Tài liệu Introduction To Statics And Dynamics P2 doc

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 vectors. In 2-D
2.3 THEORY
Using the geometric definition of the dot product to find the dot product in terms of components
Vectors are essentially a geometric concept and we have conse-
quently defined the dot product geometrically as

A ·

B = ABcosθ .
Almost 400 years ago Ren´e Descartes discovered that you could do
geometry by doing algebra on the coordinates of points.
So we should be able tofigure out thedot product of two vectors
by knowing their components. The central key to finding this com-
ponentformula isthe distributive law(

A·(

B+

C) =

A·


B+

A·

C).
If we write

A = A
x
ˆ
ı + A
y
ˆ
 + A
z
ˆ
k and

B = B
x
ˆ
ı + B
y
ˆ
 + B
z
ˆ
k
then we just repeatedly use the distributive law as follows.


A ·

B = (A
x
ˆ
ı + A
y
ˆ
 + A
z
ˆ
k) · (B
x
ˆ
ı + B
y
ˆ
 + B
z
ˆ
k)
= (A
x
ˆ
ı + A
y
ˆ
 + A
z
ˆ

k) · B
x
ˆ
ı +
(A
x
ˆ
ı + A
y
ˆ
 + A
z
ˆ
k) · B
y
ˆ
 +
(A
x
ˆ
ı + A
y
ˆ
 + A
z
ˆ
k) · B
z
ˆ
k

= A
x
B
x
ˆ
ı ·
ˆ
ı + A
y
B
x
ˆ
 ·
ˆ
ı + A
z
B
x
ˆ
k ·
ˆ
ı +
A
x
B
y
ˆ
ı ·
ˆ
 + A

y
B
y
ˆ
 ·
ˆ
 + A
z
B
y
ˆ
k ·
ˆ
 +
A
x
B
z
ˆ
ı ·
ˆ
k + A
y
B
z
ˆ
 ·
ˆ
k + A
z

B
z
ˆ
k ·
ˆ
k
= A
x
B
x
(1) + A
y
B
x
(0) + A
z
B
x
(0) +
A
x
B
y
(0) + A
y
B
y
(1) + A
z
B

y
(0) +
A
x
B
z
(0) + A
y
B
z
(0) + A
z
B
z
(1)
⇒

A ·

B = A
x
B
x
+ A
y
B
y
+ A
z
B

z
(3D).
⇒

A ·

B = A
x
B
x
+ A
y
B
y
(2D).
The demonstration above could have been carried out using a
different orthogonal coordinate system x

y

z

that was crooked with
respect to the xyz system. By identical reasoning we would find
that

A ·

B = A
x


B
x

+ A
y

B
y

+ A
z

B
z

. Even though all of the
numbersin thelist [A
x
, A
y
, A
z
]might bedifferentfromthe numbers
in the list [A
x

, A
y


, A
z

] and similarly all the list [

B]
xyz
might be
different than the list [

B]
x

y

z

, so (somewhat remarkably),
A
x
B
x
+ A
y
B
y
+ A
z
B
z

= A
x

B
x

+ A
y

B
y

+ A
z

B
z

.
If we call our coordinate x
1
, x
2
, and x
3
; and our unit base
vectors
ˆ
e
1

,
ˆ
e
2
, and
ˆ
e
3
we would have

A = A
1
ˆ
e
1
+ A
2
ˆ
e
2
+ A
3
ˆ
e
3
and

B = B
1
ˆ

e
1
+ B
2
ˆ
e
2
+ B
3
ˆ
e
3
and the dot product has the tidy
form:

A ·

B = A
1
B
1
+ A
2
B
2
+ A
3
B
3
=

3

i=1
A
i
B
i
.
2.2. The dot product of two vectors 25
assume that the dot products between the standard base vectors and the vector
ˆ


(i.e., .
ˆ
ı ·
ˆ


,
ˆ
 ·
ˆ


) are known. One can then use the dot product to find the x

y

components (A

x

, A
y

) from the xycoordinates (A
x
, A
y
). For example, as shown in
2-D in figure 2.19, we can start with the obvious equation
x
x'
y
y'
A
x
A
y
A
x'
A
y'

