted on an object by the earth’s gravitational field (or that of another
planet when the object is near the surface of that other planet) is sometimes called the weight of
the object. To stress that the gravitational force is a force that is being exerted on the object,
rather than a property of the object itself, we will refer to it as the gravitational force. The
gravitational force F exerted on an object of mass m by the gravitational field of the earth is
g
given by
F = m
(12-2)
g
g
The product of a scalar and a vector is a new vector in the same direction as the original vector.
Hence the earth’s gravitational force is in the same direction as the gravitational field, namely
downward, toward the center of the earth. The magnitude of the product of a scalar and a vector
is the product of the absolute value of the scalar and the magnitude of the vector. [Recall that the
magnitude of a vector is how big it is. A vector has both magnitude (how big) and direction
(which way). So for instance, the magnitude of the force vector F = 15 N downward, is
F = 15 newtons.] Hence,
F = m
(12-3)
g
g
relates the magnitude of the gravitational force to the magnitude of the gravitational field. The
bottom line is that every object near the surface of the earth experiences a downward-directed
gravitational force whose magnitude is given by F = m where m is the mass of the object and
g
g
N
g is 9.80
.
kg
When the Gravitational Force is the Only Force on an Object
If there is a non-zero net force on an object, that object is experiencing acceleration in the same
direction as that net force. How much acceleration depends on how big the net force is and on
the mass of the object whose acceleration we are talking about, the object upon which the net
force acts. In fact, the acceleration is directly proportional to the force. The constant of
proportionality is the reciprocal of the mass of the object.
1
a =
∑ F
(12-4)1
m
1
dp
Equation 12-4 is known as Newton’s 2nd Law. Newton’s 2nd Law can also be written ∑ F =
where p is the
dt
momentum of the object. This latter equation is valid at all possible speeds, even at speeds close to the speed of
light. To derive equation 12-4 from it, we use p =
v
m which is only valid for speeds small compared to the speed
8
of light. Hence, equation 12-4 is only valid for speeds small compared to the speed of light (3.00 ×
10 m/s). Also,
70
Chapter 12 Gravitational Force Near the Surface of the Earth, First Brush with Newton’s 2nd Law
The expression ∑ F means “the sum of the forces acting on the object.” It is a vector sum. It
is the net force acting on the object. The mass m is the inertia of the object, the object’s inherent
1
resistance to a change in its velocity. (Inherent means “of itself.”) Note that the factor
in
m
equation 12-4:
= 1
a
∑ F
m
means that the bigger the mass of the object, the smaller its acceleration will be, for a given net
force. Equation 12-4 is a concise statement of a multitude of experimental results. It is referred
to as “Newton’s 2nd Law.” Here, we want to apply it to find the acceleration of an object in
freefall near the surface of the earth.
Whenever you apply Newton’s 2nd Law, you are required to draw a free body diagram of the
object whose acceleration is under investigation. In a free body diagram, you depict the object
(in our case it is an arbitrary object, let’s think of it as a rock) free from all its surroundings, and
then draw an arrow on it for each force acting on the object. Draw the arrow with the tail
touching the object, and the arrow pointing in the direction of the force. Label the arrow with the
symbol used to represent the magnitude of the force. Finally, draw an arrow near, but not
touching, the object. Draw the arrow so that it points in the direction of the acceleration of the
object and label it with a symbol chosen to represent the magnitude of the acceleration. Here we
use the symbol ag for the acceleration to remind us that it is the acceleration due to the earth’s
gravitational field g.
Free Body Diagram for an Object in Freefall near the Surface of the Earth
m
a
g
F
g
The next step in applying Newton’s 2nd Law is to write it down.
1
a =
F
(12-5)
↓
∑ ↓
m
-9
the concept of force proves impractical on the atomic scale and smaller (distances less than about 1 ×
10
m). Such
small scales are the realm of quantum mechanics where energy and momentum still play a major role.
