Ch. 2: Force Systems
2.0 Outline
 Overview of Forces
27
27
28
2-D Force Systems
 Rectangular Coordinate Systems
 Force, Moment, and Couple
 Resultants
33
45
61
3-D Force Systems
 Rectangular Coordinate Systems
 Force, Moment, and Couple
 Resultants
85
95
109
2.0 Outline
Ch. 2: Force Systems
28
2.1 Overview of Forces
Force The measure of the attempt to move a body. It is
a fixed vector. For the rigid body problems or only the
external effects of the external force onto the objects are
of interested, force can be treated as a sliding vector.
Hence, the complete description must include
magnitude, direction, and line of action. And the
problem makes use of the principle of transmissibility.
Contact vs. Body Force
Concentrated vs. Distributed Force
2.1 Overview of Forces
Ch. 2: Force Systems
29
2.1 Overview of Forces
P
A
B
P
Contact vs. Body Force
Concentrated vs. Distributed Force
2.1 Overview of Forces
Ch. 2: Force Systems
30
2.1 Overview of Forces
Force Measurement by comparison or by deformation of
an elastic element (force sensor)
Action vs. