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CHAPTER 2
KINETICS OF A PARTICLE: FORCE AND
ACCELERATION
2.1 Newton’s Laws of Motion
Today’s Objectives:
Students will be able to:
a) Write the equation of motion
for an accelerating body.
b) Draw the free-body and
kinetic diagrams for an
accelerating body.
In-Class Activities:
• Applications
• Newton’s laws of motion
• Newton’s law of
gravitational attraction
• Equation of motion for a
particle or system of particles
APPLICATIONS
The motion of an object depends on the
forces acting on it.
A parachutist relies on the atmospheric
drag resistance force to limit his velocity.
Knowing the drag force, how can we
determine the acceleration or velocity of
the parachutist at any point in time?
APPLICATIONS (continued)
A freight elevator is lifted using a
motor attached to a cable and pulley
system as shown.
How can we determine the tension
force in the cable required to lift the
elevator at a given acceleration?
Is the tension force in the cable greater than the weight
of the elevator and its load?
NEWTON’S LAWS OF MOTION
The motion of a particle is governed by Newton’s three laws of
motion.
First Law: A particle originally at rest, or moving in a straight
line at constant velocity, will remain in this state if the resultant
force acting on the particle is zero.
Second Law: If the resultant force on the particle is not zero, the
particle experiences an acceleration in the same direction as the
resultant force. This acceleration has a magnitude proportional to
the resultant force.
Third Law: Mutual forces of action and reaction between two
particles are equal, opposite, and collinear.
NEWTON’S LAWS OF MOTION (continued)
The first and third laws were used in developing the
concepts of statics. Newton’s second law forms the
basis of the study of dynamics.
Mathematically, Newton’s second law of motion can be
written
F = ma
where F is the resultant unbalanced force acting on the
particle, and a is the acceleration of the particle. The
positive scalar m is called the mass of the particle.
Newton’s second law cannot be used when the particle’s
speed approaches the speed of light, or if the size of the
particle is extremely small (~ size of an atom).
NEWTON’S LAW OF GRAVITATIONAL ATTRACTION
Any two particles or bodies have a mutually attractive
gravitational force acting between them. Newton postulated
the law governing this gravitational force as
F = G(m1m2/r2)
where
F = force of attraction between the two bodies,
G = universal constant of gravitation ,
m1, m2 = mass of each body, and
r = distance between centers of the two bodies.
When near the surface of the earth, the only gravitational
force having any sizable magnitude is that between the earth
and the body. This force is called the weight of the body.
MASS AND WEIGHT
It is important to understand the difference between the
mass and weight of a body!
Mass is an absolute property of a body. It is independent of
the gravitational field in which it is measured. The mass
provides a measure of the resistance of a body to a change
in velocity, as defined by Newton’s second law of motion
(m = F/a).
The weight of a body is not absolute, since it depends on the
gravitational field in which it is measured. Weight is defined
as
W = mg
where g is the acceleration due to gravity.
UNITS: SI SYSTEM VS. FPS SYSTEM
SI system: In the SI system of units, mass is a base unit and
weight is a derived unit. Typically, mass is specified in
kilograms (kg), and weight is calculated from W = mg. If the
gravitational acceleration (g) is specified in units of m/s2, then
the weight is expressed in newtons (N). On the earth’s
surface, g can be taken as g = 9.81 m/s2.
W (N) = m (kg) g (m/s2) => N = kg·m/s2
FPS System: In the FPS system of units, weight is a base unit
and mass is a derived unit. Weight is typically specified in
pounds (lb), and mass is calculated from m = W/g. If g is
specified in units of ft/s2, then the mass is expressed in slugs.
On the earth’s surface, g is approximately 32.2 ft/s2.
m (slugs) = W (lb)/g (ft/s2) => slug = lb·s2/ft
2.2 The Equation of Motion
The motion of a particle is governed by Newton’s second law, relating
the unbalanced forces on a particle to its acceleration. If more than one
force acts on the particle, the equation of motion can be written
F = FR = ma
where FR is the resultant force, which is a vector summation of all the
forces.
To illustrate the equation, consider a
particle acted on by two forces.
