Transcript Vector Matrix
STATICS OF PARTICLES
Forces are
vector quantities
; they add according to the
parallelogram law. The
magnitude and direction of the resultant
R
of two forces
P
and graphically or by trigonometry.
Q
can be determined either
R P
A
Q 2 - 1
Q
A
Any given force acting on a particle can be resolved into two or more
components
, i.e.., it can be replaced by two or more forces which have the same effect on the particle.
F P
A force
F
can be resolved into two components
P
and
Q
by drawing a parallelogram which has
F
for its diagonal; the components
P
and
Q
are then represented by the two adjacent sides of the parallelogram and can be determined either graphically or by trigonometry.
2 - 2
A force
F
is said to have been resolved into two
rectangular components
if its components are directed along the coordinate axes. Introducing the
unit vectors i
and
j
along the
x
and
y
axes,
y
F
=
F
x
i
+
F
y
j
F
x
=
F
cos
q
F
y
=
F
sin
q
j F
y
=
F
y
j
q
F
tan
q
=
F
y
F
x
F
=
F
x
2
+
F
y
2
x
i F
x
=
F
x
i 2 - 3
When
three or more coplanar forces
act on a particle, the rectangular components of their resultant
R
can be obtained by adding algebraically the corresponding components of the given forces.
R
x
= S
R
x
R
y
= S
R
y
The magnitude and direction of
R
can be determined from
tan
q
=
R
y
R
x
R
=
R
x
2
+
R
y
2
2 - 4
y y
B B
F
y
F F
y
A D O O
q
x
F
x x
F
z
F
z
E E C
z z y
A force
F
in
three-dimensional space
can be resolved into components
F
x
=
F F
cos
q
x z
=
F F
y
= cos
q
z
F
cos
q
y
E
z
F
z
B
F
y
O
q
z
q
y
F
A
F F
x
C C A
F
x
D
x
D
x
2 - 5
y
l (Magnitude = 1) cos q
y
j
cos q
z
k
F
y
j
F = F l
F
x
i
x
The cosines of q
x
, q
y
, and q
z
are known as the
direction cosines
of the force
F
. Using the unit vectors
i
,
j
, and
k
, we write
F
z
k
z
cos q
x
i
F
=
F
x
i
+
F
y
j
+
F
z
k
or
F
=
F
(cos
q
x
i
+
cos
q
y
j
+
cos
q
z
k
)
2 - 6
cos q
y
j
cos q
z
k
F
z
k
z y F y
j
cos q
x
i
l (Magnitude = 1) F = F l l =
cos
+
cos
q
x
q
z
i
+
cos
q
y
k j
F
x
i
Since the magnitude of l is unity, we have
x
cos
2
q
x
+
cos
2
+
cos
2
q
z
q
y
= 1
In addition,
F
=
F
x
2
+
F
y
2
+
F
z
2
cos
q
x
=
F
x
F
cos
q
y
=
F
y
F
cos
q
z
=
F
z
F
2 - 7
y
l
M
(
x
1 ,
y
1 ,
z
1 )
F
N
(
x
2 ,
y
2 ,
z
2 )
d y
=
y
2 -
y
1
d x
=
x
2 -
x
1
dz
=
z
2 < 0 -
z
1
x
A force vector
F
in three-dimensions is defined by its magnitude
F
and two points
M
and
N
along its line of action. The vector
MN
M joining points and
N
is
z
MN
=
d
x
i
+
d
y
j
+
d
z
k
The unit vector l along the line of action of the force is l
MN
= =
MN
1
d
(
d
x
i
+
d
y
j
+
d
z
k
)
2 - 8
y N
(
x
2 ,
y
2 ,
z
2 )
d
=
d
x
2 +
d
y
2 +
d
z
2
M
(
x
1 ,
y
1 ,
z
1 )
z d y
=
y
2 -
y
1
d x
=
x
2 -
x
1
d z
=
z
2 < 0 -
z
1
x
A force
F
is defined as the product of
F
l . Therefore, and
F
=
F
l
F
=
d
(
d
x
i
+
d
y
j
+
d
z
k
)
From this it follows that
F
x
=
Fd
x
d F
y
=
Fd d
y
F
z
=
Fd d
z
2 - 9
When
two or more forces
act on a particle in
three dimensions
, the rectangular components of their resultant
R
is obtained by adding the corresponding components of the given forces.