A = A
x
ˆ
ı + A
y
ˆ



A = A
x

ˆ
ı

+ A
y

ˆ


θ
ˆ
ı
ˆ

ˆ
ı

ˆ


Figure 2.19: The dot product helps
find components in terms of crooked unit
vectors. For example, A
y


=

A·
ˆ


=
A
x
(
ˆ
ı·
ˆ


) + A
y
(
ˆ
ı·
ˆ


) = A
x
(−sin θ) +
A
y
(cos θ).
(Filename:tfigure1.3.dotprod.a)


A =

A
and dot both sides with
ˆ


to get:

A ·
ˆ


=

A ·
ˆ


(A
x

ˆ
ı

+ A
y

ˆ



)

 

A
·
ˆ


= (A
x
ˆ
ı + A
y
ˆ
)

 

A
·
ˆ


A
x

ˆ

ı

·
ˆ



0
+A
y

ˆ


·
ˆ



1
= A
x
ˆ
ı ·
ˆ


+ A
y
ˆ

 ·
ˆ


A
y

= A
x
(
ˆ
ı ·
ˆ


)

 
−sin θ
+A
y
(
ˆ
 ·
ˆ


)

 

cos θ
Similarly, one could find the component A
x

using a dot product with
ˆ
ı

.
This technique of finding components is useful when one problem uses more than
one base vector system.
Using dot products with other than ˆı, ˆ,or
ˆ
k
It is often useful to use dot products to get scalar equations using vectors other than
ˆ
ı,
ˆ
, and
ˆ
k.
Example: Getting scalar equations without dotting with
ˆ
ı,
ˆ
,or
ˆ
k
Given the vector equation.
−mg

ˆ
 + N
ˆ
n = ma
ˆ
λ
where it is known that the unit vector
ˆ
n is perpendicular to the unit vector
ˆ
λ, we can get a scalar equation by dotting both sides with
ˆ
λ which we
write as follows

(−mg
ˆ
 + N
ˆ
n) = (ma
ˆ
λ)

·
ˆ
λ
(−mg
ˆ
 + N
ˆ

n)·
ˆ
λ = (ma
ˆ
λ)·
ˆ
λ
−mg
ˆ
·
ˆ
λ + N
ˆ
n·
ˆ
λ

0
= ma
ˆ
λ·
ˆ
λ

1
−mg
ˆ
·
ˆ
λ = ma.

Then we find
ˆ
·
ˆ
λ as the cosine of the angle between
ˆ
 and
ˆ
λ.Wehave
thus turned our vector equation into a scalar equation and eliminated the
unknown N at the same time. ✷
26 CHAPTER 2. Vectors for mechanics
Using dot products to solve geometry problems
We have seen how a vector can be broken down into a sum of components each
parallel to one of the orthogonal base vectors. Another useful decomposition is this:
Given any vector

A and a unit vector
ˆ
λ the vector

A can be written as the sum of two
parts,

A =

A

+


A
⊥
where

A

is parallelto
ˆ
λ and

A
⊥
is perpendicularto
ˆ
λ (seefig. 2.20). The part parallel
to
ˆ
λ is a vector pointed in the
ˆ
λ direction that has the magnitude of the projection of

A in that direction,

A

= (

A ·
ˆ
λ)

ˆ
λ.
The perpendicular part of

A is just what you get when you subtract out the parallel
part, namely,

A
⊥
=

A −

A

=

A − (

A ·
ˆ
λ)
ˆ
λ
The claimed properties of the decomposition can now be checked, namely that

A =

A
⊥


A
ˆ
λ

A
||
Figure 2.20: For any

A and
ˆ
λ,

A can be
decomposed into a part parallel to
ˆ
λ and a
part perpendicular to
ˆ
λ
.
(Filename:tfigure.Graham1)

A

+

A
⊥
(just add the 2 equations above and see), that


A

is in the direction of
ˆ
λ
(its a scalar multiple), and that

A
⊥
is perpendicular to
ˆ
λ (evaluate

A
⊥
·
ˆ
λ and find 0).
Example. Given the positions of three points

r
A
,

r
B
, and

r

C
what is the
position of the point D on the line AB that is closest to C? The answer
is,

r
D
=

r
A
+

r
C/A

where

r
C/A

is the part of

r
C/A
that is parallel to the line segment AB.
Thus,

r
D

=

r
A
+ (

r
C
−

r
A
) ·

r
B
−

r
A
|

r
B
−

r
A
|
.

