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V. Analysis of 3DForce Systems
Most of the physical quantities in mechanics canbe expressed mathematically by means ofscalars and vectors
A scalar quantity is characterized by a positiveor negative number, i.e., mass, volume andlength
A vector quantity has both magnitude anddirection, i.e., force and moment
Unit vector
A unit vector is a vector having a magnitude of 1. If Ais a vector having a magnitude A 0, then a unitvector having the same direction as A is representedby
where A represents the magnitude of and defines the direction
and sense of
Au
Au
A
A
=
=
A
uA
1
Cartesian Unit Vectors
In threedimensions, the setof Cartesian unitvectors I, j, k, isused to designatethe directions of thex, y, and z axe
respectively,z
y
xi
j
k
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Cartesian Vector Representation
x
z
y
Ayj
Axi
Azk
j
i
k
kAjAiA zyx++=
Cartesian vector form of components:
Magnitude of a Cartesian Vector
x
z
y
Ayj
Axi
Azk
j
i
k
( ) ( ) ( )222
zyx AAAA++=
Ay
Az
Ax
A
Direction of a Cartesian Vector
x
z
y
Ayj
Axi
Azk
A
Ax=cosDirection cosines of A: , ,
A
Ay=cos A
Az=cos
kAjAiA
kAjAiA
Au
kyiu
kA
Aj
A
Ai
A
A
Au
zyx
A
A
zyxA
++=
++=
=
=++
++=
++=
=
coscoscos
1coscoscos
coscoscos
222
If magnitude and coordinate direction angles of A are given,the Cartesian vector form of A is,
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Vector Operations
x
z
y
(Ax + Bx)i
kAjAiA zyx ++=
R
B
A
(Az + Bz)k
(Ay + By)j
kBjBiBzyx ++=
Given:
( ) ( ) ( )
( ) ( ) ( )kBAjBAiBAR
kBAjBAiBAR
zzyyxx
zzyyxx
++==
+++++=+=
'
Vector Addition:
Vector Subtraction:
Resultant of Force Systems
Concurrent Force System
The concept of vector addition may be generalizedand applied to a system of several concurrentforces.
The force resultant is the vector sum of all the forcesin the system
ZkYjXiFR ++==
Problem 1
Determine themagnitude andcoordinatedirection anglesof the resultantforce acting onthe ring.
x
y
z
F1 = {300i+400j}NF2 = {- 500i+1000j+1000k}N
Problem 2
Two forces act onthe hook. Specifythe coordinatedirection angles ofF2 so that theresultant force FRacts along thepositive x axis andhas a magnitude of
800N
y
z
x
F2=700N
F1=300N
600
450
1200
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Position vector
x
z
y
( )zyxP ,,
yj
xi
zk
zkyjxir ++=
The position vector r isdefined as a fixed vectorwhich locates a point inspace relative to another
point.
O
The position vector may ne denoted from pointA to point B in space
( )AAA zyxA ,,
x
y
z
r
rA
rB
( )ixx AB
( ) ( )( ) ( ) ( )kzzjyyixxr
kzjyixkzjyixrrr
rrr
ABABAB
BBABBBAB
BA
++=
++++==
=+
x
z
y
A
B
( )jyy AB
( )kzz AB
Dot product
The dot product of vectors A and B is defined asthe product of the magnitudes of A and B andthe cosine of the angle between their tails
cosAB=
A
B
Laws of Operation
Commutative law
Multiplication by a scalar
Distributive law
=
( ) ( ) ( ) ( )aaaa ===
( ) ( ) ( )DD =
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The dot product of two general vectors A and Bexpressed in Cartesian vector form
Applications
Angle formed between two vectors or intersecting lines
Components of a vector parallel and perpendicular to a line
( ) ( )
zzyyxx
zyxzyx
BABABA
kBjBiBkAjAiA
++=
++++=
Cross Product
The cross product of two vectors A and Byields the vector C,
C = A x B
The magnitude of C is defined as the productof the magnitudes of A and B and the sine ofthe angle between their tails
C = AB sin
Vector C has a direction perpendicular to
the plane containing A and B
C = A x B = (AB sin ) ucC
BA
ucC = AB sin
Laws of Operation
A x B = -B x A
Multiplication by a scalar
a(A x B) = (aA) x B = A x (aB) = (A x B)a
Distributive law
A x (B + D) = (A x B) + (A x D)
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The cross product of two general vectors A andB expressed in Cartesian vector form
The above equation may be expressed indeterminant form
( ) ( )
( ) ( ) ( )kBABAjBABAiBABA
kBjBiBkAjAiA
xyyxxzzxyzzy
zyxzyx
+=
++++=
z
z
y
y
x
x
B
A
k
B
A
j
B
A
i
=
Moment of a Force
The moment of a force F about point or, orabout the moment axis passing through O andperpendicular to the plane containing O and F,can be expressed using the vector crossproduct,
FrMO =
F
MO
OA r
d
Moment axis
The magnitude of the above cross product
( ) FdrFrFMO
=== sinsin zyx
kyxO
FFF
rrr
ijk
rxFM =
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Equilibrium in Three Dimensions
Support reactionsCable
Smooth surface
Roller
Ball and socket
Single smooth pin
Single hinge
Fixed supportFxFz
Fy
Mx
My
Mz
Fx
Fy
Fz
Conditions of Equilibrium
0
0
0
0
0
=
=
=
=++=
=++=
z
y
x
zyxO
zyx
F
F
F
kMjMiMM
kFjFiFF
0
0
0
=
=
=
z
y
x
M
M
M
Problem
Determine thetension incables BC andBD and thereactions atthe ball-andsocket joint Afor the mast
shown.z
y
x
F = 1KN
6m
6m
6m 3m
Solution program (Vector Analysis)
Draw the FBD
Represent each force on the FBD in vector form
Apply the force equation of equilibrium
Take moment about point A to eliminate theother unknowns
Evaluate the cross product and combine liketerms and equate to zero to solve for the
unknows.