R
x
R
y
R
z
= S = S = S
F
x
F
y
F
z
The particle is in
equilibrium
when the resultant of all forces acting on it is zero.
2 - 10
To solve a problem involving a particle in equilibrium, draw a
free-body diagram
showing all the forces acting on the particle. The conditions which must be satisfied for particle equilibrium are S
F
x
= 0 S
F
y
= 0 S
F
z
= 0 In
two-dimensions
, only two of these equations are needed S
F
x
= 0 S
F
y
= 0
2 - 11
V
=
P
x
Q
The
vector product of two vectors
is defined as
Q
q
V
=
P
x
Q
P
The vector product of
P
and
Q
forms a vector which is perpendicular to both
P
and
Q
, of magnitude
V
=
PQ
sin
q This vector is directed in such a way that a person located at the tip of
V
observes as counterclockwise the rotation through q which brings vector
P
in line with vector
Q
. The three vectors
P
,
Q
, and
V
- taken in that order - form a
right-hand triad
. It follows that
Q
x
P
= - (
P
x
Q
)
2 - 12
j
It follows from the definition of the vector product of two vectors that the vector products of unit vectors
i
,
j
, and
k
are
k i i
x
i
=
j
x
j
=
k
x
k
= 0
i
x
j
=
k , j
x
k
=
i , k
x
i
=
j , i
x
k
= -
j , j
x
i
= -
k , k
x
j
= -
i
The rectangular components of the vector product
V
of two vectors
P
and
Q
are determined as follows: Given
P
=
P
x
i
+
P
y
j
+
P
z
k Q
=
Q
x
i
+
Q
y
j
+
Q
z
k
The determinant containing each component of
P
and
Q
is expanded to define the vector
V
, as well as its scalar components
2 - 13
P
=
P
x
i
+
P
y
j
+
P
z
k Q
=
Q
x
i
+
Q
y
j
+
Q
z
k V
=
P
x
Q
=
i
P
x
Q
x
j
P
y
Q
y
k
P
z
Q
z
=
V
x
i
+
V
y
j
+
V
z
k
where
V
x
V
y
V
z
=
P
y
=
P
z
=
P
x
Q
z
Q
x
Q
y
-
P
z
P
x
P
y
Q
y
Q
z
Q
x
2 - 14
M
o
The moment of force
F
about point
O
is defined as the vector product
M
O
=
r
x
F
F
O d
r
A
q where
r
is the position vector drawn from point
O
to the point of application of the force
F
. The angle between the lines of action of
r
and
F
is q . The magnitude of the moment of
F
about
O
can be expressed as
M
O
=
rF
sin
q
=
Fd
where
d
is the perpendicular distance from
O
of
F
.
to the line of action
2 - 15
y
F
y
j
A
(
x
,
y
,
z
)
y j
O
r
F
z
k x i z k
z
M
o
=
r
x
F
=
i
x F
x
y
j
F
y
k
z F
z
where
M
x
M
z
=
y F
z
=
x F
y
-
z F
y
y F
x
=
M
x
i
+
M
y
j
+
M
z
k
M F
x
i
y x
The rectangular components of the moment
M
o
force
F
are of a determined by expanding the determinant of
r
x
F
.
=
zF
x
-
x F
z
2 - 16
y
F
y
j
A
(
x A
,
y A
,
z A
)
z
B
(
x B
,
y B
,
z B
)
O
r
F
z
k
F
x
i
In the more general case of the moment about an arbitrary point
B
of a force
F
applied at
A
, we have
x
M
B
=
r
A/B
x
F
=
i
x
A/B
F
x
r
A/B
=
x
A/B
i
+
y
A/B
j
+
z
A/B
k j
y
A/B
F
y
k
z
A/B
F
z
and
x
A/B
=
x
A
-
x
B
y
A/B
=
y
A
-
y
B
z
A/B
=
z
A
-
z
B
2 - 17
y
F
y
j
z
(
y
A
-
y
B
)
j
B
O
M
B
=
M
B
k r A/B
(
x
A
-
x
B
)
i
A F
x
i
x
F
In the case of problems involving only two dimensions, the force
F
can be assumed to lie in the
xy
plane. Its moment about point
B
is perpendicular to that plane. It can be completely defined by the scalar
M
B
= (
x
A
-
x
B
)
F
y
+ (
y
A
-
y
B
)
F
x
The
right-hand rule
is useful for defining the direction of the moment as either into or out of the plane (positive or negative
k
direction).