✷
Likewise we could find the parts of a vector

A in and perpendicular to a given
plane. If the plane is defined by two vectors that are not necessarily orthogonal we
could follow these steps. First find two vectors in the plane that are orthogonal, using
the method above. Next subtract from

A the part of it that is parallel to each of the
two orthogonal vectors in the plane. In math lingo the execution of this process goes
by the intimidating name ‘Graham Schmidt orthogonalization.’
A Given vector can be written as various sums and products
A vector

A has many representations. The equivalence of different representations
of a vector is partially analogous to the case of a dimensional scalar which has the
same value no matter what units are used (e.g., the mass m = 4.41 lbm is equal to
m = 2kg). Here are some common representations of vectors.
Scalar times a unit vector in the vector’s direction.

F = F
ˆ
λ means the scalar F
multiplied by the unit vector
ˆ
λ.
2.2. The dot product of two vectors 27
Sum of orthogonal component vectors.

F =


F
x
+

F
y
is a sum of two vectors
parallel to the x and y axis, respectively. In three dimensions,

F =

F
x
+

F
y
+

F
z
.
Components times unit base vectors.

F = F
x
ˆ
ı + F
y

ˆ
 or

F = F
x
ˆ
ı + F
y
ˆ
 + F
z
ˆ
k
in three dimensions. One way to think of this sum is to realize that

F
x
= F
x
ˆ
ı,

F
y
= F
y
ˆ
 and

F

z
= F
z
ˆ
k.
Components times rotated unit base vectors.

F = F

x
i

+ F

y
j

or

F = F

x
i

+
F

y
j


+ F

z
k

in three dimensions. Here the base vectors marked with primes,
i

, j

and k

, are unit vectors parallel to some mutually orthogonal x

, y

, and
z

axes. These x

, y

, and z

axes may be crooked in relation to the x, y, and z
axis. That is, the x

axis need not be parallel to the x axis, the y


not parallel to
the y axis, and the z

axis not parallel to the z axis.
Components times other unit base vectors. If you use polar or cylindrical coordi-
nates the unit base vectors are
ˆ
e
θ
and
ˆ
e
R
, so in 2-D ,

F = F
R
ˆ
e
R
+ F
θ
ˆ
e
θ
and
in 3-D,

F = F
R

ˆ
e
R
+ F
θ
ˆ
e
θ
+ F
z
ˆ
k. If you use ‘path’ coordinates, you will use
the path-defined unit vectors
ˆ
e
t
,
ˆ
e
n
, and
ˆ
e
b
so in 2-D

F = F
t
ˆ
e

t
+ F
n
ˆ
e
n
. In 3-D

F = F
t
ˆ
e
t
+ F
n
ˆ
e
n
+ F
b
ˆ
e
b
.
A list of components. [

F ]
xy
= [F
x

, F
y
]or[

F ]
xyz
= [F
x
, F
y
, F
z
] in three dimen-
sions. This form coincides best with the way computers handle vectors. The
row vector [F
x
, F
y
] coincides with F
x
ˆ
ı + F
y
ˆ
 and the row vector [F
x
, F
y
, F
z

]
coincides with F
x
ˆ
ı + F
y
ˆ
 + F
z
ˆ
k.
In summary:

A =

A
=|

A|
ˆ
λ
A
= A
ˆ
λ
A
, where
ˆ
λ
A



A, A =|

A| and |
ˆ
λ
A
|=1
=

A
x
+

A
y
+

A
z
where

A
x
,

A
y
,


A
z
are parallel to the x, y, z axis
= A
x
ˆ
ı + A
y
ˆ
 + A
z
ˆ
k, where
ˆ
ı,
ˆ
,
ˆ
k are parallel to the x, y, z axis
= A
x

ˆ
ı

+ A
y

ˆ



+ A
z

ˆ
k

, where
ˆ
ı

,
ˆ


,
ˆ
k

are  to skewed x

, y

, z

axes
= A
R
ˆ

e
R
+ A
θ
ˆ
e
θ
+ A
z
ˆ
k, using polar coordinate basis vectors.
[

A]
xyz
= [A
x
, A
y
, A
z
][

A]
xyz
stands for the component list in xyz
[

A]
x


y

z

= [A
x

, A
y

, A
z

][

A]
x

y

z

stands for the component list in x

y

z

Vector algebra

Vectors are algebraic quantities and manipulated algebraically in equations. Therules
for vector algebra are similar to the rules for ordinary (scalar) algebra. For example,
if vector

