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2.2. The dot product of two vectors 23
2.2 The dot product of two vectors
The dot product is used to project a vector in a given direction, to reduce a vector
to components, to reduce vector equations to scalar equations, to define work and
power, and to help solve geometry problems.
The dot product of two vectors
A and
B is written
A ·
B (pronounced ‘A dot B’).
The dot product of
A and
B is the product of the magnitudes of the two vectors times
a number that expresses the degree to which
A and
B are parallel: cos θ
AB
, where
θ
AB
is the angle between
A and
B. That is,
B
A
θ
AB
B
A
cos
θ
AB
cos
θ
AB
Figure 2.17: The dot product of
A and
B is ascalar and so is not easily drawn. It is
given by
A ·
B = ABcos θ
AB
which is A
times the projection of
B in the A direction
and also B times the projection of
A in the
B direction.
(Filename:tfigure1.11)
A ·
B
def
=|
A||
B|cos θ
AB
which is sometimes written more concisely as
A ·
B = ABcos θ . One special case
is when cosθ
AB
= 1,
A and
B are parallel, and
A ·
B = AB. Another is when
cos θ
AB
= 0,
A and
B are perpendicular, and
A ·
B = 0.
1
1
If you don’t know, almost without
a thought, that cos0 = 1, cos π/2 =
0, sin 0 = 0, and sin π/2 = 1 now is as
good a time as any to draw as many trian-
gles and unit circles as it takes to cement
these special cases into your head.
The dot product of two vectors is a scalar. So the dot product is sometimes called
the scalar product. Using the geometric definition of dot product, and the rules for
vector addition we have already discussed, you can convince yourself of (or believe)
the following properties of dot products.
•
A ·
B =
B ·
A commutative law,
ABcos θ = BAcosθ
• (a
A) ·
B =
A · (a
B) = a(
A ·
B) a distributive law,
(aA)B cosθ = A(aB) cos θ
•
A · (
B +
C) =
A ·
B +
A ·
C another distributive law,
the projection of
B +
C onto
A is the
sum of the two separate projections
•
A ·
B = 0if
A ⊥
B perpendicular vectors have zero for
a dot product, ABcosπ/2 = 0
•
A ·
B =|
A||
B| if
A
B parallel vectors have the product of
their magnitudes for a dot product,
ABcos 0 = AB. In particular,
A ·
A = A
2
or |
A|=
√
A ·
A
•
ˆ
ı ·
ˆ
ı =
ˆ
·
ˆ
=
ˆ
k ·
ˆ
k = 1,
ˆ
ı ·
ˆ
=
ˆ
·
ˆ
k =
ˆ
k ·
ˆ
ı = 0
The standard base vectors used with
cartesiancoordinates are unitvectors
and they are perpendicular to each
other. In math language they are ‘or-
thonormal.’
•
ˆ
ı
·
ˆ
ı
=
ˆ
·
ˆ
=
ˆ
k
·
ˆ
k
= 1,
ˆ
ı
·
ˆ
=
ˆ
·
ˆ
k
=
ˆ
k
·
ˆ
ı
= 0
The standard crooked base vectors
are orthonormal.
The identities above lead to the following equivalent ways of expressing the dot
product of
A and
B (see box 2.2 on page 24 to see how the component formula
follows from the geometric definition above):
24 CHAPTER 2. Vectors for mechanics
A ·
B =|
A||
B|cos θ
AB
= A
x
B
x
+ A
y
B
y
+ A
z
B
z
(component formula for dot product)
= A
x
B
x
+ A
y
B
y
+ A
z
B
z
=|
A|·[projection of
B in the
A direction]
=|
B|·[projection of
A in the
B direction]
Using the dot product to find components
To find the x component of a vector or vector expression one can use the dot product
of the vector (or expression) with a unit vector in the x direction as in figure 2.18. In
particular,
v
x
=
v ·
ˆ
ı.
x
y
v
x
v
ˆ
ı
ˆ
Figure 2.18: The dot product with unit
vectors gives projection. For example,
v
x
=
v·
ˆ
ı.