71
Chapter 12 Gravitational Force Near the Surface of the Earth, First Brush with Newton’s 2nd Law
Note that equation 12-4:
= 1
a
∑ F
m
is a vector equation. As such it can be considered to be three equations in one—one equation for
each of a total of three possible mutually-orthogonal (meaning perpendicular to each other)
coordinate directions in space. In the case at hand, all the vectors, (hey, there are only two, the
gravitational force vector and the acceleration vector) are parallel to one and the same line,
namely the vertical, so we only need one of the equations. In equation 12-5,
1
a =
F
↓
∑ ↓
m
we use arrows as subscripts—the arrow shaft alignment specifies the line along which we are
summing the forces and the arrowhead specifies the direction along that line that we choose to
call the positive direction. In the case at hand, referring to equation 12-5, we note that the shafts
of the arrows are vertical, meaning that we are summing forces along the vertical and that we are
dealing with an acceleration along the vertical. Also in equation 12-5, we note that the
arrowheads are pointing downward meaning that I have chosen to call downward the positive
direction, which, by default, means that upward is the negative direction. (I chose to call
downward positive because both of the vectors in the free body diagram are downward.)
Next we replace a with what it is in the free body diagram,
↓
m
a
g
F
g
namely ag, and we replace ∑ F with the sum of the vertical forces in the free body diagram,
↓
counting downward forces as positive contributions to the sum, and upward forces as negative
contributions to the sum. This is an easy substitution in the case at hand because there is only
one force on the free body diagram, namely the gravitational force, the downward force of
magnitude F . The result of our substitutions is:
g
1
a =
F (12-6)
g
g
m
The F in equation 12-6 is the magnitude of the gravitational force, that force which you already
g
read about at the start of this chapter. It is given in terms of the mass of the object and the
magnitude of the earth’s gravitational field g by equation 12-3, F = m . Replacing the F in
g
g
g
1
equation 12-6, a =
F , with the g
m
to which it is equivalent we have
g
g
m
72
Chapter 12 Gravitational Force Near the Surface of the Earth, First Brush with Newton’s 2nd Law
1
a =
m (12-7)
g
g
m
1
Now the m that appears in the fraction
is the inertia of the object. It is the amount of inherent
m
resistance that the object has to a change in its velocity and is a measure of the total amount of
material making up the object. The m appearing in the mg part of the expression (equation 12-7)
is the gravitational mass of the object, the quantity that, in concert with the gravitational field at
the location of the object determines the force on the object. It is also a measure of the total
amount of material making up the object. As it turns out, the inertial mass and the gravitational
mass of the same object are identical (which is why we use one and the same symbol m for each)
and, in equation 12-7, they cancel. Thus,
a =
(12-8)
g
g
Now g is the magnitude of the earth’s gravitational field vector at the location of the object.
g = 9.80 N/kg and a , being an acceleration has to have units of acceleration, namely, m/s2.
g
kg ⋅ m
m
Fortunately a newton is a
so the units of g, namely N/kg, do indeed work out to be
.
2
s
2
s
Thus
m
a = 9 80
.
(12-9)
g
2
s
Now this is “wild”! The acceleration of an object in freefall does not depend on its mass. You
saw the masses cancel. The same thing that makes an object heavy makes it “sluggish.”
One-Dimensional Free-Fall, a.k.a., One-Dimensional Projectile Motion
If you throw an object straight up, or simply release it from rest, or throw it straight down;
assuming that the force of air resistance is negligibly small compared to the gravitational force:
the object will be in freefall from the instant it loses contact with your hand until the last instant
before it hits the ground (or whatever it does eventually hit), and the object will travel along a
m
straight line path with a constant acceleration of 9.80
downward.
2
s
Consider the case in which the object is thrown straight up. The whole time it is in freefall, the
m
object experiences an acceleration of 9.80
downward. While the object is on the way up, the
2
s
downward acceleration means that the object is slowing down. At the top of its motion, when
the velocity changes from being an upward velocity to being a downward velocity, and hence,
for an instant is zero, the downward acceleration means that the velocity is changing from zero to
a non-zero downward velocity. And on the way down, the downward acceleration means that
the velocity is increasing in the downward direction.