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{ }
( ) ( )
( ) ( ) ( )
++
++=
=
+
+=
=
++=
=
222
22
zx
zyx
magnitude
ctorpositionve
z
zy
magnitude
ctorpositionve
BDyBDBD
BDBDBD
BD
BD
BD
BDBD
BCBCy
BCBC
BC
BC
BC
BCBC
zyx
rrr
krjrirT
r
rTT
rr
krjrT
r
rTT
kAjAiAA
NiF
( ) ( )
03
1
2
2
0
03
2
2
2
0
03
21
0
03
1
3
2
3
2
2
2
2
2
0
0
=
=
=
=
=+
=
=
+
+
++++
=+++
=
BDBCz
BDBCy
BDx
BDBDBDBCBCzyx
BDBC
TTA
Z
TTA
Y
TA
X
kTjTiTkTjTkAjAiAi
TTAF
F
( )
( ) ( )
( )
046
0
0223
0
3
1
3
2
3
2
2
2
2
26
0
0
=
=
=
=
+
+
+
=++
=
BD
z
BDBC
x
BDBDBDBCBC
BDBCAB
A
T
M
TT
M
kTjTiTkTjTij
TTFr
M
0
5.1
0
707.02
2
5.1
:
=
=
=
==
=
z
y
x
BC
BD
A
KNA
A
KNT
KNT
Answer
Problem
Rod AB issubjected tothe 200N force.Determine thereactions at theball-and-socketjoint A and thetension incables BD and
BE.
A
E
B
C
D
1.5m
1.5m
2m
2m
200N
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Solution program (Vector Analysis)
Draw the FBD
Represent each force on the FBD in vector form
Apply the force equation of equilibrium
Take moment about point A to eliminate theother unknowns
Expand the equation and equate to zero to solvefor the unknowns
{ }
iTT
kTT
kAjAiAA
NiF
BDBD
BEBE
zyx
=
=
+=
= 200
( ) ( ) ( ) ( )
0
0
200
0200
0
0
0
0200
0
0
=+
=
=
=+
=
=+
=
=++++
=+++
=
BEz
y
y
BDx
BDBEzyx
BDBE
TA
Z
NA
A
Y
TA
X
iTkTkAjAiAj
TTAF
F( )
( ) ( ) ( ) ( )
0200
0
022
0
0200
0
0222005.0
0
0
=+
=
=+
=
=
=
=++++
=++
=
BD
z
BDBE
y
BE
x
BDBE
BDBEABAC
A
T
M
TT
M
T
M
iTkTkjijkji
TTrFr
M
NA
NATNAT
Answers
y
xBD
zBE
200
200200
:
=
==
==
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Problem Set 2Problem Set 3
Instructions for Problem Set
Use short-size bondpaper as answer
sheet
Show complete solution
Deadline of Submission: Monday, August
15, 2011
Problem
The uniformconcrete slabhas a weight of25KN. Determinethe tension ineach of the threeparallelsupportingcables when theslab is held in the
horizontal plane.
8m
y
zx
TC
A
B
C
TA
TB
Problem
A vertical load P =3600N applied tothe tripod causes acompressive forceof 1200N in leg ABand a compressiveforce of 1300N inleg AC. Determinethe force in leg ADand the coordinatesxDand zDof its
lower end D.
3m
2.4m
2.4m
1.8m
xD
zDO
C
A
B
D
P
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