First, draw the particle’s freebody diagram, showing all
forces acting on the particle.
Next, draw the kinetic diagram,
showing the inertial force ma
acting in the same direction as
the resultant force FR.
INERTIAL FRAME OF REFERENCE
This equation of motion is only valid if the acceleration is
measured in a Newtonian or inertial frame of reference.
What does this mean?
For problems concerned with motions at or near the
earth’s surface, we typically assume our “inertial frame”
to be fixed to the earth. We neglect any acceleration
effects from the earth’s rotation.
For problems involving satellites or rockets, the
inertial frame of reference is often fixed to the stars.
2.3 Equation of Motion for a System of Particles
The equation of motion can be extended to include systems of
particles. This includes the motion of solids, liquids, or gas systems.
As in statics, there are internal forces and
external forces acting on the system.
What is the difference between them?
Using the definitions of m = mi as the
total mass of all particles and aG as the
acceleration of the center of mass G of
the particles, then maG = miai .
The text shows the details, but for a system of particles: F = maG
where F is the sum of the external forces acting on the entire
system.
KEY POINTS
1) Newton’s second law is a “Law of Nature”--experimentally
proven and not the result of an analytical proof.
2) Mass (property of an object) is a measure of the resistance
to a change in velocity of the object.
3) Weight (a force) depends on the local gravitational field.
Calculating the weight of an object is an application of
F = ma, i.e., W = m g.
4) Unbalanced forces cause the acceleration of objects. This
condition is fundamental to all dynamics problems.
PROCEDURE FOR THE APPLICATION OF THE
EQUATION OF MOTION
1) Select a convenient inertial coordinate system. Rectangular,
normal/tangential, or cylindrical coordinates may be used.
2) Draw a free-body diagram showing all external forces
applied to the particle. Resolve forces into their
appropriate components.
3) Draw the kinetic diagram, showing the particle’s inertial
force, ma. Resolve this vector into its appropriate
components.
4) Apply the equations of motion in their scalar component
form and solve these equations for the unknowns.
EXAMPLE
Given: A crate of mass m is pulled by a cable attached to a truck.
The coefficient of kinetic friction between the crate and
road is mk.
Find: Draw the free-body and kinetic diagrams of the crate.
Plan: 1) Define an inertial coordinate system.
2) Draw the crate’s free-body diagram, showing all
external forces applied to the crate in the proper
directions.
3) Draw the crate’s kinetic diagram, showing the inertial
force vector ma in the proper direction.
EXAMPLE (continued)
Solution:
1) An inertial x-y frame can be defined as fixed to the ground.
2) Draw the free-body diagram of the crate:
y
W = mg
The weight force (W) acts through the
crate’s center of mass. T is the tension
x
force in the cable. The normal force (N)
is perpendicular to the surface. The
friction force (F = uKN) acts in a direction
F = uKN
opposite to the motion of the crate.
N
3) Draw the kinetic diagram of the crate:
T
30°
ma
The crate will be pulled to the right. The
acceleration vector can be directed to the
right if the truck is speeding up or to the
left if it is slowing down.
2.4 Equations of Motion: Rectangular Coordinates
Today’s Objectives:
Students will be able to apply
Newton’s second law to
determine forces and
accelerations for particles in
rectilinear motion.
In-Class Activities:
• Applications
• Equations of motion using
rectangular (Cartesian)
Coordinates
APPLICATIONS
If a man is pushing a 100 lb crate, how large a force F must
he exert to start moving the crate?
What would you have to know before you could calculate
the answer?
APPLICATIONS (continued)
Objects that move in any fluid have a drag force acting on
them. This drag force is a function of velocity.
If the ship has an initial velocity vo and the magnitude of the
opposing drag force at any instant is half the velocity, how
long it would take for the ship to come to a stop if its engines
stop?
EQUATION OF MOTION
The equation of motion, F = m a, is best used when the problem
requires finding forces (especially forces perpendicular to the
path), accelerations, velocities or mass. Remember, unbalanced
forces cause acceleration!
Three scalar equations can be written from this vector equation.
