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Three-Dimensional Force Systems
Many problems in real-life
involve 3-Dimensional Space.
How will you represent each of
the cable forces in Cartesian
vector form?
Applications
Given the forces in the cables, how will you determine the
resultant force acting at D, the top of the tower?
3D Rectangular Components
Important:
Use a righthanded
coordinate
system
y
Fy
Fx
z
j
i
x
k
z
F  Fx  Fy  Fz
F
Fz
y
 Fx i  Fy j  Fz k
x

F  F  F F  F
2
x
2
y
2
z
Direction angles &
direction cosines
y
F
b
a, b and g are called the
“coordinate direction angles”
g
a
x
z
definition:
a = angle between x-axis and F (a.k.a. = x ).
b
g
= angle between y-axis and F (a.k.a. = y ).
= angle between z-axis and F (a.k.a. = z ).
more ...
Determination of direction angles and direction cosines
Various methods are available, including:
1) Geometry and trigonometry.
y
2) If the Cartesian components of
a vector are known, then:
Fy
F
b
cos a = Fx / F
cos b = Fy / F
cos g = Fz / F
g
Fz
z
3) Use of vector dot product (more on this later ...)
a
Fx
x
Vector expressions using direction cosines
• Direction cosines satisfy: cos2 a + cos2 bcos2 g = 1
If l = cos θx m = cos θy n = cos θz
• The vector n = cos a i + cos b j + cos g k
is a unit vector in the direction of
F.
• The vector F can be written as
F = F (cos a i + cos b j + cos g k)
Example 1
z
A pole is subjected to a
force F which has
components
Fx=1.5 kN and Fz=1.25 kN.
If b = 75°, determine the
magnitudes of F and Fy.
Answer:
F = 2.02 kN
Fy = 0.523 kN
Fz
a
Fx
x
g
F
b
Fy
y
Example 2
Determine the
magnitude and
direction angles
for the resultant
force.
A:
Rx = 348 N
Ry = 75 N
Rz = -97 N
a = 19.4°
b = 78.3°
g = 105°
z
F2 = 250 N
5
3
4
y
30
60
60
x
45
F1 = 350 N
z
F2 = 250 N
5
F2z
3
F2a
F2y
a
x
4
30
y
F2x
F1 = 350 N
Relative Position Vectors: The position of a point B relative
to another point A:
B(x , y , z )
head
B
B
B
A(xA, yA, zA)
tail
r = (xB-xA) i +(yB-yA) j + (zB-zA) k
z
B
(xB -xA )i A
x
r
(zB -zA )k
y
(yB -yA )j
Position vector components = coordinates of head - coordinates of tail.
If the coordinates of
points A and B are
known, the force F can
be written as
Example 3
Express the position vector r in Cartesian vector
form. Then determine the magnitude and coordinate
direction angles.
z
1m
y
1m
2m
x
4m
r
4m
1m
A: r = (-3i - 6j + 2k) m, a = 115.4°, b = 149°, g = 73.4°
Force vectors directed along a line
Many of the structural members we will deal with will
support forces that are directed along the axis of the
member (e.g., cables, ropes, bars, often times
columns, ...). Thus, by knowing the orientation of the
member (position vector), we can conveniently
express its force in vector form.
Procedure:
1) Determine the position vector
r
for the member.
2) Compute a unit vector along the axis of the member
3) The member’s force vector is then
F= Fn.
n = r/r.
Example 4
Determine the x,y and z components of the force vector
shown. Also, determine the direction angles.
y
25
600 N
x
30
z
A: F = (220i + 544j + 127k) N, a = 68.5°, b = 25.0°, g = 77.8°
Example 5: A section of pre-cast concrete wall is
temporarily held by the cables shown. Knowing that the
tensile forces in cables AB and AC are 840 N and 1200
N, respectively, determine the magnitude and direction
of the resultant force applied to the stake at A.
C
D
B
8m
A
11 m
16 m
A: R = (-1440i + 720j - 360k) N
Example 6: Determine the magnitude of the component
of the force F along the Aa axis.
Answer: FAa = 36.0 N
Dot Product - Applications
For this geometry, can you
determine angles between the
pole and the cables?
For force F at Point A, what
component of it (F1) acts along
the pipe OA? What component
(F2) acts perpendicular to the
pipe?
Dot Product - Definition
The dot product of vectors A
and B is defined as
A•B = A B cos θ.
Angle θ is the smallest angle between the two vectors and
is always in a range of 0º to 180º.
Dot Product Characteristics:
1. The result of the dot product is a scalar (a positive or
negative number).
2. The units of the dot product will be the product of the
units of the A and B vectors.
Examples:
i•j=0
i•i=1
A • B = (Ax i + Ay j + Az k) • (Bx i + By j + Bz k)
= Ax Bx + AyBy + AzBz
Example
Given: The force acting on
the pole
Find: The magnitude of the
projection of the force along
the pole OA.
A
rOA = {2 i + 2 j – 1 k} m
rOA = (22 + 22 + 12)1/2 = 3 m
nOA = 2/3 i + 2/3 j – 1/3 k
F = {2 i + 4 j + 10 k}kN
FOA = F • nOA
= (2)(2/3) + (4)(2/3) + (10)(-1/3) = -2/3 kN
Moment in 3-D Vector Formulation
Using the vector cross product, MO = r × F .
Here r is the position vector from point O to any
point on the line of action of F.
Principle of Moments
Consider F = F1 + F2
Moment of F about point O
MO = = r ×F = r ×(F1 + F2 ) = r ×F1 + r ×F2
The moment of a force about a point is equal
to the sum of the moment of the force’s
components about the point.
Cross Product
nC
In general, the cross product of two vectors A and B
results in another vector C , i.e., C = A ×B. The
magnitude and direction of the resulting vector can be
written as
C = A × B = A B sin θ nC
Laws of operation
special commutative law
AxB=-BxA
scalar multiplication
a (A x B ) = (a A) x B = A x (a B)
distributive law
Ax(B+C)=AxB + AxC
Cross product between Cartesian vectors
Let A = Ax i + Ay j + Az k and B = Bx i + By j +Bz k
Then
AxB =
(Ay Bz - Az By ) i (Ax Bz - Az Bx ) j +
(Ax By - Ay Bx ) k
more ...
... or ... computation of cross product by
expansion of determinant

