INTRODUCTION &
RECTILINEAR KINEMATICS: CONTINUOUS MOTION
Today’s Objectives:
Students will be able to:
1. Find the kinematic quantities
(position, displacement, velocity,
and acceleration) of a particle
traveling along a straight path.
Dynamics, Fourteenth Edition
R.C. Hibbeler
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RECTILINEAR KINEMATICS: ERRATIC MOTION
Today’s Objectives (continue)
Students will be able to:
2. Determine position,
velocity, and acceleration of
a particle using graphs.
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R.C. Hibbeler
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APPLICATIONS
The motion of large objects,
such as rockets, airplanes, or
cars, can often be analyzed
as if they were particles.
Why?
If we measure the altitude
of this rocket as a function
of time, how can we
determine its velocity and
acceleration?
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R.C. Hibbeler
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APPLICATIONS (continued)
A sports car travels along a straight road.
Can we treat the car as a particle?
If the car accelerates at a constant rate, how can we
determine its position and velocity at some instant?
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R.C. Hibbeler
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An Overview of Mechanics
Mechanics: The study of how bodies
react to the forces acting on them.
Statics: The study of
bodies in equilibrium.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Dynamics:
1. Kinematics – concerned with
the geometric aspects of motion
2. Kinetics - concerned with
the forces causing the motion
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RECTILINEAR KINEMATICS:
CONTINIOUS MOTION (Section 12.2)
A particle travels along a straight-line path
defined by the coordinate axis s.
The position of the particle at any instant,
relative to the origin, O, is defined by the
position vector r, or the scalar s. Scalar s
can be positive or negative. Typical units
for r and s are meters (m) or feet (ft).
The displacement of the particle is
defined as its change in position.
Vector form:  r = r’ - r
Scalar form:  s = s’ - s
The total distance traveled by the particle, sT, is a positive scalar
that represents the total length of the path over which the particle
travels.
Dynamics, Fourteenth Edition
R.C. Hibbeler
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VELOCITY
Velocity is a measure of the rate of change in the position of a particle.
It is a vector quantity (it has both magnitude and direction). The
magnitude of the velocity is called speed, with units of m/s or ft/s.
The average velocity of a particle during a
time interval t is
vavg = r / t
The instantaneous velocity is the time-derivative of position.
v = dr / dt
Speed is the magnitude of velocity: v = ds / dt
Average speed is the total distance traveled divided by elapsed time:
(vsp)avg = sT / t
Dynamics, Fourteenth Edition
R.C. Hibbeler
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ACCELERATION
Acceleration is the rate of change in the velocity of a particle. It is a
vector quantity. Typical units are m/s2 or ft/s2.
The instantaneous acceleration is the time
derivative of velocity.
Vector form: a = dv / dt
Scalar form: a = dv / dt = d2s / dt2
Acceleration can be positive (speed
increasing) or negative (speed decreasing).
As the text shows, the derivative equations for velocity and
acceleration can be manipulated to get a ds = v dv
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SUMMARY OF KINEMATIC RELATIONS:
RECTILINEAR MOTION
• Differentiate position to get velocity and acceleration.
v = ds/dt ;
a = dv/dt or a = v dv/ds
• Integrate acceleration for velocity and position.
Position:
Velocity:
v
t
v
s
 dv =  a dt or  v dv =  a ds
s
t
 ds =  v dt
vo
o
vo
so
so
o
• Note that so and vo represent the initial position and
velocity of the particle at t = 0.
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R.C. Hibbeler
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CONSTANT ACCELERATION
The three kinematic equations can be integrated for the special case
when acceleration is constant (a = ac) to obtain very useful equations.
A common example of constant acceleration is gravity; i.e., a body
freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2
downward. These equations are:
v
t
 dv =  a dt
c
vo
o
s
t
 ds =  v dt
so
v
v = vo + act
yields
s = s o + v ot + (1/2) a c t 2
yields
v 2 = (vo )2 + 2ac(s - so)
o
s
 v dv =  ac ds
vo
yields
so
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R.C. Hibbeler
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EXAMPLE
Given: A particle travels along a straight line to the right
with a velocity of v = ( 4 t – 3 t2 ) m/s where t is
in seconds. Also, s = 0 when t = 0.
Find: The position and acceleration of the particle
when t = 4 s.
Plan: Establish the positive coordinate, s, in the direction the
particle is traveling. Since the velocity is given as a
function of time, take a derivative of it to calculate the
acceleration. Conversely, integrate the velocity
function to calculate the position.
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EXAMPLE (continued)
Solution:
1) Take a derivative of the velocity to determine the acceleration.
a = dv / dt = d(4 t – 3 t2) / dt = 4 – 6 t
 a = – 20 m/s2 (or in the  direction) when t = 4 s
2) Calculate the distance traveled in 4s by integrating the
velocity using so = 0:
s
t
v = ds / dt  ds = v dt   ds =  (4 t – 3 t2) dt
so
o
 s – so = 2 t2 – t3
 s – 0 = 2(4)2 – (4)3  s = – 32 m (or )
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CONCEPT QUIZ
3 m/s
→
5 m/s

t=2s
t=7s
1. A particle moves along a horizontal path with its velocity
varying with time as shown. The average acceleration of the
particle is _________.
A) 0.4 m/s2 →
B) 0.4 m/s2 
C) 1.6 m/s2 →
D) 1.6 m/s2 
2. A particle has an initial velocity of 30 ft/s to the left. If it
then passes through the same location 5 seconds later with a
velocity of 50 ft/s to the right, the average velocity of the
particle during the 5 s time interval is _______.
A) 10 ft/s →
B) 40 ft/s →
C) 16 m/s →
D) 0 ft/s
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R.C. Hibbeler
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GROUP PROBLEM SOLVING
Given: A sandbag is dropped from a balloon ascending
vertically at a constant speed of 6 m/s.
The bag is released with the same upward velocity of
6 m/s at t = 0 s and hits the ground when t = 8 s.
Find: The speed of the bag as it hits the ground and the altitude
of the balloon at this instant.
Plan: The sandbag is experiencing a constant downward
acceleration of 9.81 m/s2 due to gravity. Apply the
formulas for constant acceleration, with ac = - 9.81 m/s2.
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GROUP PROBLEM SOLVING (continued)
Solution:
The bag is released when t = 0 s and hits the ground when
t = 8 s.
Calculate the distance using a position equation.
+ sbag = (sbag )o + (vbag)o t + (1/2) ac t2
sbag = 0 + (-6) (8) + 0.5 (9.81) (8)2 = 265.9 m
During t = 8 s, the balloon rises
+ sballoon = (vballoon) t = 6 (8) = 48 m
Therefore, altitude is of the balloon is (sbag + sballoon).
Altitude = 265.9 + 48 = 313.9 = 314 m.
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R.C. Hibbeler
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GROUP PROBLEM SOLVING (continued)
Calculate the velocity when t = 8 s, by applying a velocity
equation.
+ vbag = (vbag )o + ac t
vbag = -6 + (9.81) 8 = 72.5 m/s 
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R.C. Hibbeler
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READING QUIZ
1. In dynamics, a particle is assumed to have _________.
A) both translation and rotational motions
B) only a mass
C) a mass but the size and shape cannot be neglected
D) no mass or size or shape, it is just a point
2. The average speed is defined as __________.
A) r/t
B) s/t
C) sT/t
D) None of the above.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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ATTENTION QUIZ
1. A particle has an initial velocity of 3 ft/s to the left at
s0 = 0 ft. Determine its position when t = 3 s if the
acceleration is 2 ft/s2 to the right.
A) 0.0 ft
C) 18.0 ft →
B) 6.0 ft 
D) 9.0 ft →
2. A particle is moving with an initial velocity of v = 12 ft/s
and constant acceleration of 3.78 ft/s2 in the same direction
as the velocity. Determine the distance the particle has
traveled when the velocity reaches 30 ft/s.
A) 50 ft
C) 150 ft
Dynamics, Fourteenth Edition
R.C. Hibbeler
B) 100 ft
D) 200 ft
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APPLICATIONS
In many experiments, a
velocity versus position (v-s)
profile is obtained.
If we have a v-s graph for the
tank truck, how can we
determine its acceleration at
position s = 1500 feet?
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R.C. Hibbeler
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APPLICATIONS (continued)
The velocity of a car is
recorded from a experiment.
The car starts from rest and
travels along a straight track.
If we know the v-t plot, how
can we determine the distance
the car traveled during the
time interval 0 < t < 30 s or
15 < t < 25 s?
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ERRATIC MOTION (Section 12.3)
Graphing provides a good way to
handle complex motions that
would be difficult to describe
with formulas.
Graphs also provide a visual
description of motion and
reinforce the calculus concepts of
differentiation and integration as
used in dynamics.
The approach builds on the facts that slope and differentiation
are linked and that integration can be thought of as finding the
area under a curve.
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S-T GRAPH
Plots of position versus time can
be used to find velocity versus
time curves. Finding the slope of
the line tangent to the motion curve
at any point is the velocity at that
point (or v = ds/dt).
Therefore, the v-t graph can be
constructed by finding the slope at
various points along the s-t graph.
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V-T GRAPH
Plots of velocity versus time can be
used to find acceleration versus time
curves. Finding the slope of the line
tangent to the velocity curve at any
point is the acceleration at that point
(or a = dv/dt).
Therefore, the acceleration versus time
(or a-t) graph can be constructed by
finding the slope at various points
along the v-t graph.
Also, the distance moved
(displacement) of the particle is the
area under the v-t graph during time t.
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A-T GRAPH
Given the acceleration versus
time or a-t curve, the change in
velocity (v) during a time
period is the area under the a-t
curve.
So we can construct a v-t graph
from an a-t graph if we know the
initial velocity of the particle.
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A-S GRAPH
A more complex case is presented by
the acceleration versus position or a-s
graph. The area under the a-s curve
represents the change in velocity
(recall  a ds =  v dv ).
s2
½ (v1² – vo²) =  a ds = area under the
s1
a-s graph
This equation can be solved for v1,
allowing you to solve for the velocity
at a point. By doing this repeatedly,
you can create a plot of velocity
versus distance.
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V-S GRAPH
Another complex case is presented
by the velocity versus distance or
v-s graph. By reading the velocity v
at a point on the curve and
multiplying it by the slope of the
curve (dv/ds) at this same point,
we can obtain the acceleration at
that point. Recall the formula
a = v (dv/ds).
Thus, we can obtain an a-s plot
from the v-s curve.
Dynamics, Fourteenth Edition
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EXAMPLE
Given: The v-t graph for a dragster moving along a straight road.
Find: The a-t graph and s-t graph over the time interval shown.
What is your plan of attack for the problem?
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R.C. Hibbeler
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EXAMPLE (continued)
Solution: The a-t graph can be constructed by finding the slope
of the v-t graph at key points. What are those?
when 0 < t < 5 s; v0-5 = ds/dt = d(30t)/dt = 30 m/s2
when 5 < t < 15 s; v5-15 = ds/dt = d(-15t+225)/dt = -15 m/s2
a(m/s2)
a-t graph
30
5
-15
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15 t(s)
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EXAMPLE (continued)
Now integrate the v - t graph to build the s – t graph.
when 0 < t < 5 s; s =  v dt = [15 t2 ] t = 15 t2 m
0
when 0 < t < 5 s; s − 15 (52) =  v dt = [(-15) (1/2) t 2 + 225 t]
s = - 7.5 t 2 + 225 t − 562.5 m
s(m)
s-t graph
1125
-7.5 t2 + 225 t − 562.5
375
t(s)
15t2
5
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15
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t
5
CONCEPT QUIZ
1. If a particle starts from rest and
accelerates according to the graph
shown, the particle’s velocity at
t = 20 s is
A) 200 m/s
B) 100 m/s
C) 0
D) 20 m/s
2. The particle in Problem 1 stops moving at t = _______.
A) 10 s
B) 20 s
C) 30 s
D) 40 s
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GROUP PROBLEM SOLVING I
Given: The v-t graph shown.
Find: The a-t graph, average
speed, and distance
traveled for the 0 - 80 s
interval.
Plan: Find slopes of the v-t curve and draw the a-t graph.
Find the area under the curve. It is the distance traveled.
Finally, calculate average speed (using basic definitions!).
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GROUP PROBLEM SOLVING I (continued)
Solution:
Find the a–t graph.
For 0 ≤ t ≤ 40
a = dv/dt = 0 m/s²
For 40 ≤ t ≤ 80 a = dv/dt = -10 / 40 = -0.25 m/s²
a(m/s²)
0
a-t graph
40
80
-0.25
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t(s)
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GROUP PROBLEM SOLVING I (continued)
Now find the distance traveled:
s0-40 =  v dt =  10 dt = 10 (40) = 400 m
s40-80 =  v dt
=  (20 − 0.25 t) dt
80
2
= [ 20 t -0.25 (1/2) t ]40 = 200 m
s0-90 = 400 + 200 = 600 m
vavg(0-90) = total distance / time
v = 10
= 600 / 80
= 7.5 m/s
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GROUP PROBLEM SOLVING II
Given: The v-t graph shown.
Find: The a-t graph and
distance traveled for
the 0 - 15 s interval.
Plan: Find slopes of the v-t curve and draw the a-t graph.
Find the area under the curve. It is the distance traveled.
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GROUP PROBLEM SOLVING II (continued)
Solution:
Find the a–t graph:
For 0 ≤ t ≤ 4
a = dv/dt = 1.25 m/s²
For 4 ≤ t ≤ 10
a = dv/dt = 0 m/s²
For 10 ≤ t ≤ 15
a = dv/dt = -1 m/s²
a(m/s²)
a-t graph
1.25
4
10
15 t(s)
-1
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GROUP PROBLEM SOLVING II (continued)
Now find the distance traveled:
s0-4 =  v dt
4
2
= [ (1.25) (1/2) t ]0 =
10 m
s4-10 =  v dt = [ 5 t ] 4 = 30 m
10
s10-15 =  v dt
15
2
= [ - (1/2) t + 15 t]10 =
12.5 m
s0-15= 10 + 30 + 12.5 = 52.5 m
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READING QUIZ
1. The slope of a v-t graph at any instant represents instantaneous
A) velocity.
B) acceleration.
C) position.
D) jerk.
2. Displacement of a particle over a given time interval equals the
area under the ___ graph during that time.
A) a-t
B) a-s
C) v-t
C) s-t
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ATTENTION QUIZ
1. If a car has the velocity curve shown, determine the time t
necessary for the car to travel 100 meters. v
A) 8 s
B) 4 s
C) 10 s
D) 6 s
75
6s
t
2. Select the correct a-t graph for the velocity curve shown.
a
a
t
A)
C)
t
B)
a
v
a
t
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D)
t
t
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e
h
Lecture
t
f
o
d
n
E
L et
Learning Continue
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CURVILINEAR MOTION:
GENERAL & RECTANGULAR COMPONENTS
Today’s Objectives:
Students will be able to:
1. Describe the motion of a
particle traveling along a
curved path.
2. Relate kinematic quantities
in terms of the rectangular
components of the vectors.
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MOTION OF A PROJECTILE
Today’s Objectives (continue):
Students will be able to:
3. Analyze the free-flight motion of a
projectile.
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APPLICATIONS
The path of motion of a plane can
be tracked with radar and its x, y,
and z-coordinates (relative to a
point on earth) recorded as a
function of time.
How can we determine the velocity
or acceleration of the plane at any
instant?
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APPLICATIONS (continued)
A roller coaster car travels down
a fixed, helical path at a constant
speed.
How can we determine its
position or acceleration at any
instant?
If you are designing the track, why is it important to be
able to predict the acceleration of the car?
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GENERAL CURVILINEAR MOTION
(Section 12.4)
A particle moving along a curved path undergoes curvilinear motion.
Since the motion is often three-dimensional, vectors are usually used
to describe the motion.
A particle moves along a curve
defined by the path function, s.
The position of the particle at any instant is designated by the vector
r = r(t). Both the magnitude and direction of r may vary with time.
If the particle moves a distance s along the
curve during time interval t, the
displacement is determined by vector
subtraction: r = r’ - r
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VELOCITY
Velocity represents the rate of change in the position of a
particle.
The average velocity of the particle
during the time increment t is
vavg = r/t .
The instantaneous velocity is the
time-derivative of position
v = dr/dt .
The velocity vector, v, is always
tangent to the path of motion.
The magnitude of v is called the speed. Since the arc length s
approaches the magnitude of r as t→0, the speed can be
obtained by differentiating the path function (v = ds/dt). Note
that this is not a vector!
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ACCELERATION
Acceleration represents the rate of change in the
velocity of a particle.
If a particle’s velocity changes from v to v’ over a
time increment t, the average acceleration during
that increment is:
aavg = v/t = (v - v’)/t
The instantaneous acceleration is the timederivative of velocity:
a = dv/dt = d2r/dt2
A plot of the locus of points defined by the arrowhead
of the velocity vector is called a hodograph. The
acceleration vector is tangent to the hodograph, but
not, in general, tangent to the path function.
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CURVILINEAR MOTION:
RECTANGULAR COMPONENTS (Section 12.5)
It is often convenient to describe the motion of a particle in
terms of its x, y, z or rectangular components, relative to a fixed
frame of reference.
The position of the particle can be
defined at any instant by the
position vector
r=xi+yj+zk .
The x, y, z-components may all be
functions of time, i.e.,
x = x(t), y = y(t), and z = z(t) .
The magnitude of the position vector is: r = (x2 + y2 + z2)0.5
The direction of r is defined by the unit vector: ur = (1/r)r
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RECTANGULAR COMPONENTS: VELOCITY
The velocity vector is the time derivative of the position vector:
v = dr/dt = d(x i)/dt + d(y j)/dt + d(z k)/dt
Since the unit vectors i, j, k are constant in magnitude and
direction, this equation reduces to v = vx i + vy j + vz k
•
•
•
where vx = x = dx/dt, vy = y = dy/dt, vz = z = dz/dt
The magnitude of the velocity
vector is
v = [(vx)2 + (vy)2 + (vz)2]0.5
The direction of v is tangent
to the path of motion.
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RECTANGULAR COMPONENTS: ACCELERATION
The acceleration vector is the time derivative of the velocity
vector (second derivative of the position vector).
a = dv/dt = d2r/dt2 = ax i + ay j + az k
•
•
••
••
v
v
x
y
where ax = x =
= dvx /dt, ay = y =
= dvy /dt,
az = v• z = z•• = dvz /dt
The magnitude of the acceleration vector is
a = (ax )2 +(ay )2 +(az )2
The direction of a is usually
not tangent to the path of the
particle.
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EXAMPLE
Given: The box slides down the slope described by the
equation y = (0.05x2) m, where x is in meters.
vx = -3 m/s, ax = -1.5 m/s2 at x = 5 m.
Find: The y components of the velocity and the acceleration
of the box at at x = 5 m.
Plan: Note that the particle’s velocity can be found by taking
the first time derivative of the path’s equation. And the
acceleration can be found by taking the second time
derivative of the path’s equation.
Take a derivative of the position to find the component
of the velocity and the acceleration.
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EXAMPLE (continued)
Solution:
Find the y-component of velocity by taking a time
derivative of the position y = (0.05x2)
 y = 2 (0.05) x x = 0.1 x x
Find the acceleration component by taking a time
derivative of the velocity y