(Filename:tfigure1.3.dotprod)
This idea can be used for finding components in any direction. If one knows the
orientation of the crooked unit vectors
ˆ
ı
,
ˆ
,
ˆ
k
relative to the standard bases
ˆ
ı,
ˆ
,
ˆ
k
then all the angles between the base vectors are known. So one can evaluate the dot
products between the standard base vectors and the crooked base Introduction to Statics and Torque Introduction to Statics and Torque Bởi: OpenStaxCollege On a short time scale, rocks like these in Australia’s Kings Canyon are static, or motionless relative to the Earth (credit: freeaussiestock.com) What might desks, bridges, buildings, trees, and mountains have in common—at least in the eyes of a physicist? The answer is that they are ordinarily motionless relative to the Earth Furthermore, their acceleration is zero because they remain motionless That means they also have something in common with a car moving at a constant velocity, because anything with a constant velocity also has an acceleration of zero Now, the important part—Newton’s second law states that net F = ma, and so the net external force is zero for all stationary objects and for all objects moving at constant velocity There are forces acting, but they are balanced That is, they are in equilibrium Statics Statics is the study of forces in equilibrium, a large group of situations that makes up a special case of Newton’s second law We have already considered a few such situations; in this chapter, we cover the topic more thoroughly, including consideration of such possible effects as the rotation and deformation of an object by the forces acting on it 1/2 Introduction to Statics and Torque How can we guarantee that a body is in equilibrium and what can we learn from systems that are in equilibrium? There are actually two conditions that must be satisfied to achieve equilibrium These conditions are the topics of the first two sections of this chapter 2/2 Introduction to
STATICS
and
DYNAMICS
F
1
F
2
N
1
N
2
F
s
M
s
ˆ
ı
ˆ
ˆ
k
Andy Ruina and Rudra Pratap
Pre-print for Oxford University Press, January 2002
Summary of Mechanics
0) The laws of mechanics apply to any collection of material or ‘body.’ This body could be the overall system of study
or any part of it. In the equations below, the forces and moments are those that show on a free body diagram. Interacting
bodies cause equal and opposite forces and moments on each other.
I) Linear Momentum Balance (LMB)/Force Balance
Equation of Motion
F
i
=
˙
L The total force on a body is equal
to its rate of change of linear
momentum.
(I)
Impulse-momentum
(integrating in time)
t
2
t
1
F
i
·dt =
L Net impulse is equal to the change in
momentum.
(Ia)
Conservation of momentum
(if
F
i
=
0 )
˙
L =
0 ⇒
L =
L
2
−
L
1
=
0
When there is no net force the linear
momentum does not change.
(Ib)
Statics
(if
˙
L is negligible)
F
i
=
0 If the inertial terms are zero the
net force on system is zero.
(Ic)
II) Angular Momentum Balance (AMB)/Moment Balance
Equation of motion
M
C
=
˙
˙
H
C
The sum of moments is equal to the
rate of change of angular momentum.
(II)
Impulse-momentum (angular)
(integrating in time)
t
2
t
1
M
C
dt =
H
C
The net angular impulse is equal to
the change in angular mo
mentum.
(IIa)
Conservation of angular momentum
(if
M
C
=
0)
˙
H
C
=
0 ⇒
H
C
=
H
C
2
−
H
C
1
=
0
If there is no net moment about point
C then the angular momentum about
point C does not change.
(IIb)
Statics
(if
˙
H
C
is negligible)
M
C
=
0 If the inertial terms are zero then the
total moment on the system is zero.
(IIc)
III) Power Balance (1st law of thermodynamics)
Equation of motion
˙
Q + P =
˙
E
K
+
˙
E
P
+
˙
E
int
˙
E
Heat flow plus mechanical power
into a system is equal to its change
in energy (kinetic + potential +
internal).
(III)
for finite time
t
2
t
1
˙
Qdt +
t
2
t
1
Pdt = E The net energy flow goingin is equal
to the net change in energy.
(IIIa)
Conservation of Energy
(if
˙
Q = P = 0)
˙
E = 0 ⇒
E = E
2
− E
1
= 0
If no energy flows into a system,
then its energy does not change.
(IIIb)
Statics
(if
˙
E
K
is negligible)
˙
Q + P =
˙
E
P
+
˙
E
int
If there is no change of kinetic energy
then the change of potential and
internal energy is due to mechanical
work and heat flow.
(IIIc)
Pure Mechanics
(if heat flow and dissipation
are negligible)
P =
˙
E
K
+
˙
E
P
In a system well modeled as purely
mechanical the change of kinetic
and potential energy is due to mechanical
work.
(IIId)
Some Definitions
r or
x Position .e.g.,
r
i
≡
r
i/O
is theposition of apoint
i relative to the origin, O)
v ≡
d
r
dt
Velocity .e.g.,
v
i
≡
v
i/O
is the velocity ofa point
i relativeto O, measured in anon-rotating
reference frame)
a ≡
d
v
dt
=
d
2
r
dt
2
Acceleration .e.g.,
a
i
≡
a
i/O
is the acceleration of a
point i relative to O, measured in a New-
tonian frame)
ω Angular
(Please also look at the tables inside the back cover.)
velocity A measure ofrotational velocityof arigid
body.