The equation of motion, being a vector equation, may be
expressed in terms of its three components in the Cartesian
(rectangular) coordinate system as
F = ma or Fx i + Fy j + Fz k = m(ax i + ay j + az k)
or, as scalar equations, Fx = max , Fy = may , and Fz = maz .
PROCEDURE FOR ANALYSIS
• Free Body Diagram
Establish your coordinate system and draw the particle’s
free body diagram showing only external forces. These
external forces usually include the weight, normal forces,
friction forces, and applied forces. Show the ‘ma’ vector
(sometimes called the inertial force) on a separate diagram.
Make sure any friction forces act opposite to the direction
of motion! If the particle is connected to an elastic spring,
a spring force equal to ks should be included on the
FBD.
PROCEDURE FOR ANALYSIS (continued)
• Equations of Motion
If the forces can be resolved directly from the free-body
diagram (often the case in 2-D problems), use the scalar
form of the equation of motion. In more complex cases
(usually 3-D), a Cartesian vector is written for every force
and a vector analysis is often best.
A Cartesian vector formulation of the second law is
F = ma or
Fx i + Fy j + Fz k = m(ax i + ay j + az k)
Three scalar equations can be written from this vector equation.
You may only need two equations if the motion is in 2-D.
PROCEDURE FOR ANALYSIS (continued)
• Kinematics
The second law only provides solutions for forces and
accelerations. If velocity or position have to be found,
kinematics equations are used once the acceleration is
found from the equation of motion.
Any of the tools learned in Chapter 12 may be needed to
solve a problem. Make sure you use consistent positive
coordinate directions as used in the equation of motion
part of the problem!
EXAMPLE
Given: WA = 10 lb
WB = 20 lb
voA = 2 ft/s
mk = 0.2
Find: vA when A has moved 4 feet.
Plan: Since both forces and velocity are involved, this
problem requires both the equation of motion and kinematics.
First, draw free body diagrams of A and B. Apply the
equation of motion . Using dependent motion equations,
derive a relationship between aA and aB and use with the
equation of motion formulas.
EXAMPLE (continued)
Solution:
2T
Free-body and kinetic
diagrams of B:
=
WB
mBaB
+   Fy = ma y
Apply the equation of
motion to B:
WB - 2 T = mB aB
20 a
20 - 2 T = 32
.2 B
(1)
EXAMPLE (continued)
Free-body and kinetic diagrams of A:
y
WA
x
T
N
=
mAaA
F = mkN
Apply the equations of motion to A:
+  F = ma

+  Fy = ma y = 0
x
x
N = WA = 10 lb
F = m N = 2 lb
k
F - T = mA aA
10
2 - T = 32 2 aA
.
(2)
EXAMPLE (continued)
Now consider the kinematics.
sA
Datums
A
sB
B
Constraint equation:
sA + 2 sB = constant
or
vA + 2 vB = 0
Therefore
aA + 2 aB = 0
aA = -2 aB
(3)
(Notice aA is considered
positive to the left and aB is
positive downward.)
EXAMPLE (continued)
Now combine equations (1), (2), and (3).
T = 22 = 7. 33 lb
3
ft 2 
ft
2
aA = -17.16 s = 17.16 s
Now use the kinematic equation:
vA
2
2
= voA + 2 aA ( sA - soA )
vA 2 = 2 2 + 2 (17.16 )(4 )
ft
vA = 11 . 9 s 
2.5 Equations of Motion: Normal and Tangential Coordinates
Today’s Objectives:
Students will be able to apply
the equation of motion using
normal and tangential
coordinates.
In-Class Activities:
• Applications
• Equation of motion in n-t
coordinates
APPLICATIONS
Race tracks are often banked in the
turns to reduce the frictional forces
required to keep the cars from sliding
at high speeds.
If the car’s maximum velocity and a
minimum coefficient of friction
between the tires and track are
specified, how can we determine the
minimum banking angle (q) required
to prevent the car from sliding?
APPLICATIONS (continued)
Satellites are held in orbit around
the earth by using the earth’s
gravitational pull as the centripetal
force – the force acting to change
the direction of the satellite’s
velocity.