A is the same as the vector

B,

A =

B. For any scalar a and any vector

C,
we then

A +

C =

B +

C,
a

A = a

B, and

A ·


C =

B ·

C,
because performing the same operation on equal quantities maintains the equality.
The vectors

A,

B, and

C might themselves be expressions involving other vectors.
The equations above show the allowable manipulations of vector equations:
adding a common term to both sides, multiplying both sides by a common scalar,
taking the dot product of both sides with a common vector.
All the distributive, associative, and commutative laws of ordinary addition and
multiplication hold.
1

.
1

Caution: Butyou cannot divide a vector
by a vector or a scalar by a vector: 7/
ˆ
ı
and

A/


C are nonsense expressions. And it
does not make sense to add a vector and a
scalar, 7 +

A is a nonsense expression.
28 CHAPTER 2. Vectors for mechanics
Vector calculations on the computer
Most computer programs deal conveniently with lists of numbers, but not with vec-
tor notation and units. Thus our computer calculations will be in terms of vector
components with the units left off. For example, when we write on the computer
F=[35-7]
we take that to be the plain computer typing for [

F ]
xyz
= [3N, 5N, −7 N]. This
assumes that we are clear about what units and what coordinate system we are using.
In particular, at this point in the course, you should only use one coordinate system
in one problem in computer calculations.
Most computer languages will allow vector addition by a sequence of lines some-
thing like this:
A=[125]
B=[-2 419]
C=A+B
scaling (stretching) like this:
A=[125]
C = 3*A
and dot products like this:
A =[125]

B =[-2 419]
D = A(1)*B(1) + A(2)*B(2) + A(3)*B(3).
In our pseudo code we write D = A dot B. Many computer languages have a
shorter way to write the dot product like dot(A,B). In a language built for linear
algebra D = A*B’
1

will work because the rules of matrix multiplication are then
1

B’ is a common notation for the trans-
pose of B, which means, in this case, to turn
the rowof numbers B into a column of num-
bers.
the same as the component formula for the dot product.
2.2. The dot product of two vectors 29
SAMPLE 2.12 Calculating dot products: Find the dot product of the two vectors

a = 2
ˆ
ı +3
ˆ
 − 2
ˆ
k and

r = 5m
ˆ
ı −2m
ˆ

.
Solution The dot product of the two vectors is

a ·

r = (2
ˆ
ı +3
ˆ
 − 2
ˆ
k) · (5m
ˆ
ı −2m
ˆ
)
= (2 · 5m)
ˆ
ı ·
ˆ
ı

1
−(2 · 2m)
ˆ
ı ·
ˆ


0

+(3 · 5m)
ˆ
 ·
ˆ
ı

0
−(3 · 2m)
ˆ
 ·
ˆ



1
−(2 · 5m)
ˆ
k ·
ˆ
ı

0
+(2 · 2m)
ˆ
k ·
ˆ



0

= 10 m −6m
= 4m.

a ·

r = 4m
Comments: Note that with just a little bit of foresight, we could totally ignore the
ˆ
k component of

a since

r has no
ˆ
k component, i.e.,
ˆ
k ·

r = 0. Also, if we keep in
mind that
ˆ
ı ·
ˆ
 =
ˆ
 ·
ˆ
ı = 0, we could compute the above dot product in one line:

a ·


r = (2
ˆ
ı +3
ˆ
) · (5m
ˆ
ı −2m
ˆ
) = (2 · 5m)
ˆ
ı ·
ˆ
ı

1
−(3 · 2m)
ˆ
 ·
ˆ



1
= 4m.
SAMPLE 2.13 What is the y-component of

F = 5N
ˆ
ı +3N

ˆ
 + 2N
ˆ
k ?
Solution Although it is perhaps obvious that the y-component of

F is 3 N, the scalar
multiplying the unit vector
ˆ
, we calculate it below in a formal way using the dot
product between two vectors. We will use this method later to find components of
vectors in arbitrary directions.
F
y
=