α ≡
˙
ω Angular acceleration A measure of rotational acceleration of a
rigid body.
L ≡
m
i
v
i
discrete
vdm continuous
Linear momentum A measure of a system’s net translational
rate (weighted by mass).
= m
tot
v
cm
˙
L ≡
m
i
a
i
discrete
adm continuous
Rate of change of linear
momentum
The aspect of motion thatbalances thenet
force on a system.
= m
tot
a
cm
H
C
≡
r
i/C
× m
i
v
i
discrete
r
/C
×
vdm continuous
Angular momentum about
point C
A measure of the rotational rate of a sys-
tem about a point C (weighted by mass
and distance from C).
˙
H
C
≡
r
i/C
× m
i
a
i
discrete
r
/C
×
adm continuous
Rate of change of angular mo-
mentum about point C
The aspect of motion thatbalances thenet
torque on a Introduction to
STATICS
and
DYNAMICS
F
1
F
2
N
1
N
2
F
s
M
s
ˆ
ı
ˆ
ˆ
k
Andy Ruina and Rudra Pratap
Pre-print for Oxford University Press, January 2002
Summary of Mechanics
0) The laws of mechanics apply to any collection of material or ‘body.’ This body could be the overall system of study
or any part of it. In the equations below, the forces and moments are those that show on a free body diagram. Interacting
bodies cause equal and opposite forces and moments on each other.
I) Linear Momentum Balance (LMB)/Force Balance
Equation of Motion
F
i
=
˙
L The total force on a body is equal
to its rate of change of linear
momentum.
(I)
Impulse-momentum
(integrating in time)
t
2
t
1
F
i
·dt =
L Net impulse is equal to the change in
momentum.
(Ia)
Conservation of momentum
(if
F
i
=
0 )
˙
L =
0 ⇒
L =
L
2
−
L
1
=
0
When there is no net force the linear
momentum does not change.
(Ib)
Statics
(if
˙
L is negligible)
F
i
=
0 If the inertial terms are zero the
net force on system is zero.
(Ic)
II) Angular Momentum Balance (AMB)/Moment Balance
Equation of motion
M
C
=
˙
˙
H
C
The sum of moments is equal to the
rate of change of angular momentum.
(II)
Impulse-momentum (angular)
(integrating in time)
t
2
t
1
M
C
dt =
H
C
The net angular impulse is equal to
the change in angular mo
mentum.
(IIa)
Conservation of angular momentum
(if
M
C
=
0)
˙
H
C
=
0 ⇒
H
C
=
H
C
2
−
H
C
1
=
0
If there is no net moment about point
C then the angular momentum about
point C does not change.
(IIb)
Statics
(if
˙
H
C
is negligible)
M
C
=
0 If the inertial terms are zero then the
total moment on the system is zero.
(IIc)
III) Power Balance (1st law of thermodynamics)
Equation of motion
˙
Q + P =
˙
E
K
+
˙
E
P
+
˙
E
int
˙
E
Heat flow plus mechanical power
into a system is equal to its change
in energy (kinetic + potential +
internal).
(III)
for finite time
t
2
t
1
˙
Qdt +
t
2
t
1
Pdt = E The net energy flow goingin is equal
to the net change in energy.
(IIIa)
Conservation of Energy
(if
˙
Q = P = 0)
˙
E = 0 ⇒
E = E
2
− E
1
= 0
If no energy flows into a system,
then its energy does not change.
(IIIb)
Statics
(if
˙
E
K
is negligible)
˙
Q + P =
˙
E
P
+
˙
E
int
If there is no change of kinetic energy
then the change of potential and
internal energy is due to mechanical
work and heat flow.
(IIIc)
Pure Mechanics
(if heat flow and dissipation
are negligible)
P =
˙
E
K
+
˙
E
P
In a system well modeled as purely
mechanical the change of kinetic
and potential energy is due to mechanical
work.
(IIId)
Some Definitions
r or
x Position .e.g.,
r
i
≡
r
i/O
is theposition of apoint
i relative to the origin, O)
v ≡
d
r
dt
Velocity .e.g.,
v
i
≡
v
i/O
is the velocity ofa point
i relativeto O, measured in anon-rotating
reference frame)
a ≡
d
v
dt
=
d
2
r
dt
2
Acceleration .e.g.,
a
i
≡
a
i/O
is the acceleration of a
point i relative to O, measured in a New-
tonian frame)
ω Angular
(Please also look at the tables inside the back cover.)
velocity A measure ofrotational velocityof arigid
body.