 

 y = 0.1 x x + 0.1 x x
Substituting the x-component of the acceleration, velocity

at x=5 into y and y.
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EXAMPLE (continued)


Since x = vx = -3 m/s, x = ax = -1.5 m/s2 at x = 5 m


 y = 0.1 x x = 0.1 (5) (-3) = -1.5 m/s

y = 0.1 x x + 0.1 x x
= 0.1 (-3)2 + 0.1 (5) (-1.5)
= 0.9 – 0.75
= 0.15 m/s2
At x = 5 m
vy = – 1.5 m/s = 1.5 m/s 
ay = 0.15 m/s2 
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CONCEPT QUIZ
1. If the position of a particle is defined by r = [(1.5t2 + 1) i +
(4t – 1) j ] (m), its speed at t = 1 s is ________.
A) 2 m/s
B) 3 m/s
C) 5 m/s
D) 7 m/s
2. The path of a particle is defined by y = 0.5x2. If the
component of its velocity along the x-axis at x = 2 m is
vx = 1 m/s, its velocity component along the y-axis at this
position is ____.
A) 0.25 m/s
B) 0.5 m/s
C) 1 m/s
D) 2 m/s
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R.C. Hibbeler
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GROUP PROBLEM SOLVING
Given: The particle travels along the path y = 0.5 x2.
When t = 0, x = y = z = 0.
Find: The particle’s distance and the magnitude of its
acceleration when t = 1 s, if vx = (5 t) ft/s, where t is in
seconds.
Plan:
1) Determine x and ax by integrating and differentiating
vx, respectively, using the initial conditions.
2) Find the y-component of velocity & acceleration by
taking a time derivative of the path.
3) Determine the magnitude of the acceleration &
position.
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GROUP PROBLEM SOLVING (continued)
Solution:
1) x-components:
Velocity known as:
Position:

•
vx = x = (5 t ) ft/s  5 ft/s at t=1s
t

vxdt = (5t) dt  x = 2.5 t2  2.5 ft at t=1s
0
••
Acceleration: ax = x = d/dt (5 t)  5 ft/s2 at t=1s
2) y-components:
 3.125 ft at t=1s
Position known as : y = 0.5 x2
•
•
•
Velocity: y = 0.5 (2) x x = x x
••
• •
••
Acceleration: ay = y = x x + x x
Dynamics, Fourteenth Edition
R.C. Hibbeler
 12.5 ft/s at t=1s
 37.5 ft/s2 at t=1s
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GROUP PROBLEM SOLVING (continued)
3) The position vector and the acceleration vector are
Position vector: r = [ x i + y j ] ft
where x= 2.5 ft, y= 3.125 ft
Magnitude: r = 2.52 + 3.1252 = 4.00 ft
Acceleration vector: a = [ ax i + ay j] ft/s2
where ax = 5 ft/s2, ay = 37.5 ft/s2
Magnitude: a = 52 + 37.52 = 37.8 ft/s2
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R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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READING QUIZ
1. In curvilinear motion, the direction of the instantaneous
velocity is always
A) tangent to the hodograph.
B) perpendicular to the hodograph.
C) tangent to the path.
D) perpendicular to the path.
2. In curvilinear motion, the direction of the instantaneous
acceleration is always
A) tangent to the hodograph.
B) perpendicular to the hodograph.
C) tangent to the path.
D) perpendicular to the path.
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R.C. Hibbeler
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ATTENTION QUIZ
1. If a particle has moved from A to B along the circular path in
4s, what is the average velocity of the particle?
y
A) 2.5 i m/s
B) 2.5 i +1.25 j m/s
C) 1.25  i m/s
R=5m
A
x
B
D) 1.25  j m/s
2. The position of a particle is given as r = (4t2 i - 2x j) m.
Determine the particle’s acceleration.
A) (4 i +8 j ) m/s2
B) (8 i -16 j ) m/s2
C) (8 i ) m/s2
D) (8 j ) m/s2
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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APPLICATIONS
A good kicker instinctively knows at what angle, , and initial
velocity, vA, he must kick the ball to make a field goal.
For a given kick “strength”, at what angle should the ball be
kicked to get the maximum distance?
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R.C. Hibbeler
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APPLICATIONS (continued)
A basketball is shot at a certain angle. What parameters should
the shooter consider in order for the basketball to pass through
the basket?
Distance, speed, the basket location, … anything else?
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R.C. Hibbeler
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APPLICATIONS (continued)
A firefighter needs to know the maximum height on the wall
she can project water from the hose. What parameters would
you program into a wrist computer to find the angle, , that
she should use to hold the hose?
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R.C. Hibbeler
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MOTION OF A PROJECTILE (Section 12.6)
Projectile motion can be treated as two rectilinear
motions, one in the horizontal direction experiencing
zero acceleration and the other in the vertical
direction experiencing constant acceleration (i.e.,
from gravity).
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MOTION OF A PROJECTILE (Section 12.6)
For illustration, consider the two balls on
the left. The red ball falls from rest,
whereas the yellow ball is given a
horizontal velocity. Each picture in this
sequence is taken after the same time
interval. Notice both balls are subjected to
the same downward acceleration since
they remain at the same elevation at any
instant. Also, note that the horizontal
distance between successive photos of the
yellow ball is constant since the velocity
in the horizontal direction is constant.
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KINEMATIC EQUATIONS: HORIZONTAL MOTION
Since ax = 0, the velocity in the horizontal direction remains
constant (vx = vox) and the position in the x direction can be
determined by:
x = xo + (vox) t
Why is ax equal to zero (what assumption must be made if the
movement is through the air)?
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R.C. Hibbeler
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KINEMATIC EQUATIONS: VERTICAL MOTION
Since the positive y-axis is directed upward, ay = – g.
Application of the constant acceleration equations yields:
vy = voy – g t
y = yo + (voy) t – ½ g t2
vy2 = voy2 – 2 g (y – yo)
For any given problem, only two of these three equations
can be used. Why?
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R.C. Hibbeler
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EXAMPLE I
Given: vA and θ
Find: Horizontal distance it
travels and vC.
Plan: Apply the kinematic relations
in x- and y-directions.
Solution: Using vAx = 10 cos 30 and vAy = 10 sin 30
We can write vx = 10 cos 30
vy = 10 sin 30 – (9.81) t
x = (10 cos 30) t
y = (10 sin 30) t – ½ (9.81) t2
Since y = 0 at C
0 = (10 sin 30) t – ½ (9.81) t2  t = 0, 1.019 s
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EXAMPLE I (continued)
Only the time of 1.019 s makes sense!
Velocity components at C are;
vCx = 10 cos 30
= 8.66 m/s →
vCy = 10 sin 30 – (9.81) (1.019)
= -5 m/s = 5 m/s 
vC =
8.662 + (−5)2 =10 m/s
Horizontal distance the ball travels is;
x = (10 cos 30) t
x = (10 cos 30) 1.019 = 8.83 m
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R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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EXAMPLE II
Given: Projectile is fired with vA=150 m/s
at point A.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Find:
The horizontal distance it travels (R)
and the time in the air.
Plan:
How will you proceed?
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EXAMPLE II
Given: Projectile is fired with vA=150 m/s
at point A.
Find:
The horizontal distance it travels (R)
and the time in the air.
Plan: Establish a fixed x, y coordinate system (in this solution,
the origin of the coordinate system is placed at A).
Apply the kinematic relations in x- and y-directions.
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EXAMPLE II (continued)
Solution:
1) Place the coordinate system at point A.
Then, write the equation for horizontal motion.
+ → xB = xA + vAx tAB
where xB = R, xA = 0, vAx = 150 (4/5) m/s
Range, R, will be R = 120 tAB
2) Now write a vertical motion equation. Use the distance equation.
+ yB = yA + vAy tAB – 0.5 g tAB2
where yB = – 150, yA = 0, and vAy = 150(3/5) m/s
We get the following equation: –150 = 90 tAB + 0.5 (– 9.81) tAB2
Solving for tAB first, tAB = 19.89 s.
Then, R = 120 tAB = 120 (19.89) = 2387 m
Dynamics, Fourteenth Edition
R.C. Hibbeler
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CONCEPT QUIZ
1. In a projectile motion problem, what is the maximum
number of unknowns that can be solved?
2.
A) 1
B) 2
C) 3
D) 4
The time of flight of a projectile, fired over level ground, with initial
velocity Vo at angle θ, is equal to?
A) (vo sin )/g
B) (2vo sin )/g
C) (vo cos )/g
D) (2vo cos )/g
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R.C. Hibbeler
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GROUP PROBLEM SOLVING I
y
x
Given: A skier leaves the ski
jump ramp at A = 25o
and hits the slope at B.
Find: The skier’s initial speed vA.
Plan:
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R.C. Hibbeler
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GROUP PROBLEM SOLVING I
y
x
Given: A skier leaves the ski
jump ramp at A = 25o
and hits the slope at B.
Find: The skier’s initial speed vA.
Plan: Establish a fixed x,y coordinate system (in this solution,
the origin of the coordinate system is placed at A).
Apply the kinematic relations in x- and y-directions.
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R.C. Hibbeler
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GROUP PROBLEM SOLVING I (continued)
Solution:
Motion in x-direction:
Using xB = xA + vox(tAB)  (4/5)100 = 0 + vA (cos 25) tAB
tAB=
80
vA (cos 25)
=
88.27
vA
Motion in y-direction:
Using
yB = yA + voy(tAB) – ½ g(tAB)2
88.27
88.27 }2
– 64 = 0 + vA(sin 25) {
} – ½ (9.81) {
vA
vA
vA = 19.42 m/s
tAB= (88.27 / 19.42) = 4.54 s
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING II
Given: The golf ball is struck
with a velocity of 80
ft/s as shown.
y
x
Find: Distance d to where it
will land.
Plan:
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R.C. Hibbeler
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GROUP PROBLEM SOLVING II
Given: The golf ball is struck
with a velocity of 80
ft/s as shown.
y
x
Find: Distance d to where it
will land.
Plan: Establish a fixed x, y coordinate system (in this solution,
the origin of the coordinate system is placed at A).
Apply the kinematic relations in x- and y-directions.
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R.C. Hibbeler
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GROUP PROBLEM SOLVING II (continued)
Solution:
Motion in x-direction:
Using xB = xA + vox(tAB)
 d cos10 = 0 + 80 (cos 55) tAB
y
x
tAB = 0.02146 d
Motion in y-direction:
Using
yB = yA + voy(tAB) – ½ g(tAB)2
 d sin10 = 0 + 80(sin 55)(0.02146 d) – ½ 32.2 (0.02146 d)2
 0 = 1.233 d – 0.007415 d2
d = 0, 166 ft
Only the non-zero answer is meaningful.
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R.C. Hibbeler
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READING QUIZ
1. The downward acceleration of an object in free-flight
motion is
A) zero.
B) increasing with time.
C) 9.81 m/s2.
D) 9.81 ft/s2.
2. The horizontal component of velocity remains _________
during a free-flight motion.
A) zero
B) constant
C) at 9.81 m/s2
D) at 32.2 ft/s2
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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ATTENTION QUIZ
1. A projectile is given an initial velocity
vo at an angle  above the horizontal.
The velocity of the projectile when it
hits the slope is ____________ the
initial velocity vo.
A) less than
C) greater than
B) equal to
D) None of the above.
2. A particle has an initial velocity vo at angle  with respect to the
horizontal. The maximum height it can reach is when
A)  = 30°
B)  = 45°
C)  = 60°
D)  = 90°
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R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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e
h
Lecture
t
f
o
d
n
E
L et
Learning Continue
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R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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CURVILINEAR MOTION:
NORMAL AND TANGENTIAL COMPONENTS
Today’s Objectives:
Students will be able to:
1. Determine the normal and
tangential components of
velocity and acceleration of a
particle traveling along a
curved path.
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R.C. Hibbeler
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CURVILINEAR MOTION: CYLINDRICAL
COMPONENTS
Today’s Objectives (continue):
2. Determine velocity and
acceleration components
using cylindrical coordinates.
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R.C. Hibbeler
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APPLICATIONS
Cars traveling along a clover-leaf
interchange experience an
acceleration due to a change in
velocity as well as due to a change
in direction of the velocity.
If the car’s speed is increasing at a
known rate as it travels along a
curve, how can we determine the
magnitude and direction of its total
acceleration?
Why would you care about the total acceleration of the car?
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R.C. Hibbeler
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APPLICATIONS (continued)
As the boy swings upward with a
velocity v, his motion can be
analyzed using n–t coordinates.
y
x
As he rises, the magnitude of his
velocity is changing, and thus his
acceleration is also changing.
How can we determine his velocity and acceleration at the
bottom of the arc?
Can we use different coordinates, such as x-y coordinates,
to describe his motion? Which coordinate system would
be easier to use to describe his motion? Why?
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R.C. Hibbeler
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APPLICATIONS (continued)
A roller coaster travels down a
hill for which the path can be
approximated by a function
y = f(x).
The roller coaster starts from rest
and increases its speed at a
constant rate.
How can we determine its velocity
and acceleration at the bottom?
Why would we want to know
these values?
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R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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NORMAL AND TANGENTIAL COMPONENTS
(Section 12.7)
When a particle moves along a curved path, it is sometimes convenient
to describe its motion using coordinates other than Cartesian. When the
path of motion is known, normal (n) and tangential (t) coordinates are
often used.
In the n-t coordinate system, the
origin is located on the particle
(thus the origin and coordinate
system move with the particle).
The t-axis is tangent to the path (curve) at the instant considered,
positive in the direction of the particle’s motion.
The n-axis is perpendicular to the t-axis with the positive direction
toward the center of curvature of the curve.
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NORMAL AND TANGENTIAL COMPONENTS
(continued)
The positive n and t directions are
defined by the unit vectors un and ut,
respectively.
The center of curvature, O’, always
lies on the concave side of the curve.
The radius of curvature, , is defined
as the perpendicular distance from
the curve to the center of curvature at
that point.
The position of the particle at any
instant is defined by the distance, s, along the curve from a
fixed reference point.
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VELOCITY IN THE n-t COORDINATE SYSTEM
The velocity vector is always
tangent to the path of motion
(t-direction).
The magnitude is determined by taking the time derivative of
the path function, s(t).
.
v = v ut where v = s = ds/dt
Here v defines the magnitude of the velocity (speed) and
ut defines the direction of the velocity vector.
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ACCELERATION IN THE n-t COORDINATE SYSTEM
Acceleration is the time rate of change
of velocity:
.
.
a = dv/dt = d(vut)/dt = vut + vut
.
Here v represents the change in
.
the magnitude of velocity and ut
represents the rate of change in
the direction of ut.
After mathematical manipulation,
the acceleration vector can be
expressed as:
.
a = v ut + (v2/) un = at ut + an un.
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ACCELERATION IN THE n-t COORDINATE SYSTEM
(continued)
So, there are two components to the
acceleration vector:
a = at ut + an un
• The tangential component is tangent to the curve and in the
direction of increasing or decreasing velocity.
.
at = v or at ds = v dv
• The normal or centripetal component is always directed
toward the center of curvature of the curve. an = v2/
• The magnitude of the acceleration vector is
a = (an )2 +(at )2
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R.C. Hibbeler
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SPECIAL CASES OF MOTION
There are some special cases of motion to consider.
1) The particle moves along a straight line.
.
2

 => an = v / =  = a = at = v
The tangential component represents the time rate of change in
the magnitude of the velocity.
2) The particle moves along a curve at constant speed.
.
at = v = 0 => a = an = v2/
The normal component represents the time rate of change in the
direction of the velocity.
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SPECIAL CASES OF MOTION (continued)
3) The tangential component of acceleration is constant, at = (at)c.
In this case,
s = so + vo t + (1/2) (at)c t2
v = vo + (at)c t
v2 = (vo)2 + 2 (at)c (s – so)
As before, so and vo are the initial position and velocity of the
particle at t = 0. How are these equations related to projectile
motion equations? Why?
4) The particle moves along a path expressed as y = f(x).
The radius of curvature,  at any point on the path can be
calculated from
[ 1 + (dy/dx)2 ]3/2
 = ________________
d2y/dx2
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R.C. Hibbeler
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THREE-DIMENSIONAL MOTION
If a particle moves along a space curve,
the n-t axes are defined as before. At
any point, the t-axis is tangent to the
path and the n-axis points toward the
center of curvature. The plane
containing the n-t axes is called the
osculating plane.
A third axis can be defined, called the binomial axis, b. The
binomial unit vector, ub, is directed perpendicular to the osculating
plane, and its sense is defined by the cross product ub = ut × un.
There is no motion, thus no velocity or acceleration, in the
binomial direction.
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R.C. Hibbeler
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EXAMPLE I
Given: A car travels along the road
with a speed of v = (2s) m/s,
where s is in meters.
 = 50 m
Find: The magnitudes of the car’s
acceleration at s = 10 m.
Plan:
1) Calculate the velocity when s = 10 m using v(s).
2) Calculate the tangential and normal components of
acceleration and then the magnitude of the
acceleration vector.
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EXAMPLE I (continued)
Solution:
1) The velocity vector is v = v ut , where the magnitude is
given by v = (2s) m/s.
When s = 10 m: v = 20 m/s
.
2) The acceleration vector is a = atut + anun = vut + (v2/)un
Tangential component:
.
Since at = v = dv/dt = (dv/ds) (ds/dt) = v (dv/ds)
where v = 2s  at = d(2s)/ds (v)= 2 v
At s = 10 m: at = 40 m/s2
Normal component: an = v2/
When s = 10 m: an = (20)2 / (50) = 8 m/s2
The magnitude of the acceleration is
a = (an )2 +(at )2 = 402 + 82 = 40.8 m/s2
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R.C. Hibbeler
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EXAMPLE II
Given: A boat travels around a
circular path,  = 40 m, at a
speed that increases with
time, v = (0.0625 t2) m/s.
Find: The magnitudes of the boat’s
velocity and acceleration at
the instant t = 10 s.
Plan:
The boat starts from rest (v = 0 when t = 0).
1) Calculate the velocity at t = 10 s using v(t).
2) Calculate the tangential and normal components of
acceleration and then the magnitude of the
acceleration vector.
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EXAMPLE II (continued)
Solution:
1) The velocity vector is v = v ut , where the magnitude is
given by v = (0.0625t2) m/s. At t = 10s:
v = 0.0625 t2 = 0.0625 (10)2 = 6.25 m/s
.
2) The acceleration vector is a = atut + anun = vut + (v2/)un.
.
Tangential component: at = v = d(.0625 t2 )/dt = 0.125 t m/s2
At t = 10s: at = 0.125t = 0.125(10) = 1.25 m/s2
Normal component: an = v2/ m/s2
At t = 10s: an = (6.25)2 / (40) = 0.9766 m/s2
The magnitude of the acceleration is
a = (an )2 +(at )2 = 1.252 + 0.97662 = 1.59 m/s2
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R.C. Hibbeler
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CONCEPT QUIZ
1. A particle traveling in a circular path of radius 300 m has an
instantaneous velocity of 30 m/s and its velocity is
increasing at a constant rate of 4 m/s2. What is the
magnitude of its total acceleration at this instant?
A) 3 m/s2
B) 4 m/s2
C) 5 m/s2
D) -5 m/s2
2. If a particle moving in a circular path of radius 5 m has a
velocity function v = 4t2 m/s, what is the magnitude of its
total acceleration at t = 1 s?
A) 8 m/s
B) 8.6 m/s
C) 3.2 m/s
D) 11.2 m/s
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING I
Given: The train engine at E has a
speed of 20 m/s and an
acceleration of 14 m/s2 acting
in the direction shown.
at
an
Find: The rate of increase in the
train’s speed and the radius of
curvature  of the path.
Plan:
1. Determine the tangential and normal components of the
acceleration.
2. Calculate vሶ from the tangential component of the
acceleration.
3. Calculate  from the normal component of the acceleration.
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Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING I (continued)
Solution:
1) Acceleration
Tangential component :
at =14 cos(75) = 3.623 m/s2
Normal component :
an = 14 sin(75) = 13.52 m/s2
2) The tangential component of acceleration is the rate of
increase of the train’s speed, so
at = vሶ = 3.62 m/s2.
3) The normal component of acceleration is
an = v2/  13.52 = 202 / 
 =  m
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING II
Given: Starting from rest, a bicyclist travels around a
horizontal circular path,  = 10 m, at a speed of
v = (0.09 t2 + 0.1 t) m/s.
Find: The magnitudes of her velocity and acceleration when
she has traveled 3 m.
Plan:
The bicyclist starts from rest (v = 0 when t = 0).
1) Integrate v(t) to find the position s(t).
2) Calculate the time when s = 3 m using s(t).
3) Calculate the tangential and normal components of
acceleration and then the magnitude of the
acceleration vector.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING II (continued)
Solution:
1) The velocity vector is v = (0.09 t2 + 0.1 t) m/s, where t is in
seconds. Integrate the velocity and find the position s(t).
Position:

v dt =

(0.09 t2 + 0.1 t) dt
s (t) = 0.03 t3 + 0.05 t2
2) Calculate the time, t when s = 3 m.
3 = 0.03 t3 + 0.05 t2
Solving for t, t = 4.147 s
The velocity at t = 4.147 s is,
v = 0.09 (4.147 ) 2 + 0.1 (4.147 ) = 1.96 m/s
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING II (continued)
.
3) The acceleration vector is a = atut + anun = vut + (v2/)un.
Tangential component:
.
at = v = d(0.09 t2 + 0.1 t) / dt = (0.18 t + 0.1) m/s2
At t = 4.147 s : at = 0.18 (4.147) + 0.1 = 0.8465 m/s2
Normal component:
an = v2/ m/s2
At t = 4.147 s : an = (1.96)2 / (10) = 0.3852 m/s2
The magnitude of the acceleration is
a = (an )2 +(at )2 = 0.84652 + 0.38522 = 0.930 m/s2
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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READING QUIZ
1. If a particle moves along a curve with a constant speed, then
its tangential component of acceleration is
A) positive.
B) negative.
C) zero.
D) constant.
2. The normal component of acceleration represents
A) the time rate of change in the magnitude of the velocity.
B) the time rate of change in the direction of the velocity.
C) magnitude of the velocity.
D) direction of the total acceleration.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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ATTENTION QUIZ
1. The magnitude of the normal acceleration is
A) proportional to radius of curvature.
B) inversely proportional to radius of curvature.
C) sometimes negative.
D) zero when velocity is constant.
2. The directions of the tangential acceleration and velocity are
always
A) perpendicular to each other. B) collinear.
C) in the same direction.
Dynamics, Fourteenth Edition
R.C. Hibbeler
D) in opposite directions.
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APPLICATIONS
A cylindrical coordinate
system is used in cases
where the particle moves
along a 3-D curve.
In the figure shown, the box
slides down the helical ramp.
How would you find the
box’s velocity components to
check to see if the package
will fly off the ramp?
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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APPLICATIONS (continued)
The cylindrical coordinate
system can be used to describe
the motion of the girl on the
slide.
Here the radial coordinate is
constant, the transverse
coordinate increases
with time as the girl rotates
about the vertical axis, and her
altitude, z, decreases with time.
How can you find her acceleration components?
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
CYLINDRICAL COMPONENTS
(Section 12.8)
We can express the location of P in polar coordinates as r = r ur.
Note that the radial direction, r, extends outward from the fixed
origin, O, and the transverse coordinate,  is measured counterclockwise (CCW) from the horizontal.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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VELOCITY in POLAR COORDINATES)
The instantaneous velocity is defined as:
v = dr/dt = d(rur)/dt
dur
.
v = rur + r
dt
Using the chain rule:
dur/dt = (dur/d)(d/dt)

We can prove that dur/d
  = uθ so dur/dt = uθ

Therefore: v = rur + ruθ

Thus, the velocity vector has two components:
r,

called the radial component, and r called the
transverse component. The speed of the particle at
any given instant is the sum of the squares of both
components or
 2

v = (r  ) + ( r )2
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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ACCELERATION (POLAR COORDINATES)
The instantaneous acceleration is defined as:
.
.
a = dv/dt = (d/dt)(rur + ruθ)
After manipulation, the acceleration can be
expressed as
.. . .
.
..
2
a = (r – r )ur + (r + 2r )uθ
.
..
The term (r – r 2) is the radial acceleration
or ar .
..
The term (r + 2r ) is the transverse
acceleration or a 
.. . .
.2 2
..
The magnitude of acceleration is a = (r – r ) + (r + 2r ) 2
..
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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CYLINDRICAL COORDINATES
If the particle P moves along a space
curve, its position can be written as
rP = rur + zuz
Taking time derivatives and using
the chain rule:
Velocity:
.
.
.
vP = rur + ruθ + zuz
.
.. . .
..
..
2
Acceleration: aP = (r – r )ur + (r + 2r )uθ + zuz
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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EXAMPLE
Given: The platform is rotating such
that, at any instant, its angular
position is  = (4t3/2) rad, where
t is in seconds.
A ball rolls outward so that its
position is r = (0.1t3) m.
Find: The magnitude of velocity and acceleration of the
ball when t = 1.5 s.
Plan: Use a polar coordinate system and related
kinematic equations
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R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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EXAMPLE (continued)
Solution:
𝑟 = 0.1𝑡 3 , rሶ = 0.3 t 2 , rሷ = 0.6 t
𝜃 = 4 t3/2, 𝜃ሶ = 6 t1/2, 𝜃ሷ = 3 t−1/2
At t=1.5 s,
r = 0.3375 m, rሶ = 0.675 m/s, rሷ = 0.9 m/s2
𝜃 = 7.348 rad, 𝜃ሶ = 7.348 rad/s, 𝜃ሷ = 2.449 rad/s2
Substitute into the equation
for velocity
.
.
v = r ur + r uθ = 0.675 ur + 0.3375 (7.348) uθ
= 0.675 ur + 2.480 uθ
v = (0.675)2 + (2.480)2 = 2.57 m/s
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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EXAMPLE (continued)
Substitute in the equation for acceleration:
.. . .
.
..
2
a = (r – r )ur + (r + 2r)uθ
a = [0.9 – 0.3375(7.348)2] ur
+ [0.3375(2.449) + 2(0.675)(7.348)] uθ
a = – 17.33 ur + 10.75 uθ m/s2
a = (– 17.33)2 + (10.75)2 = 20.4 m/s2
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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CONCEPT QUIZ
.
1. If r is zero for a particle, the particle is
A) not moving.
B) moving in a circular path.
C) moving on a straight line.
D) moving with constant velocity.
2. If a particle moves in a circular path with constant velocity, its
radial acceleration is
A) zero.
.
C) − r 2.
Dynamics, Fourteenth Edition
R.C. Hibbeler
..
B) r .
. .
D) 2r 
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GROUP PROBLEM SOLVING
Given: The arm of the robot is
extending at a constant rate
𝑟ሶ = 1.5 ft/s when r = 3 ft,
z = (4t2) ft, and  = (0.5 t) rad,
where t is in seconds.
Find: The velocity and acceleration
of the grip A when t = 3 s.
Plan: Use cylindrical coordinates.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING (continued)
Solution:
When t = 3 s, r = 3 ft and the arm is extending at a constant
rate 𝑟ሶ = 1.5 ft/s. Thus 𝑟ሷ = 0 ft/s2
𝜃 = 1.5 t = 4.5 rad, 𝜃ሶ = 1.5 rad/s,
𝜃ሷ = 0 rad/s2
z = 4 t2 = 36 ft,
zሶ = 8 t = 24 ft/s,
zሷ = 8 ft/s2
Substitute in the equation
for velocity
.
.
.
v = r ur + r uθ + z ur
= 1.5 ur + 3 (1.5) uθ + 24 uz
= 1.5 ur + 4.5 uθ + 24 uz
Magnitude v = (1.5)2 + (4.5)2 + (24)2 = 24.5 ft/s
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING (continued)
Acceleration equation in cylindrical coordinates
.
.. . .
..
..
2
a = (r – r )ur + (r + 2r)uθ + zuz
= {0 – 3 (1.5)2}ur +{3 (0) + 2 (1.5) 1.5 } uθ + 8 uz
a = [6.75 ur + 4.5 uθ + 8 uz] ft/s2
a = (6.75)2 + (4.5)2 + (8)2 = 11.4 ft/s2
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
READING QUIZ
1. In a polar coordinate system, the velocity
vector can
be
.
.
.
written as v = vrur + vθuθ = rur + ru. The term  is called
A) transverse velocity.
B) radial velocity.
C) angular velocity.
D) angular acceleration.
2. The speed of a particle in a cylindrical coordinate system is
.
.
A) r
C)
B) r
.2 .2
(r) + (r)
Dynamics, Fourteenth Edition
R.C. Hibbeler
D)
.2 . 2 . 2
(r) + (r) + (z)
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ATTENTION QUIZ
1. The radial component of velocity of a particle moving in a
circular path is always
A) zero.
B) constant.
C) greater than its transverse component.
D) less than its transverse component.
2. The radial component of acceleration of a particle moving in
a circular path is always
A) negative.
B) directed toward the center of the path.
C) perpendicular to the transverse component of acceleration.
D) All of the above.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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e
h
Lecture
t
f
o
d
n
E
L et
Learning Continue
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
RELATIVE-MOTION ANALYSIS OF TWO PARTICLES
USING TRANSLATING AXES
Today’s Objectives :
Students will be able to
1. Understand translating frames of reference.
2. Use translating frames of reference to analyze
relative motion.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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APPLICATIONS
When you try to hit a
moving object, the position,
velocity, and acceleration of
the object all have to be
accounted for by your mind.
You are smarter than you
thought!
Here, the boy on the ground is at d = 10 ft when the girl in
the window throws the ball to him.
If the boy on the ground is running at a constant speed of 4
ft/s, how fast should the ball be thrown?
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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APPLICATIONS (continued)
When fighter jets take off or
land on an aircraft carrier,
the velocity of the carrier
becomes an issue.
If the aircraft carrier is underway with a forward velocity of 50
km/hr and plane A takes off at a horizontal air speed of 200
km/hr (measured by someone on the water), how do we find the
velocity of the plane relative to the carrier?
How would you find the same thing for airplane B?
How does the wind impact this sort of situation?
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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RELATIVE POSITION (Section 12.10)
The absolute positions of two
particles A and B with respect to
the fixed x, y, z-reference frame are
given by rA and rB. The position of
B relative to A is represented by
rB/A = rB – rA
Therefore, if rB = (10 i + 2 j ) m
and
rA = (4 i + 5 j ) m,
then
rB/A = rB – rA = (6 i – 3 j ) m.
Dynamics, Fourteenth Edition
R.C. Hibbeler
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RELATIVE VELOCITY
To determine the relative velocity of B
with respect to A, the time derivative of
the relative position equation is taken.
vB/A = vB – vA
or
vB = vA + vB/A
In these equations, vB and vA are called absolute velocities
and vB/A is the relative velocity of B with respect to A.
Note that vB/A = - vA/B .
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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RELATIVE ACCELERATION
The time derivative of the relative
velocity equation yields a similar
vector relationship between the
absolute and relative accelerations
of particles A and B.
These derivatives yield: aB/A = aB – aA
or
aB = aA + aB/A
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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SOLVING PROBLEMS
Since the relative motion equations are vector equations,
problems involving them may be solved in one of two ways.
For instance, the velocity vectors in vB = vA + vB/A could be
written as two dimensional (2-D) Cartesian vectors and the
resulting 2-D scalar component equations solved for up to
two unknowns.
Alternatively, vector problems can be solved “graphically” by
use of trigonometry. This approach usually makes use of the
law of sines or the law of cosines.
Could a CAD system be used to solve these types of problems?
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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LAWS OF SINES AND COSINES
Since vector addition or subtraction forms
a triangle, sine and cosine laws can be
applied to solve for relative or absolute
velocities and accelerations. As a review,
their formulations are provided below.
C
b
a
B
A
c
Law of Sines:
a
sin A
Law of Cosines:
b
=
sin B
c
=
sin C
a 2 = b 2 + c 2 − 2 bc cos A
b = a + c
2
2
− 2 ac cos B
2
c = a + b − 2 ab cos C
2
Dynamics, Fourteenth Edition
R.C. Hibbeler
2
2
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EXAMPLE
Given:
Two aircraft as shown.
vA = 650 km/h
vB = 800 km/h
Find:
vB/A
Plan:
1) Vector Method: Write vectors vA and vB in Cartesian
form, then determine vB – vA
2) Graphical Method: Draw vectors vA and vB from a
common point. Apply the laws of sines and cosines to
determine vB/A.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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EXAMPLE (continued)
Solution:
1) Vector Method
vA = (650 i ) km/h
vB = –800 cos 60 i – 800 sin 60 j
= ( –400 i – 692.8 j) km/h
vB/A = vB – vA = (–1050 i – 692.8 j) km/h
vB /A = (−) +(−) =  km/h
 = tan-1(