 i

A  B  det Ax
B
 x

A  B  Ay Bzi  Az Bx j  Ax By k - (Ay Bx k  Az Byi  Ax Bz j)
j
Ay
By
k 
i

Az   Ax
Bz 
Bx

j
Ay
k i
Az Ax
j
Ay
By
Bz Bx
By
... or ... expand determinat using co-factors
 i

A  B  det Ax
Bx

. .
= i . Ay
. By
j
Ay
By
.
.
Az - j Ax
Bz
Bx
k 

Az 
Bz 

. .
.
. Az  k Ax
. Bz
Bx
. .
Ay .
By .

i(A
B
B
A
)
j(
A
B
B
A
)

k(
A
B
B
A
)
y
z
y
z
x
z
x
z
x
y
x
y

Cross Product
The right hand rule is a useful tool for determining the direction
of the vector resulting from a cross product.
For example: i × j = k
Note that a vector crossed into itself is zero, e.g., i × i = 0
z
demonstration of
orthogonality
properties:
k
x
Let A = i and B = j. Then

j
y
i
i j
i  j  det 
1 0

0 1
k
0 
  k !
0 

Similarly:
ixi=0
j x i = -k
kxi=j
ixj=k
jxj=0
k x j = -i
i x k = -j
jxk=i
kxk=0
more ...
Demonstration of i x j = k using C = A x B = (A B sin ) uC :
Magnitude of the cross product is
| i x j | = (1)(1) sin 90° = 1.
The direction of the cross product result C is:
z
C
x
j
y
i
Thus, we observe that C = k.
Example
Determine the magnitude of the moment of the 600 N
force about A.
Solution
rAB = (-0.6i – 0.5j + 0.4k) m
MA = rAB x F = (-0.6i – 0.5j + 0.4k) x (600i)
MA = 240 j + 300 k N.m
MA = 384 N.m
Moment of a force about an axis
Our goal is to find the moment of F
(the tendency to rotate the body)
about the axis a’-a.
First compute the moment of F
about any arbitrary point O that
lies on the a’- a axis using the cross product.
MO = r × F
Moment of a force about an axis
Now, find the component of Ma along the axis a’-a using the
dot product.
Ma = na • MO
Ma can also be obtained as
na x
M a  na  r  F  rx
Fx
na y
ry
Fy
na z
rz
Fz
The above equation is also called the triple scalar product.
Moment of a force about an axis
Ma = na • ( r ×F )
na represents the unit vector along the axis a’-a
axis,
r is the position vector from any point on the a’-a
axis to any point A on the line of action of the
force, and
F is the force vector.
Example
Given: A force is applied to
the tool to open a gas valve.
Find: The magnitude of the
moment of this force about
the z axis of the valve.
Solution
n=1k
rAB = {0.25 sin 30° i + 0.25 cos30° j} m
= {0.125 i + 0.2165 j} m
F = {-60 i + 20 j + 15 k} N
Mz = n • ( rAB × F )
0
0
M z  0.125 0.2165
- 60
20
1
0 = 1{0.125(20) – 0.2165(-60)} N·m
= 15.5 N·m
15
Example
The magnitude of the force F is 0.2 N and its direction cosines
are cos θx = 0.727, cos θy = -0.364, and cos θz = 0.582.
Determine the magnitude of the moment of F about the axis
AB of the spool.
Solution
rAB = (0.3i – 0.1j – 0.4k) m
rAB 
0.32  0.12  0.42 
0.26 m
nAB = (0.3i – 0.1j – 0.4k) / (0.26)1/2 m
F = 0.2 (0.72i – 0.364j – 0.582k) m
rAP = (0.26i – 0.025j – 0.11k) m
i