α ≡
˙
ω Angular acceleration A measure of rotational acceleration of a
rigid body.
L ≡
m
i
v
i
discrete
vdm continuous
Linear momentum A measure of a system’s net translational
rate (weighted by mass).
= m
tot
v
cm
˙
L ≡
m
i
a
i
discrete
adm continuous
Rate of change of linear
momentum
The aspect of motion thatbalances thenet
force on a system.
= m
tot
a
cm
H
C
≡
r
i/C
× m
i
v
i
discrete
r
/C
×
vdm continuous
Angular momentum about
point C
A measure of the rotational rate of a sys-
tem about a point C (weighted by mass
and distance from C).
˙
H
C
≡
r
i/C
× m
i
a
i
discrete
r
/C
×
adm continuous
Rate of change of angular mo-
mentum about point C
The aspect of motion thatbalances thenet
torque on a Introduction to STATICS and DYNAMICS Chapters 1-10 Rudra Pratap and Andy Ruina Spring 2001 c Rudra Pratap and Andy Ruina, 1994-2001. All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, or otherwise, without prior written permission of the authors. This book is a pre-release version of a book in progress for Oxford University Press. The following are amongst those who have helped with this book as editors, artists, advisors, or critics: Alexa Barnes, Joseph Burns, Jason Cortell, Ivan Dobrianov, Gabor Domokos, Thu Dong, Gail Fish, John Gibson, Saptarsi Hal- dar, Dave Heimstra, Theresa Howley, Herbert Hui, Michael Marder, Elaina Mc- Cartney, Arthur Ogawa, Kalpana Pratap, Richard Rand, Dane Quinn, Phoebus Rosakis, Les Schaeffer, David Shipman, Jill Startzell, Saskya van Nouhuys, Bill Zobrist. Mike Coleman worked extensively on the text, wrote many of the ex- amples and homework problems and created many of the figures. David Ho has brought almost all of the artwork to its present state. Some of the home- work problems are modifications from the Cornell’s Theoretical and Applied Mechanics archives and thus are due to T&AM faculty or their libraries in ways that we do not know how to give proper attribution. Many unlisted friends, colleagues, relatives, students, and anonymous reviewers have also made helpful suggestions. Software used to prepare this book includes TeXtures, BLUESKY’s implemen- tation of LaTeX, Adobe Illustrator and MATLAB. Most recent text modifications on January 21, 2001. Summary of Mechanics 0) The laws of mechanics apply to any collection of material or ‘body.’ This body could be the overall system of study or any part of it. In the equations below, the forces and moments are those that show on a free body diagram. Interacting bodies cause equal and opposite forces and moments on each other. I) Linear Momentum Balance (LMB)/Force Balance Equation of Motion F i = ˙ L The total force on a body is equal to its rate of change of linear momentum. (I) Impulse-momentum (integrating in time) t 2 t 1 F i ·dt = L Net impulseis equalto thechange in momentum. (Ia) Conservation of momentum (if F i = 0 ) ˙ L = 0 ⇒ L = L 2 − L 1 = 0 When there is no net force the linear momentum does not change. (Ib) Statics (if ˙ L is negligible) F i = 0 If the inertial terms are zero the net force on system is zero. (Ic) II) Angular Momentum Balance (AMB)/Moment Balance Equation of motion M C = ˙ ˙ H C The sum of moments is equal to the rate of change of angular momentum. (II) Impulse-momentum (angular) (integrating in time) t 2 t 1 M C dt = H C The net angular impulse is equal to the change in angular mo mentum. (IIa) Conservation of angular momentum (if M C = 0) ˙ H C = 0 ⇒ H C = H C 2 − H C 1 = 0 If there is no net moment about point C then the angular momentum about point C does not change. (IIb) Statics (if ˙ H C is negligible) M C = 0 If the inertial terms are zero then the total moment on the system is zero. (IIc) III) Power Balance (1st law of thermodynamics) Equation of motion ˙ Q + P = ˙ E K + ˙ E P + ˙ E int ˙ E Heat flow plus mechanical power into a system is equal to its change in energy (kinetic + potential + internal). (III) for finite time t 2 t 1 ˙ Qdt + t 2 t 1 Pdt =E The net energy flow going in is equal to the net change in energy. (IIIa) Conservation of Energy (if ˙ Q = P = 0) ˙ E = 0 ⇒ E = E 2 − E 1 = 0 If no energy flows into a system, then its energy does not change. (IIIb) Statics (if ˙ E K is negligible) ˙ Q + P = ˙ E P + ˙ E int If there is no