F · (a unit vector along y-axis)
= (5N
ˆ
ı +3N
ˆ
 + 2N
ˆ
k) ·
ˆ

= 5N
ˆ
ı ·
ˆ



0
+3N
ˆ
 ·
ˆ



1
+2N
ˆ
k ·
ˆ



0
= 3N.
F
y
=

F ·
ˆ
 = 3N.
30 CHAPTER 2. Vectors for mechanics
SAMPLE 2.14 Finding angle between two vectors using dot product: Find the angle
between the vectors


r
1
= 2
ˆ
ı +3
ˆ
 and

r
2
= 2
ˆ
ı −
ˆ
.
Solution From the definition of dot product between two vectors

r
1
·

r
2
=|

r
1
||


r
2
|cos θ
or cos θ =

r
1
·

r
2
|

r
1
||

r
2
|
=
(2
ˆ
ı +3
ˆ
) · (2
ˆ
ı −
ˆ
)

(
√
2
2
+ 3
2
)(
√
2
2
+ 1
2
)
=
4 − 3
√
13
√
5
= 0.124
Therefore, θ = cos
−1
(0.124) = 82.87
o
.
θ = 83
o
SAMPLE 2.15 Finding direction cosines from unit vectors: Find the angles (from
direction cosines) between


F = 4N
ˆ
ı +6N
ˆ
 + 7N
ˆ
k and each of the three axes.
Solution

F = F
ˆ
λ
ˆ
λ =

F
F
=
4N
ˆ
ı +6N
ˆ
 + 7N
ˆ
k
√
4
2
+ 6
2

+ 7
2
N
= 0.4
ˆ
ı +0.6
ˆ
 + 0.7
ˆ
k.
Let the angles between
ˆ
λ and the x, y, and z axes be θ, φ and ψ respectively. Then
cos θ =
ˆ
ı ·
ˆ
λ
|
ˆ
ı||
ˆ
λ|
=
0.4
|1||1|
= 0.4.
⇒ θ = cos
−1
(0.4) = 66.4

o
.
Similarly,
cos φ = 0.6orφ = 53.1
o
cos ψ = 0.7orψ = 45.6
o
.
θ = 66.4
o
,φ= 53.1
o
,ψ= 45.6
o
Comments: The components of a unit vector give the direction cosines with the
respective axes. That is, if the angle between the unit vector and the x, y, and z axes
are θ, φ and ψ, respectively (as above), then
ˆ
λ = cosθ


λ
x
ˆ
ı +cosφ


λ
y
ˆ

 + cosψ


λ
z
ˆ
k.
2.2. The dot product of two vectors 31
SAMPLE 2.16 Projection of a vector in the direction of another vector: Find the
component of

F = 5N
ˆ
ı +3N
ˆ
 + 2N
ˆ
k along the vector

r = 3m
ˆ
ı −4m
ˆ
.
Solution The dot product of a vector

a with a unit vector
ˆ
λ gives the projection of
the vector


a in the direction of the unit vector
ˆ
λ. Therefore, to find the component of

F along

r ,wefirst find a unit vector
ˆ
λ
r
along

r and dot it with

F .
ˆ
λ
r
=

r
|

r |
=
3m
ˆ
ı −4m
ˆ


√
3
2
+ 4
2
m
= 0.6
ˆ
ı −0.8
ˆ

F
r
=

F ·
ˆ
λ
r
= (5N
ˆ
ı +3N
ˆ
 + 2N
ˆ
k) · (0.6
ˆ
ı −0.8
ˆ

)
= 3.0N+2.4N= 5.4N.
F
r
= 5.4N
SAMPLE 2.17 Assume that after writing the equation


F = m

a in a particular
problem, astudent finds


F = (20 N−P
1
)
ˆ
ı +7N
ˆ
 −P
2
ˆ
k and

a = 2.4m/s
2
ˆ
ı +a
3

ˆ
.
Separate the scalar equations in the
ˆ
ı,
ˆ
, and
ˆ
k directions.
Solution


F = m

a
Taking the dot product of both sides of this equation with
ˆ
ı, we write
ˆ
ı ·


F =
ˆ
ı ·m

a
ˆ
ı ·


(20 N − P
1
)
ˆ
ı +7N
ˆ
 − P
2
ˆ
k = m(2.4m/s
2
ˆ
ı +a
3
ˆ
)