M A  0.26
j
k
- 0.025 - 0.11
0.727 - 0.364 0.582
Solution
MAB = MA . nAB
Example
What is the moment of the force F in figure about
the bar BC?
Solution
Determine a Vector r:
We need a vector from any point on the line BC to
any point on the line of action of the force.
We can let r be the vector from B to the point of
application of F:
rBA = (4 - 0)i + (2 - 0)j + (2 - 3)k
= 4i + 2j - k (m)
nBC
Solution
Determine a Vector n:
To obtain a unit vector along the bar BC, we
determine the vector from B to C:
(0 - 0)i + (4 - 0)j + (0 - 3)k = 4j - 3k (m)
Divide it by its magnitude:
n BC 
4 j - 3k (m)
4 m  - 3 m
2
2
 0.8 j - 0.6k
Solution
Evaluate MBC (in scalar form):
0
n BC  r  F   4
-2
0.8 - 0.6
2
- 1  -24.8 kN - m
6
3
the moment of F about the bar BC in vectorial form:
MBC = [nBC  (r  F)]nBC = -24.8 nBC
Couples
• It is possible to exert a moment on an object without
subjecting it to a net force:
– E.g. when a compact disk begins rotating or a
screw is turned by a screwdriver
– Forces are exerted on these objects in such a way
that the net force is zero while the net moment is
not zero
Couples
• Couple: 2 forces that have equal magnitudes,
opposite directions & different lines of action
– Tends to cause rotation of an object even though
the vector sum of the forces is zero & has the same
remarkable property that the moment it exerts is
the same about any point
Couples
• The moment of a couple is simply the sum of the
moments of the forces about a point P:
M = [r1× F] + [r2 × (-F)] = (r1 - r2) × F
– The vector r1 - r2 is equal to the vector r:
M=r×F
– Since r doesn’t depend on the position of P, the
moment M is the same for any point P
• A couple is often represented in diagrams by showing
the moment:
Couples
• Notice that M = r × F is the moment of F about a
point on the line of action of -F:
– The magnitude of the moment of a force about a
point equals the product of the magnitude of the
force & the perpendicular distance from the point
to the line of action of the force:
|M| = D|F|
where D is the perpendicular distance between the lines of
action of the 2 forces
Couples
– The cross product r × F is perpendicular to r &
F, which means that M is perpendicular to the
plane containing F & -F
– Pointing the thumb of the right hand in the
direction of M, the arc of the fingers indicates
the direction of the moment
Couples
• A plane containing 2 forces is perpendicular to our view:
– The distance between the lines of action of the forces is 4
m, so the magnitude of the moment of the couple is:
|M| = (4 m)(2 kN) = 8 kN-m
Couples
– The moment M is perpendicular to the plane
containing the 2 forces:
• Pointing the arc of the fingers of the right hand
counterclockwise, the right-hand rule indicates that
M points out of the page
• Therefore, the moment of the couple is:
M = 8k (kN-m)
– We can also determine the moment of the
couple by calculating the sum of the moments
of the 2 forces about any point
Couples
• The sum of the moments of the forces about the
origin O is:
M = [r1× (2j)] + [r2 × (-2j)]
= [(7i + 2j) × (2j)] + [(3i + 7j) × (-2j)]
= 8k (kN-m)
Couples
– In a 2-D situation like this example, we represent
the couple by showing its magnitude & a circular
arrow that indicates its direction:
• It is not convenient to represent the couple by showing
the moment vector because the vector is
perpendicular to
the page
Couples
• By grasping a bar & twisting it, a moment can be
exerted about its axis:
– Although the system of forces exerted is
distributed over the surface of the bar in a
complicated way, the effect is the same as if 2
equal & opposite forces are exerted
Couples
– When we represent a couple as in the figure or by
showing the moment vector M, we imply that
some system
of forces exerts
that moment:
– The system of forces is nearly always more
complicated than 2 equal & opposite forces but
the effect is the same:
• Model the actual system as a simple system of 2 forces
Example
The force F in Figure is 10i - 4j (N). Determine the
moment of the couple & represent it as shown in next
figure
Example
Strategy