) = 33.4

Dynamics, Fourteenth Edition
R.C. Hibbeler

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EXAMPLE (continued)
2) Graphical Method:
Note that the vector that measures the tip of B relative to A is vB/A.
650 km/h
vA
120
vB
vB/A
Law of Cosines:
(vB/A)2 = (800) 2 + (650) 2 − (800) (650) cos 120
vB/A = 1258 km/h
Law of Sines:
vB/A
sin(120 )
Dynamics, Fourteenth Edition
R.C. Hibbeler
=
vA
sin 
or  = 
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CONCEPT QUIZ
1. Two particles, A and B, are moving in
the directions shown. What should be
the angle  so that vB/A is minimum?
A) 0°
B) 180°
C) 90°
D) 270°
vB = 4 ft/s

B
vA= 3 ft/s
A
2. Determine the velocity of plane A with respect to plane B.
A) (400 i + 520 j ) km/hr
B) (1220 i - 300 j ) km/hr
30
C) (-181 i - 300 j ) km/hr
D) (-1220 i + 300 j ) km/hr
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING
Given: Car A moves in a straight line
while Car B moves along a
curve having a radius of
curvature of 200 m.
vA = 40 m/s
vB = 30 m/s
aA = 4 m/s2
aB = -3 m/s2
Find:
vB/A
aB/A
Plan: Write the velocity and acceleration vectors for
Cars A and B. Determine vB/A and aB/A by using
vector relationships.
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R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING (continued)
y
Solution:
The velocity of B is:
vB = -30 sin(30) i + 30 cos(30) j
= (-15 i + 25.98 j) m/s
The velocity of A is:
vA = 40 j (m/s)
x
The relative velocity of B with respect to A is (vB/A):
vB/A = vB – vA = (-15 i + 25.98 j) – (40 j) = (-15 i – 14.02 j) m/s
or
vB/A =
(-15)2 + (-14.02)2 = 20.5 m/s
 = tan-1(
14.02
) = 43.1°
15
Dynamics, Fourteenth Edition
R.C. Hibbeler

Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING (continued)
Since car B is traveling along a curve, the acceleration of B is:
aB = (at)B + (an)B = [-(-3) sin(30) i + (-3) cos(30) j]
2
302
30
+ [- (
) cos(30) i - (
) sin(30) j ]
200
200
y
aB = ( -2.397 i – 4.848 j ) m/s2
The acceleration of A is: aA = (4 j ) m/s2
x
The relative acceleration of B with respect to A is:
aB/A = aB – aA = ( -2.397 i – 8.848 j ) m/s2
aB/A =
(-2.397)2 + (-8.848)2 = 9.17 m/s2
 = tan-1(8.848 / 2.397) = 74.8°
Dynamics, Fourteenth Edition
R.C. Hibbeler

Copyright ©2016 by Pearson Education, Inc.
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READING QUIZ
1. The velocity of B relative to A is defined as
A) vB – vA .
B) vA – vB .
C) vB + vA .
D) vA + vB .
2. Since two-dimensional vector addition forms a triangle,
there can be at most _________ unknowns (either
magnitudes and/or directions of the vectors).
A) one
B) two
C) three
D) four
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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ATTENTION QUIZ
1. Determine the relative velocity of particle B with respect to
particle A.
y
A) (48i + 30j) km/h
B
vB=100 km/h
B) (- 48i + 30j ) km/h
C) (48i - 30j ) km/h
D) (- 48i - 30j ) km/h
30
A
x
vA=60 km/h
2. If theta equals 90° and A and B start moving from the same
point, what is the magnitude of rB/A at t = 5 s?
A) 20 ft
vB = 4 ft/s
B) 15 ft
C) 18 ft
D) 25 ft
Dynamics, Fourteenth Edition
R.C. Hibbeler

B
A
vA = 3 ft/s
Copyright ©2016 by Pearson Education, Inc.
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e
h
Lecture
t
f
o
d
n
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L et
Learning Continue
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
NEWTON’S LAWS OF MOTION, EQUATIONS OF MOTION, &
EQUATIONS OF MOTION FOR A SYSTEM OF PARTICLES
Today’s Objectives:
Students will be able to:
1. Write the equation of motion
for an accelerating body.
2. Draw the free-body and kinetic
diagrams for an accelerating
body.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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EQUATIONS OF MOTION:
RECTANGULAR COORDINATES
Today’s Objectives (continue):
3. Apply Newton’s second law
to determine forces and
accelerations for particles in
rectilinear motion.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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APPLICATIONS
The motion of an object depends on the
forces acting on it.
A parachutist relies on the atmospheric
drag resistance force generated by her
parachute to limit her velocity.
Knowing the drag force, how can we
determine the acceleration or velocity of
the parachutist at any point in time? This
has some importance when landing!
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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APPLICATIONS (continued)
The baggage truck A tows a cart B,
and a cart C.
If we know the frictional force
developed at the driving wheels of
the truck, could we determine the
acceleration of the truck?
How?
Can we also determine the horizontal force acting on the
coupling between the truck and cart B? This is needed when
designing the coupling (or understanding why it failed).
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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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 and load at a given
acceleration? This is needed to decide
the size of the cable that should be
used.
Is the tension force in the cable greater than the weight
of the elevator and its load?
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
NEWTON’S LAWS OF MOTION (Section 13.1)
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.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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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 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).
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
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
m m
F = G 12 2
r
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.
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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.
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UNITS: SI SYSTEM VERSUS 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
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UNITS: SI SYSTEM VERSUS FPS SYSTEM
(continued)
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
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EQUATION OF MOTION (Section 13.2)
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 free-body
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.
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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.
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EQUATION OF MOTION FOR A SYSTEM OF PARTICLES
(Section 13.3)
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 m aG = mi ai .
The text shows the details, but for a system of particles: F = m aG
where F is the sum of the external forces acting on the entire
system.
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KEY POINTS
1) Newton’s second law is a “law of nature”-- experimentally
proven, not the result of an analytical proof.
2) Mass (a 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 = mg.
4) Unbalanced forces cause the acceleration of objects.
This condition is fundamental to all dynamics problems!
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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.
5) It may be necessary to apply the proper kinematic relations
to generate additional equations.
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EXAMPLE
Given: A 25-kg block is subjected to
the force F=100 N. The
spring has a stiffness of k =
200 N/m and is unstretched
when the block is at A. The
contact surface is smooth.
Find: Draw the free-body and kinetic diagrams of the block
when s=0.4 m.
Plan: 1) Define an inertial coordinate system.
2) Draw the block’s free-body diagram, showing all
external forces.
3) Draw the block’s kinetic diagram, showing the inertial
force vector in the proper direction.
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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 block:
W = 25g
y
F=100 (N)
x
3
4
Fs= 200  (N)
= 40 (N)
N
Dynamics, Fourteenth Edition
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The weight force (W) acts through the
block’s center of mass. F is the applied
load and Fs = 200 (N) is the spring
force, where  is the spring deformation.
When s = 0.4,
 = 0.5 − 0.3 = 0.2 m.
The normal force (N) is perpendicular to
the surface. There is no friction force
since the contact surface is smooth.
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EXAMPLE (continued)
3) Draw the kinetic diagram of the block.
25 a
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The block will be moved to the right.
The acceleration can be directed to
the right if the block is speeding up or
to the left if it is slowing down.
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CONCEPT QUIZ
1. The block (mass = m) is moving upward with a speed v.
Draw the FBD if the kinetic friction coefficient is k.
mg
mg
A)
kN
v
B)
kN
N
N
mg
kmg
C)
D) None of the above.
N
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CONCEPT QUIZ
2. Packaging for oranges is tested using a machine that exerts
ay = 20 m/s2 and ax = 3 m/s2, simultaneously. Select the
y
correct FBD and kinetic diagram for this condition.
A)
may
W
=
•
x
W
B)
=
max
Rx
•
max
Rx
Ry
Ry
C)
may
=
D)
may
W
•
Ry
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=
•
max
Ry
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GROUP PROBLEM SOLVING
Given: A 10-kg block is subjected to a
force F=500 N. A spring of
stiffness k=500 N/m is mounted
against the block. When s = 0, the
block is at rest and the spring is
uncompressed. The contact surface
is smooth.
Find: Draw the free-body and kinetic diagrams of the block.
Plan: 1) Define an inertial coordinate system.
2) Draw the block’s free-body diagram, showing all
external forces applied to the block in the proper
directions.
3) Draw the block’s kinetic diagram, showing the inertial
force vector ma in the proper direction.
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GROUP PROBLEM SOLVING (continued)
Solution:
1) An inertial x-y frame can be defined as fixed to the ground.
2) Draw the free-body diagram of the block:
The weight force (W) acts through the
block’s center of mass. F is the applied
Fs=500 s (N) load and Fs =500 s (N) is the spring force,
where s is the spring deformation. The
normal force (N) is perpendicular to the
surface. There is no friction force since
the contact surface is smooth.
F=500 (N) W = 10 g
3
4
y
x
N
3) Draw the kinetic diagram of the block:
10 a
Dynamics, Fourteenth Edition
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The block will be moved to the right.
The acceleration can be directed to the
right if the block is speeding up or to the
left if it is slowing down.
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READING QUIZ
1. Newton’s second law can be written in mathematical form
as F = ma. Within the summation of forces, F,
________ are(is) not included.
A) external forces
B) weight
C) internal forces
D) All of the above.
2. The equation of motion for a system of n-particles can be
written as Fi =  miai = maG, where aG indicates _______.
A) summation of each particle’s acceleration
B) acceleration of the center of mass of the system
C) acceleration of the largest particle
D) None of the above.
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ATTENTION QUIZ
1. Internal forces are not included in an equation of motion
analysis because the internal forces are_____
A) equal to zero.
B) equal and opposite and do not affect the calculations.
C) negligibly small.
D) not important.
2. A 10 lb block is initially moving down a ramp
with a velocity of v. The force F is applied to
bring the block to rest. Select the correct FBD.
F
10
A)
k10
F
10
B)
N
Dynamics, Fourteenth Edition
R.C. Hibbeler
k10
N
F
10
C)
F
v
kN
N
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APPLICATIONS
If a man is trying to move a 100 lb crate, how large a force F
must he exert to start moving the crate? What factors influence
how large this force must be to start moving the crate?
If the crate starts moving, is there acceleration present?
What would you have to know before you could find these
answers?
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APPLICATIONS (continued)
Objects that move in air (or other fluid) have a drag force
acting on them. This drag force is a function of velocity.
If the dragster is traveling with a known velocity and the
magnitude of the opposing drag force at any instant is given
as a function of velocity, can we determine the time and
distance required for dragster to come to a stop if its engine is
shut off? How ?
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RECTANGULAR COORDINATES
(Section 13.4)
The equation of motion, F = ma, 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 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.
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PROCEDURE FOR ANALYSIS
• Free Body Diagram (is always critical!!)
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 kinetic diagram.
Make sure any friction forces act opposite to the direction
of motion! If the particle is connected to an elastic linear
spring, a spring force equal to ‘k s’ should be included on
the FBD.
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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 the best approach.
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.
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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 kinematics 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!
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EXAMPLE
Given: The motor winds in the cable
with a constant acceleration
such that the 20-kg crate moves
a distance s = 6 m in 3 s,
starting from rest. k = 0.3.
Find: The tension developed in
the cable.
Plan:
1) Draw the free-body and kinetic diagrams of the crate.
2) Using a kinematic equation, determine the acceleration of the
crate.
3) Apply the equation of motion to determine the cable tension.
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EXAMPLE (continued)
Solution:
1) Draw the free-body and kinetic diagrams of the crate.
W = 20 g
20 a
T
y
x
Fk= 0.3 N
°
=
N
Since the motion is up the incline, rotate the x-y axes so the
x-axis aligns with the incline. Then, motion occurs only in
the x-direction.
There is a friction force acting between the surface and the
crate. Why is it in the direction shown on the FBD?
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EXAMPLE (continued)
2) Using kinematic equation
s = v0 t + ½ a t 2
 6 = (0) 3 + ½ a (32)
 a = 1.333 m/s2
s = 6 m at t=3 s
v0 = 0 m/s
3) Apply the equations of motion
+  Fy = 0  -20 g (cos°) + N = 0
 N = 169.9 N
+  Fx = m a  T – 20g(sin°) –0.3 N = 20 a
 T = 20 (981) (sin°) + 0.3(169.9) + 20 (1.333)
 T = 176 N
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CONCEPT QUIZ
1. If the cable has a tension of 3 N,
determine the acceleration of block B.
A) 4.26 m/s2 
B) 4.26 m/s2 
C) 8.31 m/s2 
D) 8.31 m/s2 
10 kg
k=0.4
4 kg
2. Determine the acceleration of the block.
A) 2.20 m/s2 
B) 3.17 m/s2 
C) 11.0 m/s2 
D) 4.26 m/s2 
•
30
60 N
5 kg
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GROUP PROBLEM SOLVING
Given: The 300-kg bar B, originally at
rest, is towed over a series of
small rollers. The motor M is
drawing in the cable at a rate of
v = (0.4 t2) m/s, where t is in
seconds.
Find: Force in the cable and distance
s when t = 5 s.
Plan: Since both forces and velocity are involved, this
problem requires both kinematics and the equation of motion.
1) Draw the free-body and kinetic diagrams of the bar.
2) Apply the equation of motion to determine the acceleration
and force.
3) Using a kinematic equation, determine distance.
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GROUP PROBLEM SOLVING (continued)
Solution:
1) Free-body and kinetic diagrams of the bar:
W = 300 g
y
T
x
300 a
=
N
Note that the bar is moving along the x-axis.
2) Apply the scalar equation of motion in the x-direction
+ →  Fx = 300 a  T = 300 a
Since v = 0.4 t2, a = ( dv/dt ) = 0.8 t
T = 240 t  T = 1200 N when t = 5s.
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GROUP PROBLEM SOLVING (continued)
3) Using kinematic equation to determine distance;
Since v = (0.4 t2) m/s
t
s = s0 + ‫ ׬‬v dt = 0 + ‫׬‬0 (0.4 t2) dt
s=
0.4 3
t
3
At t = 5 s,
0.4 3
s = 5 = 16.7 m
3
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READING QUIZ
1. In dynamics, the friction force acting on a moving object is
always ________
A) in the direction of its motion.
B) a kinetic friction.
C) a static friction.
D) zero.
2. If a particle is connected to a spring, the elastic spring force is
expressed by F = ks . The “s” in this equation is the
A) spring constant.
B) un-deformed length of the spring.
C) difference between deformed length and un-deformed
length.
D) deformed length of the spring.
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ATTENTION QUIZ
1. Determine the tension in the cable when the
400 kg box is moving upward with a 4 m/s2
acceleration.
T
60
A) 2265 N
B) 3365 N
a = 4 m/s2
C) 5524 N
D) 6543 N
2. A 10 lb particle has forces of F1= (3i + 5j) lb and
F2= (-7i + 9j) lb acting on it. Determine the acceleration of
the particle.
A) (-0.4 i + 1.4 j) ft/s2
B) (-4 i + 14 j) ft/s2
C) (-12.9 i + 45 j) ft/s2
D) (13 i + 4 j) ft/s2
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e
h
Lecture
t
f
o
d
n
E
L et
Learning Continue
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EQUATIONS OF MOTION:
NORMAL AND TANGENTIAL COORDINATES
Today’s Objectives:
Students will be able to:
1. Apply the equation of motion
using normal and tangential
coordinates.
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EQUATIONS OF MOTION:
CYLINDRICAL COORDINATES
Today’s Objectives (continue):
2. Analyze the kinetics of a
particle using cylindrical coordinates.
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APPLICATIONS
Race track turns are often banked to reduce the frictional forces
required to keep the cars from sliding up to the outer rail 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 () required to prevent
the car from sliding up the track?
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APPLICATIONS (continued)
This picture shows a ride at the amusement park. The
hydraulically-powered arms turn at a constant rate, which creates
a centrifugal force on the riders.
We need to determine the smallest angular velocity of cars A
and B such that the passengers do not lose contact with their
seat. What parameters are needed for this calculation?
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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, we need to
determine the required speed of the satellite to maintain this
orbit. What equation governs this situation?
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NORMAL & TANGENTIAL COORDINATES
(Section 13.5)
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.
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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+ Fbub = 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.
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NORMAL AND TANGENTIAL ACCELERATION
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/, 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
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R.C. Hibbeler
1+
ρ=
dy
dx
2 3/2
d2 y
dx 2
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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 n-direction).
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
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an = v2/
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EXAMPLE
Given: The 10-kg ball has a velocity of
3 m/s when it is at A, along the
vertical path.
Find: The tension in the cord and the
increase in the speed of the ball.
Plan: 1) Since the problem involves a curved path and requires
finding the force perpendicular to the path, use n-t
coordinates. Draw the ball’s free-body and kinetic
diagrams.
2) Apply the equation of motion in the n-t directions.
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EXAMPLE (continued)
Solution:
1) The n-t coordinate system can
be established on the ball at
Point A, thus at an angle of °.
Draw the free-body and kinetic
diagrams of the ball.
Kinetic diagram
n
man
Free-body diagram
n
T W