⇒ (20 N − P
1
)

 
F
x
ˆ
ı ·
ˆ
ı

1

+7N
ˆ
 ·
ˆ
ı

0
−P − 2
ˆ
k ·
ˆ
ı

0
= m(2.4m/s
2
  
a
x
ˆ
ı ·
ˆ
ı

1
+a
3
ˆ
 ·
ˆ

ı

0
)
⇒

F
x
= ma
x
⇒ 20 N − P
1
= m(2.4m/s
2
)
Similarly,
ˆ
 ·



F = m

a

⇒

F
y
= ma

y
(2.1)
ˆ
k ·



F = m

a

⇒

F
z
= ma
z
. (2.2)
Substituting the given components of

F and

a in the remaining Eqns. (2.1) and (2.2)
we get
7N = ma
y
−P
2
= 0.
Comments: Aslongasbothsides ofavectorequationareinthe samebasis, separating

the scalar equations is trivial—simply equate the respective components from both
sides. The technique of taking the dot product of both sides with a vector is quite
general and powerful. It gives a scalar equation valid in any direction that one desires.
You will appreciate this technique more if the vector equation uses more than one
basis.
32 CHAPTER 2. Vectors for mechanics
2.3 Cross product, moment, and mo-
ment about an axis
When you try to move something you can push it and you can turn it. In mechanics,
the measure of your pushing is the net force you apply. The measure of your turning
is the net moment, also sometimes called the net torque or net couple. In this section
we will define the moment of a force intuitively, geometrically, and finally using
vector algebra. We will do this first in 2 dimensions and then in 3. The main
mathematical tool here is the vector cross product, a second way of multiplying
vectors together. The cross product is used to define (and calculate) moment and to
calculate various quantities in dynamics. The cross product also sometimes helps
solve three-dimensional geometry problems.
Although concepts involving moment(and rotation) areoften harder forbeginners
than force (and translation), they were understood first. The ancient principle of the
lever is the basic idea incorporated by moments. The principle of the lever can be
viewed as the root of all mechanics.
Ultimately you can take on faith the vector definition of moment (given opposite
the inside cover) and its role in eqs. II. But we can more or less deduce the definition
by generalizing from common experience.
Teeter totter mechanics
(not a free body diagram)
Figure 2.21:
On a balanced teeter totter
the bigger person gets the short end of the
stick. A sideways force directed towards

the hinge has no effect on the balance.
(Filename:tfigure.teeter)
The two people weighing down on the teeter totter in Fig. 2.21 tend to rotate it about
its hinge, the right one clockwise and the left one counterclockwise. We will now
cook up a measure of the tendency of each force to cause rotation about the hinge and
call it the moment of the force about the hinge.
As is verified a million times a year by young future engineering students, to
balance a teeter-totter the smaller person needs to be further from the hinge. If two
people are on one side then the teeter totter is balanced by two similar people an equal
distance from the hinge on the other side. Two people can balance one similar person
by scooting twice as close to the hinge. These proportionalities generalize to this:
the tendency of a force to cause rotation is proportional to the size of the force and to
its distance from the hinge (for forces perpendicular to the teeter totter).
If someone standingnearbyadds a forcethat is directedtowards the hingeitcauses
no tendency to rotate. Because any force can be decomposed into a sum of forces,
one perpendicular to the teeter totter and the other towards the hinge, and because
we assume that the affect of the sum of these forces is the sum of the affects of each
separately, and because the force towards the hinge has no tendency to rotate, we
have deduced:
The moment of a force about a hinge is the product of its distance from
the hinge and the component of the force perpendicular to the line from
the hinge to the force.
Here then is the formula for 2D moment about C or moment with respect to C.
1

1

The ‘/’ in the subscript of

M reads as

‘relative to’ or ‘about’. For simplicity we
often leave the / out and just write

M
C
.
M
/C
=|

r | (|

F |sinθ) = (|

r |sinθ) |

F |.(2.3)
Here, θ is the angle between

r (the position of the point of force application relative
to the hinge) and

F (see fig. 2.22). This formula for moment has all the teeter
totter deduced properties. Moment is proportional to r, and to the part of

F that is
perpendicular to

r . The re-grouping as (|


r |sinθ) shows that a force

F has the same
effect if it is applied at a new location that is displaced in the direction of

F . That is,