We can determine the moment in 2 ways:
1st Method:
Calculate the sum of the moments of the forces
about a point
2nd Method:
Sum the moments of the 2 couples formed by the
x & y components of the forces
Example
Solution
1st Method:
If we calculate the sum of the moments about a point on
the line of action of 1 of the forces, the moment of that
force is zero & we only need to calculate the moment of
the other force.
Choosing the point of application of F, the moment is:
M = r × (-F)
= (-2i + 3j) × (-10i + 4j)
= 22k (N-m)
Example
Solution
2nd Method:
The x & y components of the forces from 2 couples.
Determine the moment of the original couple by
summing the moments of the couples formed by the
components:
Example
Solution
Consider the 10-N couple. The magnitude of its moment
is (3 m)(10 N) = 30 N-m & its direction is
counterclockwise, indicating that the moment vector
points out of the page. Therefore, the moment is 30k Nm.
The 4-N couple causes a moment of magnitude (2 m)(4
N) = 8 N-m & its direction is clockwise, so the moment is
-8k N-m.
Example
Solution
The moment of the original couple is:
M = 30k - 8k = 22k (N-m)
Its magnitude is 22 N-m & its direction is
counterclockwise:
Example - Sum of the Moments Due to 2
Couples
Determine the sum of the moments exerted on the pipe
in figure by the 2 couples.
Example
Strategy
Express the moment exerted by each couple as a
vector:
To express the 30-N couple in terms of a vector, express
the forces in terms of their components.
Then sum the moment vectors to determine the sum of
the moments exerted by the couples.
Example
Solution
Consider the 20-N couple.
The magnitude of the moment of the couple is:
(2 m)(20 N) = 40 N-m
The direction of the moment vector is perpendicular
to the y-z plane & the right-hand rule indicates that
it points in the positive x axis direction.
The moment of the 20-N couple is 40i (N-m).
Example
Solution
By resolving the 30-N forces into y & z components,
we obtain the 2 couples:
Example
Solution
The moment of the couple formed by the y
components is -(30 sin 60°)(4)k (N-m) & the
moment of the couple formed by the z components
is (30 cos 60°)(4)j (N-m).
The sum of the moments is therefore:
Σ M = 40i + (30 cos 60°)(4)j - (30 sin 60°)(4)k (N-m)
= 40i + 60j - 104k (N-m)
Example
The two forces acting on the
handles of the pipe wrenches
constitute a coupe M. Express
the couple vector.
Answer: M = -75 i + 22.5 j N.m
Extra Question 1
The tension in cable AB is 2 kN. What is the
magnitude of the moment about shaft CD due to
the force exerted by the cable at A?
Answer: MCD = 1.633 kN.m
Extra Question 2
Determine the distance d between A and B so that the resultant
couple moment has a magnitude of MR = 20 N.m
Extra Question 3
Determine the total moment on
point O.
Resultant
Example: Representing a Force by a
Force and Couple
System 1 in Figure consists of a force
FA = 10i + 4j - 3k (N) acting at A.
Represent it by a force acting at B and a couple.
Example
Strategy
Represent the force FA by a force F acting at B & a
couple M (system 2).
Determine F & M by
using the 2 conditions
for equivalence.
Example
Solution
The sums of the forces must be equal:
(Σ F)2 = (Σ F)1:
F = FA = 10i + 4j - 3k (N)
The sums of the moments about an arbitrary point
must be equal:
The vector from B to A is:
rBA = (4 - 8)i + (4 - 0)j + (2 - 6)k
= -4i + 4j - 4k (m)
Example
Solution
So the moment about B in system 1 is:
i
j k
rBA  FA  - 4 4 - 4  4i - 52 j - 56k (N - m)
10 4 - 3
The sums of the moments about B must be equal:
(MB)2 = (MB)1:
M = 4i - 52j - 56k (N-m)
Example
The building slab has four
columns. F1 and F2 = 0.
Find the equivalent resultant
force and couple moment at
the origin O.
Also find the location (x,y)
of the single equivalent
resultant force.
Solution
FRO = {-50 k – 20 k} = {-70 k} kN
MRO = (10 i) × (-20 k) + (4 i + 3 j)x(-50 k)
= {200 j + 200 j – 150 i} kN·m
= {-150 i + 400 j } kN·m
The location of the single equivalent resultant force is given as,
x = -MRyo/FRzo = -400/(-70) = 5.71 m
y = MRxo/FRzo = (-150)/(-70) = 2.14 m
1/--pages
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