t
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R.C. Hibbeler
=
t
mat
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EXAMPLE (continued)
2) Apply the equations of motion in the n-t directions.
(a) Fn = man  T – W sin ° = m an
Using an = v2/ = 32/2, W = 10(9.81) N, and m = 10 kg
 T – 98.1 sin ° = (10) (32/2)
 T = 114 N
(b) Ft = mat
 W cos ° = mat
 98.1 cos ° = 10 at
 at = (dv/dt) = 6.94 m/s2
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CONCEPT QUIZ
1. A 10 kg sack slides down a smooth surface. If the normal
force at the flat spot on the surface, A, is 98.1 N () , the
radius of curvature is ____.
A) 0.2 m
B) 0.4 m
v=2m/s
C) 1.0 m
D) None of the above.
A
2. A 20 lb block is moving along a smooth surface. If the
normal force on the surface at A is 10 lb, the velocity is
________.
A
A) 7.6 ft/s
B) 9.6 ft/s
C) 10.6 ft/s
D) 12.6 ft/s
Dynamics, Fourteenth Edition
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=7 ft
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GROUP PROBLEM SOLVING I
Given: The boy has a weight of 60 lb.
At the instant  = 60, the boy’s
center of mass G experiences a
speed v = 15 ft/s.
Find: The tension in each of the two
supporting cords of the swing
and the rate of increase in his
speed at this instant.
Plan: 1) Use n-t coordinates and treat the boy as a particle.
Draw the free-body and kinetic diagrams.
2) Apply the equation of motion in the n-t directions.
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GROUP PROBLEM SOLVING I (continued)
Solution:
1) The n-t coordinate system can
be established on the boy at
angle °. Approximating the
boy as a particle, the free-body
and kinetic diagrams can be
drawn:
Free-body diagram
W

n
2T
t
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R.C. Hibbeler
Kinetic diagram
=

n
man
mat
t
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GROUP PROBLEM SOLVING I (continued)
Free-body diagram
W

n
2T
Kinetic diagram
=

n
man
mat
t
t
2) Apply the equations of motion in the n-t directions.
Fn = man  2T − W sin ° = man
Using an = v2/ = 152/10, W = 60 lb,
we get: T = 46.9 lb
Ft = mat  60 cos ° = (60 / 32.2) at

at = v = 16.1 ft/s2
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GROUP PROBLEM SOLVING II
Given: A 800 kg car is traveling over
a hill with the shape of a
parabola. When the car is at
point A, its v = 9 m/s and
a = 3 m/s2. (Neglect the size
of the car.)
Find: The resultant normal force and resultant frictional force
exerted on the road at point A by the car.
Plan:
1) Treat the car as a particle. Draw its free-body and
kinetic diagrams.
2) Apply the equations of motion in the n-t directions.
3) Use calculus to determine the slope and radius of
curvature of the path at point A.
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GROUP PROBLEM SOLVING II (continued)
Solution:
1) The n-t coordinate system can
be established on the car at
point A. Treat the car as a
particle and draw the freebody and kinetic diagrams:
W
F
N

=

n
t
man
mat
n
t
W = mg = weight of car
N = resultant normal force on road
F = resultant friction force on road
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GROUP PROBLEM SOLVING II (continued)
2) Apply the equations of motion in the n-t directions:
 Fn = man  W cos  – N = man
Using W = mg and an = v2/ = (9)2/
 (800)(9.81) cos  – N = (800) (81/)
 N = 7848 cos  – 64800 / 
Eq. ()
 Ft = mat  W sin  – F = mat
Using W = mg and at = 3 m/s2 (given)
 (800)(9.81) sin  – F = (800) (3)
 F = 7848 sin  – 2400
Dynamics, Fourteenth Edition
R.C. Hibbeler
Eq. (2)
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GROUP PROBLEM SOLVING II (continued)
3) Determine  by differentiating y = f(x) at x = 80 m:
y = 20(1 – x2/6400)  dy/dx = (–40) x / 6400
 d2y/dx2 = (–40) / 6400
dy 2 3/2
[1 + (–0.5)2]3/2
[1 + ( ) ]
dx
 =
=
= 223.6 m
2
dy
0.00625
x = 80 m
2
dx
Determine  from the slope of the curve at A:
dy
tan  = dy/dx

dx
x = 80 m
 = tan-1 (dy/dx) = tan-1 (-0.5) = 26.6°
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GROUP PROBLEM SOLVING II (continued)
From Eq. (1): N = 7848 cos  – 64800 / 
= 7848 cos (26.6°) – 64800 / 223.6 = 6728 N
From Eq. (2): F = 7848 sin  – 2400
= 7848 sin (26.6°) – 2400 = 1114 N
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READING QUIZ
1. The “normal” component of the equation of motion is written
as Fn=man, where Fn is referred to as the _______.
A) impulse
B) centripetal force
C) tangential force
D) inertia force
2. The positive n direction of the normal and tangential
coordinates is ____________.
A) normal to the tangential component
B) always directed toward the center of curvature
C) normal to the bi-normal component
D) All of the above.
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ATTENTION QUIZ
1. The tangential acceleration of an object
A) represents the rate of change of the velocity vector’s
direction.
B) represents the rate of change in the magnitude of the
velocity.
C) is a function of the radius of curvature.
D) Both B and C.
2. The block has a mass of 20 kg and a speed of
v = 30 m/s at the instant it is at its lowest point.
Determine the tension in the cord at this instant.
10 m

A) 1596 N
C) 1996 N
Dynamics, Fourteenth Edition
R.C. Hibbeler
B) 1796 N
D) 2196 N
v = 30m/s
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APPLICATIONS
The forces acting on the 100lb boy can be analyzed using
the cylindrical coordinate
system.
How would you write the
equation describing the
frictional force on the boy as
he slides down this helical
slide?
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APPLICATIONS (continued)
When an airplane executes the vertical loop shown above,
the centrifugal force causes the normal force (apparent
weight) on the pilot to be smaller than her actual weight.
How would you calculate the velocity necessary for the pilot
to experience weightlessness at A?
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CYLINDRICAL COORDINATES
(Section 13.6)
This approach to solving problems has
some external similarity to the normal &
tangential method just studied.
However, the path may be more complex
or the problem may have other attributes
that make it desirable to use cylindrical
coordinates.
Equilibrium equations or “Equations of Motion” in
cylindrical coordinates (using r,  , and z coordinates) may be
expressed in scalar form as:
 2

 Fr = mar = m (r – r  )
  
 F = ma = m (r  – 2 r  )

 Fz = maz = m z
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CYLINDRICAL COORDINATES
(continued)
If the particle is constrained to move only in the r – 
plane (i.e., the z coordinate is constant), then only the
first two equations are used (as shown below). The
coordinate system in such a case becomes a polar
coordinate system. In this case, the path is only a
function of 