73
Chapter 13 Freefal , a.k.a. Projectile Motion
13 Freefall, a.k.a. Projectile Motion
The constant acceleration equations apply from the first instant in time after the
projectile leaves the launcher to the last instant in time before the projectile hits
something, such as the ground. Once the projectile makes contact with the
ground, the ground exerts a huge force on the projectile causing a drastic change
in the acceleration of the projectile over a very short period of time until, in the
case of a projectile that doesn’t bounce, both the acceleration and the velocity
become zero. To take this zero value of velocity and plug it into constant
acceleration equations that are devoid of post-ground-contact acceleration
information is a big mistake. In fact, at that last instant in time during which the
constant acceleration equations still apply, when the projectile is at ground level
but has not yet made contact with the ground, (assuming that ground level is the
lowest elevation achieved by the projectile) the magnitude of the velocity of the
projectile is at its biggest value, as far from zero as it ever gets!
Consider an object in freefall with a non-zero initial velocity directed either horizontally
forward; or both forward and vertically (either upward or downward). The object will move
forward, and upward or downward—perhaps upward and then downward—while continuing to
move forward. In all cases of freefall, the motion of the object (typically referred to as the
projectile when freefall is under consideration) all takes place within a single vertical plane. We
can define that plane to be the x-y plane by defining the forward direction to be the x direction
and the upward direction to be the y direction.
One of the interesting things about projectile motion is that the horizontal motion is independent
of the vertical motion. Recall that in freefall, an object continually experiences a downward
m
acceleration of 9.
80
but has no horizontal acceleration. This means that if you fire a
2
s
projectile so that it is approaching a wall at a certain speed, it will continue to get closer to the
wall at that speed, independently of whether it is also moving upward and/or downward as it
approaches the wall. An interesting consequence of the independence of the vertical and
horizontal motion is the fact that, neglecting air resistance, if you fire a bullet horizontally from,
say, shoulder height, over flat level ground, and at the instant the bullet emerges from the gun,
you drop a second bullet from the same height, the two bullets will hit the ground at the same
time. The forward motion of the fired bullet has no effect on its vertical motion.
The most common mistake that folks make in solving projectile motion problems
is combining the x and y motion in one standard constant-acceleration equation.
Don’t do that. Treat the x-motion and the y-motion separately.
In solving projectile motion problems, we take advantage of the independence of the horizontal
(x) motion and the vertical (y) motion by treating them separately. The one thing that is common
to both the x motion and the y motion is the time. The key to the solution of many projectile
motion problems is finding the total time of “flight.” For example, consider the following
sample problem:
74
Chapter 13 Freefal , a.k.a. Projectile Motion
Example 13-1: A projectile is launched with a velocity of 11 m/s at an angle of
28° above the horizontal over flat level ground from a height of 2.0 m above
ground level. How far forward does it go before hitting the ground? (Assume
that air resistance is negligible.)
Before getting started, we better clearly establish what we are being asked to find. We define the
forward direction as the x direction so what we are looking for is a value of x. More specifically,
we are looking for the distance, measured along the ground, from that point on the ground
directly below the point at which the projectile leaves the launcher, to the point on the ground
where the projectile hits. This distance is known as the range of the projectile. It is also known
as the range of the launcher for the given angle of launch and the downrange distance traveled by
the projectile.
Okay, now that we know what we’re solving for, let’s get started. An initial velocity of 11 m/s at
28° above the horizontal, eh? Uh oh! We’ve got a dilemma. The key to solving projectile
motion problems is to treat the x motion and the y motion separately. But we are given an initial
velocity v which is a mix of the two of them. We have no choice but to break up the initial
o
velocity into its x and y components.
y
v = 11 m/s
v
θ = 25°
y
v
x
x
v
v
x
= cosθ
y
v
= sinθ
v
v
= v cosθ
v = v sinθ
x
y
m
o
v = 11
cos 25
m
o
x
v
s
= 11
sin 25
y
s
m
v = 9 97
.
m
x
s
v = 4 65
.
y
s
Now we’re ready to get started. We’ll begin with a sketch which defines our coordinate system,
thus establishing the origin and the positive directions for x and y.
75
Chapter 13 Freefal , a.k.a. Projectile Motion
y [m]
3
vo
v
2