As this windsurfer is propelled through the air, his motion is determined by
forces due to the wind and his weight. The relationship between the forces
acting on an object and the resulting motion is discussed in this chapter.
(Tom King/The Image Bank/Getty Images)
Statement:

An object continues in a state of rest or
in a state of motion at a constant speed along a straight
line, unless compelled to change that state by a net
force.

Definition:
Inertia is the natural tendency of an
object to remain at rest or in motion at a constant speed
along a straight line. The mass of an object is a
quantitative measure of inertia.

SI Unit of Inertia and Mass: kilogram (kg)

Definition:
An inertial reference frame is
one in which Newton’s law of inertia is valid.

The acceleration of an inertial reference
frame is zero, so it moves with a constant
velocity. All of Newton’s laws of motion are
valid in inertial reference frames, and when
we apply these laws, we will be assuming
such a reference frame. In particular, the
earth itself is a good approximation of an
inertial reference frame.


F
When a net external force
acts on an object
of mass m, the acceleration a that results is
directly proportional to the net force and has a
magnitude that is inversely proportional to the
mass. The direction of the acceleration is the
same as the direction
 of the net force.

 F

a
or  F  m a
m

SI Unit of Force: kg·m/s2 = newton (N)
Two people are pushing a stalled car,
(figure). The mass of the car is 1850 kg. One
person applies a force of 275 N to the car, while
the other applies a force of 395 N. Both forces
act in the same direction. A third force of 560 N
also acts on the car, but in a direction opposite
to that in which the people are pushing. This
force arises because of friction and the extent
to which the pavement opposes the motion of
the tires. Find the acceleration of the car.
A man is stranded on a raft
(mass of man and raft = 1300 kg).
By paddling, he causes an average
force of P = 17 N to be applied to
the raft in a direction due east (the
+x direction). The wind also
exerts a force on the raft that has a
magnitude of A = 15 N and points
67° north of east. Ignoring any
resistance from the water, find the
x and y components of the raft’s
acceleration.
Whenever one body exerts a force on a
second body, the second body exerts an
oppositely directed force of equal
magnitude on the first body.
Fundamental
1. Gravitational force.
2. Strong nuclear
force.
3. Electroweak force.
are the ones that are
truly unique, in the
sense that all other
forces can be
explained in terms of
them.
Non-fundamental
1. The normal force.
2. Static and kinetic
frictional forces.
3. The tension force.
Every particle in the universe exerts an attractive
force on every other particle. A particle is a piece of
matter, small enough in size to be regarded as a
mathematical point. For two particles that have masses
m1 and m2 and are separated by a distance r, the force
that each exerts on the other is directed along the line
joining the particles and has a magnitude given by

m1 m2
F G
r2
The symbol G denotes the universal gravitational
constant, whose value is found experimentally to be
G = 6.673 x 10-11 N.m2/kg2
What is the magnitude of the gravitational force
that acts on each particle in the above figure,
assuming m1 = 12 kg, m2 = 25 kg, and r = 1.2 m?

m1 m2
12 kg 25 kg 
11
2
8
F  G 2  6.67 10 N m / kg

14

10
N
2
r
1.2 m 
(You exert a force of about 1 N when pushing a
doorbell, so F is exceedingly small in such
circumstances). This is due to the fact that G itself
is very small. However, for large masses, like that
of the earth, the gravitational force can be large.
 Definition:
The weight of an object on or above the earth is
the gravitational force that the earth exerts on
the object.
ME G
F m g m 2 
 r 
 The
weight always acts downward, toward the
center of the earth.
 On or above another astronomical body, the
weight is the gravitational force exerted on the
object by that body.
SI Unit of Weight: newton (N)
Definition of the Normal Force
 The normal force FN is one component of the
force that a surface exerts on an object with
which it is in contact.
 namely, the component that is perpendicular
to the surface.
•
Figure (b) shows the freebody diagram for the
standing performer’s head
and neck before the act.
The only forces acting are
the normal force and the
50-N weight.
Therefore, the seventh cervical vertebra exerts a normal force of
FN = 50 N.
 Figure (c) shows the free-body diagram that applies during the
act. Now, the total downward force exerted on the standing
performer’s head and neck is 490 N + 50 N = 540 N, which must
be balanced by the upward normal force, so that FN = 540 N.
(a) When the elevator is not accelerating, the scale registers the
true weight (W = 700 N) of the person. (b) When the elevator
accelerates upward, the apparent weight (1000 N) exceeds the true
weight. (c) When the elevator accelerates downward, the apparent
weight (400 N) is less than the true weight. (d) The apparent
weight is zero if the elevator falls freely—that is, if it falls with the
FN  mg  ma
acceleration due to gravity.
Definition of Equilibrium:
An object is in equilibrium when it has zero
acceleration.
When an object is accelerating, it is not in
equilibrium.
i.e. the net force is not zero in Newton’s second law
airplane has a mass of 3.1 x 104 kg
and takes off under the influence of a constant
net force of 3.7 x 104 N . What is the net force
that acts on the plane’s 78-kg pilot?
 1/451) An
 2/451)
A boat has a mass of 6800 kg. Its engines
generate a drive force of 4100 N, due west, while
the wind exerts a force of 800 N, due east, and
the water exerts a resistive force of 1200 N due
east. What is the magnitude and direction of the
boat’s acceleration?
 3/451)
In the amusement park ride known as
Magic Mountain Superman, powerful magnets
accelerate a car and its riders from rest to 45 m/s
(about 100 mi/h) in a time of 7.0 s. The mass of
the car and riders is 5.5 x 103 kg. Find the
average net force exerted on the car and riders by
the magnets.
Solution:
v = vo + a t
45 = 0 + (a) (7.0 s)
a = 6.42 m/s2
F=ma
F = 5.5 x 103 x 6.42
F = 3.51 x 104 N

5/451) A 15-g bullet is fired from a rifle. It takes
2.50 X 10-3 s for the bullet to travel the length of the
barrel, and it exits the barrel with a speed of 715 m/s.
Assuming that the acceleration of the bullet is
constant, find the average net force exerted on the
bullet.
11/452) Two forces, F1 and F2, act on the 7.00kg block shown in the drawing. The magnitudes
of the forces are F1 = 59.0 N and F2 = 33.0 N.
What is the horizontal acceleration (magnitude
and direction) of the block?
12/452) When a parachute opens, the air
exerts a large drag force on it. This upward
force is initially greater than the weight of
the sky diver and, thus, slows him down.
Suppose the weight of the sky diver is 915
N and the drag force has a magnitude of
1027 N. The mass of the sky diver is 93.4
kg. What are the magnitude and direction
of his acceleration?