2
 F = ma = m(r – r )
r

r
 F = ma = m(r – 2r )
Note that a fixed coordinate system is used, not a “bodycentered” system as used in the n – t approach.
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TANGENTIAL AND NORMAL FORCES
If a force P causes the particle to move along a path defined
by r = f ( ), the normal force N exerted by the path on the
particle is always perpendicular to the path’s tangent. The
frictional force F always acts along the tangent in the
opposite direction of motion. The directions of N and F can
be specified relative to the radial coordinate by using angle 
.
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DETERMINATION OF ANGLE 
The angle  defined as the angle
between the extended radial line
and the tangent to the curve, can
be required to solve some
problems.
It can be determined from the
following relationship.
𝑟𝑑𝜃
𝑟
tan 𝜓 =
=
𝑑𝑟
𝑑𝑟/𝑑𝜃
If  is positive, it is measured counterclockwise from the radial
line to the tangent. If it is negative, it is measured clockwise.
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EXAMPLE
Given: The 0.2 kg pin (P) is
constrained to move in the
smooth curved slot, defined
by r = (0.6 cos 2 ) m.
The slotted arm OA has a
constant angular velocity of
𝜃ሶ = −3 rad/s. Motion is in
the vertical plane.
Find:
Force of the arm OA on the
pin P when  = 0.
Plan:
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EXAMPLE
Given: The 0.2 kg pin (P) is
constrained to move in the
smooth curved slot, defined
by r = (0.6 cos 2) m.
The slotted arm OA has a
constant angular velocity of
𝜃ሶ = −3 rad/s. Motion is in
the vertical plane.
Find:
Force of the arm OA on the
pin P when  = 0.
Plan: 1) Draw the FBD and kinetic diagrams.
2) Develop the kinematic equations using cylindrical
coordinates.
3) Apply the equation of motion to find the force.
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EXAMPLE (continued)
Solution :
1) Free Body and Kinetic Diagrams:
Establish the r,  coordinate
system when  = 0, and draw
the free body and kinetic
diagrams.
Free-body diagram
Kinetic diagram

W
r
=
ma
mar
N
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EXAMPLE (continued)
2) Notice that r = 0.6 cos(2θ), therefore:
rሶ = −1.2 sin 2θ θሶ
rሷ = −2.4 cos(2θ) θሶ 2 − 1.2 sin(2θ) θሷ
Kinematics: at  = 0, θሶ = −3 rad/s, 𝜃ሷ = 0 rad/s2.
r = 0.6 cos 0 = 0.6 m
rሶ = −1.2 sin 0 −3 = 0 m/s
rሷ = −2.4 cos(0) −3 2 − 1.2 sin(0)(0) = −21.6 m/s 2
Acceleration components are
ar = 𝑟ሷ − 𝑟𝜃ሶ 2 = - 21.6 – (0.6)(-3)2 = – 27 m/s2
a = 𝑟𝜃ሷ + 2𝑟ሶ 𝜃ሶ = (0.6)(0) + 2(0)(-3) = 0 m/s2
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EXAMPLE (continued)
3) Equation of motion:  direction
(+)  Fq = maq
N – 0.2 (9.81) = 0.2 (0)
N = 1.96 N 
Free-body diagram
Kinetic diagram

W
r
N
Dynamics, Fourteenth Edition
R.C. Hibbeler
=
ma
mar
ar = –27 m/s2
aq = 0 m/s2
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CONCEPT QUIZ
1. When a pilot flies an airplane in a
vertical loop of constant radius r at
constant speed v, his apparent weight
is maximum at
A) Point A
C) Point C
B
C
r
A
D
B) Point B (top of the loop)
D) Point D (bottom of the loop)
2. If needing to solve a problem involving the pilot’s weight at
Point C, select the approach that would be best.
A) Equations of Motion: Cylindrical Coordinates
B) Equations of Motion: Normal & Tangential Coordinates
C) Equations of Motion: Polar Coordinates
D) No real difference – all are bad.
E) Toss up between B and C.
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R.C. Hibbeler
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GROUP PROBLEM SOLVING I
Given: The smooth can C is lifted
from A to B by a rotating rod.
The mass of can is 3 kg.
Neglect the effects of friction
in the calculation and the size
of the can so that
r = (1.2 cos ) m.
Find: Forces of the rod on the can when  = 30 and
𝜃ሶ = 0.5 rad/s, which is constant.
Plan: 1) Find the acceleration components using the kinematic
equations.
2) Draw free body diagram & kinetic diagram.
3) Apply the equation of motion to find the forces.
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GROUP PROBLEM SOLVING (continued)
Solution:
1) Kinematics:
r = 1.2 cos θ
rሶ = −1.2 (sin θ)θሶ
rሷ = −1.2 cos θ θሶ 2 − 1.2 (sin θ)θሷ
When  = 30, θሶ = 0.5 rad/s and θሷ = 0 rad/s2.
r = 1.039 m
rሶ = −0.3 m/s
rሷ = −0.2598 m/s2
Accelerations:
ar = rሷ − rθሶ 2 = − 0.2598 − (1.039) 0.52 = − 0.5196 m/s2
a = rθሷ + 2rሶ θሶ = (1.039) 0 + 2 (−0.3) 0.5 = − 0.3 m/s2
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GROUP PROBLEM SOLVING (continued)
2) Free Body Diagram

30
Kinetic Diagram
3(9.81) N
r
30
N
ma
=
mar
F
3) Apply equation of motion:
 Fr = mar  -3(9.81) sin30 + N cos30 = 3 (-0.5196)
 F = ma  F + N sin30 − 3(9.81) cos30 = 3 (-0.3)
N = 15.2 N, F = 17.0 N
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R.C. Hibbeler
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e
h
Lecture
t
f
o
d
n
E
L et
Learning Continue
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R.C. Hibbeler
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THE WORK OF A FORCE, THE PRINCIPLE OF
WORK AND ENERGY & SYSTEMS OF PARTICLES
Today’s Objectives:
Students will be able to:
1. Apply the principle of work and
energy to a particle or system of
particles.
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R.C. Hibbeler
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POWER AND EFFICIENCY
Today’s Objectives (continue):
Students will be able to:
2. Calculate the mechanical
efficiency of a machine.
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CONSERVATIVE FORCES, POTENTIAL ENERGY
AND CONSERVATION OF ENERGY
Today’s Objectives (continue):
3. Apply the principle of conservation of energy.
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R.C. Hibbeler
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APPLICATIONS
A roller coaster makes use of gravitational forces to assist the
cars in reaching high speeds in the “valleys” of the track.
How can we design the track (e.g., the height, h, and the radius
of curvature, ) to control the forces experienced by the
passengers?
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R.C. Hibbeler
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WORK AND ENERGY
Another equation for working kinetics problems involving
particles can be derived by integrating the equation of motion
(F = ma) with respect to displacement.
By substituting at = v (dv/ds) into
Ft = mat, the result is integrated to
yield an equation known as the
principle of work and energy.
This principle is useful for solving problems that involve
force, velocity, and displacement. It can also be used to
explore the concept of power.
To use this principle, we must first understand how to
calculate the work of a force.
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WORK OF A FORCE (Section 14.1)
A force does work on a particle when the particle undergoes a
displacement along the line of action of the force.
Work is defined as the product of force
and displacement components acting in
the same direction. So, if the angle
between the force and displacement
vector is , the increment of work dU
done by the force is
dU = F ds cos 
r2
By using the definition of the dot product
and integrating, the total work can be U =
F • dr

1-2
written as
r
1
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WORK OF A FORCE (continued)
If F is a function of position (a common
case) this becomes
s2
U1-2 =  F cos  ds
s1
If both F and  are constant (F = Fc), this equation further
simplifies to
U1-2 = Fc cos  (s2 - s1)
Work is positive if the force and the movement are in the
same direction. If they are opposing, then the work is
negative. If the force and the displacement directions are
perpendicular, the work is zero.
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WORK OF A WEIGHT
The work done by the gravitational force acting on a particle
(or weight of an object) can be calculated by using
y2
U1-2 =
 - W dy
y1
U1-2 = - W (y2 − y1) = - W y
The work of a weight is the product of the magnitude of
the particle’s weight and its vertical displacement. If y
is upward, the work is negative since the weight force
always acts downward.
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WORK OF A SPRING FORCE
When stretched, a linear elastic spring
develops a force of magnitude Fs = ks, where
k is the spring stiffness and s is the
displacement from the unstretched position.
The work of the spring force moving from position s1 to position
s2
s2
s2 is
U1-2 = Fs ds =  k s ds = 0.5 k (s2)2 – 0.5 k (s1)2
s1
s1
If a particle is attached to the spring, the force Fs exerted on the
particle is opposite to that exerted on the spring. Thus, the work
done on the particle by the spring force will be negative or
U1-2 = – [ 0.5 k (s2)2 – 0.5 k (s1)2 ].
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SPRING FORCES
It is important to note the following about spring forces.
1. The equations above are for linear springs only! Recall
that a linear spring develops a force according to
F = ks (essentially the equation of a line).
2. The work of a spring is not just spring force times distance
at some point, i.e., (ksi)(si). Beware, this is a trap that
students often fall into!
3. Always double check the sign of the spring work after
calculating it. It is positive work if the force on the object by
the spring and the movement are in the same direction.
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PRINCIPLE OF WORK AND ENERGY
(Section 14.2 & Section 14.3)
By integrating the equation of motion,  Ft = mat = mv(dv/ds), the
principle of work and energy can be written as
 U1-2 = 0.5 m (v2)2 – 0.5 m (v1)2 or T1 +  U1-2 = T2
U1-2 is the work done by all the forces acting on the particle as it
moves from point 1 to point 2. Work can be either a positive or
negative scalar.
T1 and T2 are the kinetic energies of the particle at the initial and final
position, respectively. Thus, T1 = 0.5 m (v1)2 and T2 = 0.5 m (v2)2.
The kinetic energy is always a positive scalar (velocity is squared!).
So, the particle’s initial kinetic energy plus the work done by all the
forces acting on the particle as it moves from its initial to final position
is equal to the particle’s final kinetic energy.
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PRINCIPLE OF WORK AND ENERGY (continued)
Note that the principle of work and energy (T1 +  U1-2 = T2) is
not a vector equation! Each term results in a scalar value.
Both kinetic energy and work have the same units, that of
energy! In the SI system, the unit for energy is called a joule (J),
where 1 J = 1 N·m. In the FPS system, units are ft·lb.
The principle of work and energy cannot be used, in general, to
determine forces directed normal to the path, since these forces
do no work.
The principle of work and energy can also be applied to a system
of particles by summing the kinetic energies of all particles in the
system and the work due to all forces acting on the system.
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WORK OF FRICTION CAUSED BY SLIDING
The case of a body sliding over a rough surface merits special
consideration.
Consider a block which is moving over a
rough surface. If the applied force P just
balances the resultant frictional force k N,
a constant velocity v would be maintained.
The principle of work and energy would be
applied as
0.5m (v)2 + P s – (k N) s = 0.5m (v)2
This equation is satisfied if P = k N. However, we know from
experience that friction generates heat, a form of energy that does
not seem to be accounted for in this equation. It can be shown that
the work term (k N)s represents both the external work of the
friction force and the internal work that is converted into heat.
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EXAMPLE
Given: When s = 0.6 m, the spring is
not stretched or compressed,
and the 10 kg block, which is
subjected to a force of 100 N,
has a speed of 5 m/s down
the smooth plane.
Find: The distance s when the block stops.
Plan: Since this problem involves forces, velocity and
displacement, apply the principle of work and energy to
determine s.
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EXAMPLE (continued)
Solution:
Apply the principle of work and energy between position 1
(s = 0.6 m) and position 2 (s). Note that the normal force (N)
does no work since it is always perpendicular to the
displacement.
S =0.6 m
T + U = T
1
1-2
2
1
S2
There is work done by three different forces;
1) work of a the force F =100 N;
UF = 100 (s− s1) = 100 (s − 0.6)
2) work of the block weight;
UW = 10 (9.81) (s− s1) sin 30 = 49.05 (s − 0.6)
3) and, work of the spring force.
US = - 0.5 (200) (s−0.6)2 = -100 (s − 0.6)2
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EXAMPLE (continued)
The work and energy equation will be
T1 + U1-2 = T2
0.5 (10) 52 + 100(s − 0.6) + 49.05(s − 0.6) − 100(s − 0.6)2 = 0
 125 + 149.05(s − 0.6) − 100(s − 0.6)2 = 0
Solving for (s − 0.6),
(s − 0.6) = {-149.05 ± (149.052 – 4×(-100)×125)0.5} / 2(-100)
Selecting the positive root, indicating a positive spring deflection,
(s − 0.6) = 2.09 m
Therefore, s = 2.69 m
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R.C. Hibbeler
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GROUP PROBLEM SOLVING I
Given: The 2 lb brick slides
down a smooth roof,
with vA=5 ft/s.
C
C
Find: The speed at B,
the distance d from the
wall to where the brick
strikes the ground, and
its speed at C.
Plan: 1) Apply the principle of work and energy to the brick,
and determine the speeds at B and C.
2) Apply the kinematic relations in x and y-directions.
Dynamics, Fourteenth Edition
R.C. Hibbeler
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GROUP PROBLEM SOLVING I (continued)
Solution:
1) Apply the principle of work and energy
TA + UA-B = TB
1 2
1 2
2
5 + 2 15 =
2 32.2
2 32.2
vB 2
C
C
Solving for the unknown velocity yields vB = 31.48 ft/s
Similarly, apply the work and energy principle between A and C
TA + UA-C = TC
1 2
1 2
2
5 + 2 45 =
2 32.2
2 32.2
vC 2
vC = 54.1 ft/s
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING I (continued)
2) Apply the kinematic relations in x and y-directions:
Equation for horizontal motion
+→ xC = xB + vBx tBC
d = 0 + 31.48 (4/5) tBC
 d = 6.996 tBC
Equation for vertical motion
+ yC = yB + vBy tBC – 0.5 g tBC2
•C
 -30 = 0 + (-31.48)(3/5) tBC – 0.5 (32.2) tBC2
Solving for the positive tBC yields tBC = 0.899 s.
 d = 6.996 tBC = 6.996 (0.899) = 22.6 ft
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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APPLICATIONS
Engines and motors are often
rated in terms of their power
output. The power output of the
motor lifting this elevator is
related to the vertical force F
acting on the elevator, causing it
to move upwards.
Given a desired lift velocity for the
elevator (with a known maximum
load), how can we determine the
power requirement of the motor?
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
POWER AND EFFICIENCY (Section 14.4)
Power is defined as the amount of work performed per unit
of time.
If a machine or engine performs a certain amount of work,
dU, within a given time interval, dt, the power generated can
be calculated as
P = dU/dt
Since the work can be expressed as dU = F • dr, the power
can be written
P = dU/dt = (F • dr)/dt = F • (dr/dt) = F • v
Thus, power is a scalar defined as the product of the force
and velocity components acting in the same direction.
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POWER
Using scalar notation, power can be written
P = F • v = F v cos 
where  is the angle between the force and velocity vectors.
So if the velocity of a body acted on by a force F is known,
the power can be determined by calculating the dot product
or by multiplying force and velocity components.
The unit of power in the SI system is the Watt (W) where
1 W = 1 J/s = 1 (N · m)/s .
In the FPS system, power is usually expressed in units of
horsepower (hp) where
1 hp = 550 (ft · lb)/s = 746 W.
Dynamics, Fourteenth Edition
R.C. Hibbeler
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EFFICIENCY
The mechanical efficiency of a machine is the ratio of the
useful power produced (output power) to the power supplied
to the machine (input power) or
 = (power output) / (power input)
If energy input and removal occur at the same time, efficiency
may also be expressed in terms of the ratio of output energy
to input energy or
 = (energy output) / (energy input)
Machines will always have frictional forces. Since frictional
forces dissipate energy, additional power will be required to
overcome these forces. Consequently, the efficiency of a
machine is always less than 1.
Dynamics, Fourteenth Edition
R.C. Hibbeler
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PROCEDURE FOR ANALYSIS
• Find the resultant external force acting on the body causing
its motion. It may be necessary to draw a free-body diagram.
• Determine the velocity of the point on the body at which the
force is applied. Energy methods or the equation of motion
and appropriate kinematic relations, may be necessary.
• Multiply the force magnitude by the component of velocity
acting in the direction of F to determine the power supplied
to the body (P = F v cos  ).
• In some cases, power may be found by calculating the work
done per unit of time (P = dU/dt).
• If the mechanical efficiency of a machine is known, either
the power input or output can be determined.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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CONCEPT QUIZ
1. A motor pulls a 10 lb block up a smooth
incline at a constant velocity of 4 ft/s.
Find the power supplied by the motor.
A) 8.4 ft·lb/s
B) 20 ft·lb/s
C) 34.6 ft·lb/s
D) 40 ft·lb/s
30º
2. A twin engine jet aircraft is climbing at a 10 degree angle at
260 ft/s. The thrust developed by a jet engine is 1000 lb.
The power developed by the aircraft is
A) (1000 lb)(260 ft/s)
B) (2000 lb)(260 ft/s) cos 
C) (1000 lb)(260 ft/s) cos 
D) (2000 lb)(260 ft/s)
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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READING QUIZ
1. The formula definition of power is ___________.
A) dU / dt
B) F • v
C) F • dr/dt
D) All of the above.
2. Kinetic energy results from _______.
A) displacement
B) velocity
C) gravity
D) friction
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
ATTENTION QUIZ
1. The power supplied by a machine will always be
_________ the power supplied to the machine.
A) less than
B) equal to
C) greater than
D) A or B
2. A car is traveling a level road at 88 ft/s. The power being
supplied to the wheels is 52,800 ft·lb/s. Find the
combined friction force on the tires.
A) 8.82 lb
B) 400 lb
C) 600 lb
D) 4.64 x 106 lb
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
APPLICATIONS
The roller coaster is released from rest at the top of the hill A.
As the coaster moves down the hill, potential energy is
transformed into kinetic energy.
What is the velocity of the coaster when it is at B and C?
Also, how can we determine the minimum height of hill A so
that the car travels around both inside loops without leaving
the track?
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
CONSERVATIVE FORCE (Section 14.5)
A force F is said to be conservative if the work done is
independent of the path followed by the force acting on a particle
as it moves from A to B. This also means that the work done by
the force F in a closed path (i.e., from A to B and then back to A)
is zero.
z F
=
F
d
r
0
·
B