2 - 18
q
P Q
The
scalar product of two vectors
P
and
Q
is denoted as
P Q
,and is defined as
P Q = PQ cos
q where q is the angle between the two vectors The scalar product of
P
and
Q
is expressed in terms of the rectangular components of the two vectors as
P Q = P
x
Q
x
+
P
y
Q
y
+
P
z
Q
z
2 - 19
y z O
l q
y
q
z
A
q
x
P
L
x
The projection of a vector
P
on an axis
OL
can be obtained by forming the scalar product of
P
unit vector and the l along
OL
.
P
OL
=
P
l Using rectangular components,
P
OL
=
P
x
cos
q
x
+
P
y
cos
q
y
+
P
z
cos
q
z
2 - 20
The mixed triple product of three vectors
S
,
P
, and
Q
is
S
(
P
x
Q
) =
S
x
P
x
Q
x
S
y
P
y
Q
y
S
z
P
z
Q
z
The elements of the determinant are the rectangular components of the three vectors.
2 - 21
M
O z y
l
O
C L
r
M
OL
=
l
M
O
=
l
F
A
(
x, y, z
)
x
(
r
x
F
) =
The
moment of a force F about an axis OL
is the projection
OC
on
OL
of the moment
M
O
force
F
of the . This can be written as a mixed triple product.
l
x
x F
x
l
y y F
y
l
z
z F
z
l
x
x
, ,
y
l
y
,
z
, l
z
= direction cosines of axis = components of
r
OL F
x
,
F
y
,
F
z
= components of
F 2 - 22
M
-
F
d
F
Two forces
F
and
-
F
having the same magnitude, parallel lines of action, and opposite sense are said to form a couple
. The moment of a couple is independent of the point about which it is computed; it is a vector
M
perpendicular to the plane of the couple and equal in magnitude to the product
Fd
.
2 - 23
z y
-
F
O
d
x
F
y z O
M
(
M
=
Fd
)
x z
M
y y O
M
z
M
M
x x
Two couples having the same moment
M
are
equivalent
(they have the same effect on a given rigid body).
2 - 24
F
O
r
A
M
O
O
F
A Any force
F
acting at a point
A
of a rigid body can be replaced by a
force-couple system
at an arbitrary point O , consisting of the force
F
applied at O the moment about point O and a couple of moment
M O
of the force
F
equal to in its original position. The force vector
F
and the couple vector
M O
perpendicular to each other.
are always
2 - 25
F 3
A 3
F 1 M 1 F 3 F 1 R r
3 O
r
1
r
2 A A 1 2
M 3
O
M 2
O
F 2 F 2
Any system of forces can be reduced to a force-couple
M
R O system at a given point O
. First, each of the forces of the system is replaced by an equivalent force-couple system at
O
. Then all of the forces are added to obtain a resultant force
R
, and all of couples are added to obtain a resultant couple vector
R M O
. In general, the resultant force
R R
and the couple vector
M O
will not be perpendicular to each other.
2 - 26
r
3
F 3
A 3 O
r
1
r
2 A 1 A 2
F 1 R
O
F 2 M
R O
As far as rigid bodies are concerned,
two systems of forces
,
F 1
,
F 2
,
F 3 . . . , and F’ 1
,
F’ 2
,
F’ 3 . . . , are equivalent if, and
only if,
S
F
=
S
F’
and S
M
o
=
S
M
o
’
2 - 27
F 3
O A
r
1 3
r
2 A A 1 2
F 1 R
O If the resultant force
R
and the resultant
R
couple vector
M O
are perpendicular to each other, the force-couple system at
O
can be further reduced to a single resultant force.
F 2 M
R O
This is the case for systems consisting of either (a) concurrent forces, (b) coplanar forces, or (c) parallel forces. If the resultant force and couple are directed along the same line, the force-couple system is termed a
wrench
.
2 - 28