Knowing the radius of orbit of
the satellite, how can we
determine the required speed of
the satellite to maintain this orbit?
NORMAL & TANGENTIAL COORDINATES
When a particle moves along a
curved path, it may be more
convenient to write the equation of
motion in terms of normal and
tangential coordinates.
The normal direction (n) always points toward the path’s
center of curvature. In a circle, the center of curvature is the
center of the circle.
The tangential direction (t) is tangent to the path, usually set
as positive in the direction of motion of the particle.
EQUATIONS OF MOTION
Since the equation of motion is a
vector equation , F = ma,
it may be written in terms of the
n & t coordinates as
Ftut + Fnun = mat + man
Here Ft & Fn are the sums of the force components acting in
the t & n directions, respectively.
This vector equation will be satisfied provided the individual
components on each side of the equation are equal, resulting in
the two scalar equations: Ft = mat and
Fn = man .
Since there is no motion in the binormal (b) direction, we can also
write Fb = 0.
NORMAL AND TANGENTIAL ACCERLERATIONS
The tangential acceleration, at = dv/dt, represents the time rate of
change in the magnitude of the velocity. Depending on the direction
of Ft, the particle’s speed will either be increasing or decreasing.
The normal acceleration, an = v2/r, represents the time rate of change
in the direction of the velocity vector. Remember, an always acts
toward the path’s center of curvature. Thus, Fn will always be
directed toward the center of the path.
Recall, if the path of motion is defined
as y = f(x), the radius of curvature at
any point can be obtained from
r =
dy 2 3/2
[1 + ( ) ]
dx
d2y
dx2
SOLVING PROBLEMS WITH n-t COORDINATES
• Use n-t coordinates when a particle is moving along a
known, curved path.
• Establish the n-t coordinate system on the particle.
• Draw free-body and kinetic diagrams of the particle. The
normal acceleration (an) always acts “inward” (the positive ndirection). The tangential acceleration (at) may act in either the
positive or negative t direction.
• Apply the equations of motion in scalar form and solve.
• It may be necessary to employ the kinematic
relations:
at = dv/dt = v dv/ds
an = v2/r
EXAMPLE
Given:At the instant q = 60°, the boy’s
center of mass G is momentarily
at rest. The boy has a weight of
60 lb. Neglect his size and the
mass of the seat and cords.
Find: The boy’s speed and the tension
in each of the two supporting
cords of the swing when q = 90°.
Plan:
1) Since the problem involves a curved path and finding the
force perpendicular to the path, use n-t coordinates. Draw
the boy’s free-body and kinetic diagrams.
2) Apply the equation of motion in the n-t directions.
3) Use kinematics to relate the boy’s acceleration to his speed.
EXAMPLE (continued)
Solution:
1) The n-t coordinate system can be established on the
boy at some arbitrary angle q. Approximating the boy
and seat together as a particle, the free-body and
kinetic diagrams can be drawn.
Free-body diagram
Kinetic diagram
n
n
2T
q
=
man
mat
W
t
t
T = tension in each cord
W = weight of the boy
EXAMPLE (continued)
2) Apply the equations of motion in the n-t directions.
(a) Fn = man => 2T – W sin q = man
Using an = v2/r = v2/10, w = 60 lb, and m = w/g = (60/32.2),
we get: 2T – 60 sin q = (60/32.2)(v2/10) (1)
(b) Ft = mat => W cos q = mat
=> 60 cos q = (60/32.2) at
Solving for at: at = 32.2 cos q (2)
Note that there are 2 equations and 3 unknowns (T, v, at).
One more equation is needed.
EXAMPLE (continued)
3) Apply kinematics to relate at and v.
v dv = at ds where ds = r dq = 10 dq
=> v dv = 32.2 cosq ds = 32.2 cosq (10 dq )
v
=>  v dv =
0
=>
90
 32.2 cosq dq
60
v2
2
90
= 32.2 sinq
=> v = 9.29 ft/s
60
This v is the speed of the boy at q = 90. This value can be
substituted into equation (1) to solve for T.
2T – 60 sin(90°) = (60/32.2)(9.29)2/10
T = 38.0 lb (the tension in each cord)