Thus, we say the work is conserved.
The work done by a conservative
force depends only on the positions
of the particle, and is independent of
its velocity or acceleration.
Dynamics, Fourteenth Edition
R.C. Hibbeler
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y
CONSERVATIVE FORCE (continued)
A more rigorous definition of a conservative force makes
use of a potential function (V) and partial differential
calculus, as explained in the text. However, even without
the use of the these more complex mathematical
relationships, much can be understood and accomplished.
The “conservative” potential energy of a particle/system is
typically written using the potential function V. There are two
major components to V commonly encountered in mechanical
systems, the potential energy from gravity and the potential
energy from springs or other elastic elements.
Vtotal = Vgravity + Vsprings
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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POTENTIAL ENERGY
Potential energy is a measure of the amount of work a
conservative force will do when a body changes position.
In general, for any conservative force system, we can define
the potential function (V) as a function of position. The work
done by conservative forces as the particle moves equals the
change in the value of the potential function (e.g., the sum of
Vgravity and Vsprings).
It is important to become familiar with the two types of
potential energy and how to calculate their magnitudes.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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POTENTIAL ENERGY DUE TO GRAVITY
The potential function (formula) for a gravitational force, e.g.,
weight (W = mg), is the force multiplied by its elevation from a
datum. The datum can be defined at any convenient location.
Vg = ± W y
Vg is positive if y is above the
datum and negative if y is
below the datum. Remember,
YOU get to set the datum.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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ELASTIC POTENTIAL ENERGY
Recall that the force of an elastic spring is F = ks. It is
important to realize that the potential energy of a spring, while
it looks similar, is a different formula.
Ve (where ‘e’ denotes an
elastic spring) has the distance
“s” raised to a power (the
result of an integration) or
1 2
=
Ve
ks
2
Notice that the potential
function Ve always yields
positive energy.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
CONSERVATION OF ENERGY (Section 14.6)
When a particle is acted upon by a system of conservative
forces, the work done by these forces is conserved and the
sum of kinetic energy and potential energy remains
constant. In other words, as the particle moves, kinetic
energy is converted to potential energy and vice versa.
This principle is called the principle of conservation of
energy and is expressed as
T1 + V1 = T2 + V2 = Constant
T1 stands for the kinetic energy at state 1 and V1 is the
potential energy function for state 1. T2 and V2
represent these energy states at state 2. Recall, the
kinetic energy is defined as T = ½ mv2.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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EXAMPLE
Given: The 4 kg collar, C, has a
velocity of 2 m/s at A.
The spring constant is 400
N/m. The unstretched length
of the spring is 0.2 m.
Find:
The velocity of the collar at
B.
Plan: Apply the conservation of energy equation between A and
B. Set the gravitational potential energy datum at point A
or point B (in this example, choose point A—why?).
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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EXAMPLE (continued)
Solution:
.
0.3 m
Datum
0.5 m
.
Note that the potential energy at B has two
parts.
VB = (VB)e + (VB)g
VB = 0.5 (400) (0.5 – 0.2)2 – 4 (9.81) 0.4
The kinetic energy at B is
TB = 0.5 (4) vB2
Similarly, the potential and kinetic energies at A will be
VA = 0.5 (400) (0.1 – 0.2)2, TA = 0.5 (4) 22
The energy conservation equation becomes TA + VA = TB + VB.
[ 0.5(400) (0.5 – 0.2)2 – 4(9.81)0.4 ] + 0.5 (4) vB2
= [0.5 (400) (0.1 – 0.2)2 ]+ 0.5 (4) 22
 vB = 1.96 m/s
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
CONCEPT QUIZ
1. If the work done by a conservative force on a particle as it
moves between two positions is –10 ft·lb, the change in its
potential energy is _______
A) 0 ft·lb.
B) -10 ft·lb.
C) +10 ft·lb.
D) None of the above.
2. Recall that the work of a spring is U1-2 = -½ k(s22 – s12) and
can be either positive or negative. The potential energy of a
spring is V = ½ ks2. Its value is __________
A) always negative.
B) either positive or negative.
C) always positive.
D) an imaginary number!
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
GROUP PROBLEM SOLVING I
Given: The 800 kg roller
coaster car is
released from rest
at A.
Find: The minimum height, h, of Point A so that the car travels
around inside loop at B without leaving the track. Also find the
velocity of the car at C for this height, h, of A.
Plan: Note that only kinetic energy and potential energy due
to gravity are involved. Determine the velocity at B using the
equation of motion and then apply the conservation of energy
equation to find minimum height h .
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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GROUP PROBLEM SOLVING I (continued)
Datum
Solution:
1) Placing the datum at A:
TA + VA = TB + VB
 0.5 (800) 02 + 0
= 0.5 (800) (vB)2 − 800(9.81) (h − 20)
(1)
2) Find the required velocity of the coaster at B so it doesn’t
leave the track.
Equation of motion applied at B:
NB  0
2
v
 Fn = man = m 
(vB)2
=
800 (9.81) = 800
7.5
man
mg
 vB = 8.578 m/s
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
GROUP PROBLEM SOLVING I (continued)
Datum
Now using the energy
conservation, eq. (1), the
minimum h can be determined.
0.5 (800) 02 + 0 = 0.5 (800) (8.578)2 − 800(9.81) (h − 20)
 h = 23.75 m
3) Find the velocity at C applying the energy conservation.
TA + VA = TC + VC
 0.5 (800) 02 + 0 = 0.5 (800) (vC)2 − 800(9.81) (23.75)
 VC = 21.6 m/s
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
GROUP PROBLEM SOLVING II
Given: The arm is pulled back such that
s = 100 mm and released.
When s = 0, the spring is
unstretched.
Assume all surfaces of contact to
be smooth. Neglect the mass of
the spring and the size of the ball.
Find: The speed of the 0.3-kg ball and the normal reaction of the
circular track on the ball when  = 60.
Plan: Determine the velocity at  = 60 using the conservation
of energy equation and then apply the equation of motion
to find the normal reaction on the ball.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
GROUP PROBLEM SOLVING II (continued)
Solution:
1) Placing the datum at A:
TA + VA = TB + VB
where
TA = 0.5 (0.3) 02
VA = 0 + 0.5 (1500) 0.12
TB = 0.5 (0.3) 02
VB = 0.3 (9.81) 1.5 (1 − cos 60)
60
Datum
•A
• B
The conservation of energy equation is
0 + 0.5 (1500) 0.12 = 0.5 (0.3) (vB)2
+ 0.3 (9.81) 1.5 (1 − cos 60)
vB = 5.94 m/s
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
GROUP PROBLEM SOLVING II (continued)
2) Find the normal reaction on the ball when  = 60.
Free-body diagram
Kinetic diagram
n
n
W
mat


man
t
t
N
=
Equation of motion applied at  = 60 :
2
v
 Fn = man = m B

5.942
N − 0.3 (9.81) cos 60 = 0.3
1.5
N = 8.53 N
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
READING QUIZ
1. The potential energy of a spring is ________
A) always negative.
B) always positive.
C) positive or negative.
D) equal to ks.
2. When the potential energy of a conservative system
increases, the kinetic energy _________
A) always decreases.
B) always increases.
C) could decrease or
increase.
D) does not change.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.
ATTENTION QUIZ
1. The principle of conservation of energy is usually ______ to
apply than the principle of work & energy.
A) harder
B) easier
C) the same amount of work
D) It is a mystery!
2. If the pendulum is released from the
horizontal position, the velocity of its
bob in the vertical position is _____
A) 3.8 m/s.
B) 6.9 m/s.
C) 14.7 m/s.
D) 21 m/s.
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
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e
h
Lecture
t
f
o
d
n
E
L et
Learning Continue
Dynamics, Fourteenth Edition
R.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.
All rights reserved.