change of kinetic energy then the change of potential 11 Mechanics of planar mechanisms Many parts of practical machines and structures move in ways that can be idealized as straight-line motion (Chapter 6) or circular motion (Chapters 7 and 8). But often an engineer most analyze parts with more general motions, like a plane in unsteady flight, and a connecting rod in a car engine. Of course, the same basic laws of mechanics still apply. The chapter starts with the kinematics of a rigid body in two dimensions and then progresses to the mechanics and analysis of motions of a planar body. 11.1 Dynamics of particles in the context of 2-D mechanisms Now that we know more kinematics, we can deal with the mechanics of more mech- anisms. Although it is not efficient for problem solving, we take a simple example to illustrate some comparison between some of the ways of keeping track of the motion. For one point mass it is easy to write balance of linear momentum. It is: F = m a. The mass of the particle m times its vector acceleration a is equal to the total force on the particle F . Big deal. Now, however, we can write this equation in four different ways. (a) In general abstract vector form: F = m a. (b) In cartesian coordinates: F x ˆ ı + F y ˆ + F z ˆ k = m[ ¨x ˆ ı +¨y ˆ +¨z ˆ k]. 581 582 CHAPTER 11. Mechanics of planar mechanisms (c) In polar coordinates: F R ˆ e R + F θ ˆ e θ + F z ˆ k = m[( ¨ R − R ˙ θ 2 ) ˆ e R + (2 ˙ R ˙ θ + R ¨ θ) ˆ e θ +¨z ˆ k]. (d) In path coordinates: F t ˆ e t + F n ˆ e n = m[ ˙v ˆ e t + (v 2 /ρ) ˆ e n ] All of these equations are always right. They are summarized in the table on the inside cover. Additionally, for a given particle moving under the action of a given force there are many more correct equations that can be found by shifting the origin and orientation of the coordinate systems. A particle that moves under the influence of no force. In the special case that a particle has no force on it we know intuitively, or from the verbal statement of Newton’s First Law, that the particle travels in a straight line at constant speed. As a first example, let’s look at this result using the vector equations of motion four different ways: in the general abstract form, in cartesian coordinates, in polar coordinates, and in path coordinates. General abstract form The equation of linear momentum balance is F = m a or, if there is no force, a = 0, which means that d v/dt = 0.So v is a constant. We can call this constant v 0 .So after some time the particle is where it was at t = 0, say, r 0 , plus its velocity v 0 times time. That is: r = r 0 + v 0 t.(11.1) This vector relation is a parametric equation for a straight line. The particle moves in a straight line, as expected. Cartesian coordinates If instead we break the linear momentum balance equation into cartesian coordinates we get F x ˆ ı + F y ˆ + F z ˆ k = m( ¨x ˆ ı +¨y ˆ +¨z ˆ k). Because the net force is zero and the net mass is not negligible, ¨x = 0, ¨y = 0, and ¨z = 0. These equations imply that ˙x, ˙y, and ˙z are all constants, lets call them v x0 , v y0 , v z0 . So x, y, and z are given by x = x 0 + v x0 t, y = y 0 + v y0 t, & z = z 0 + v z0 t. We can put these components into their place in vector form to get: r = x ˆ ı + y ˆ + z ˆ k = (x 0 + v x0 t) ˆ ı + (y 0 + v y0 t) ˆ + (z 0 + v z0 t) ˆ k.(11.2) Note that there are six free constants in this equation representing the initial position and velocity. Equation 11.2 is a cartesian representation of equation 11.1; it describes a straight line being traversed at constant rate. 11.1. Dynamics of particles in the context of 2-D mechanisms 583 Particle with no force: Polar/cylindrical coordinates When there is no force, in polar coordinates we have: F R 0 ˆ e R + F θ 0 ˆ e θ + F z 0 ˆ k = m[( ¨ R − R ˙ θ 2 ) ˆ e R + (2 ˙ R ˙ θ + R ¨ θ) ˆ e θ +¨z ˆ k]. This .. .Introduction to Statics and Torque How can we guarantee that a body is in equilibrium and what can we learn from systems that are in equilibrium?... in equilibrium? There are actually two conditions that must be satisfied to achieve equilibrium These conditions are the topics of the first two sections of this chapter 2/2
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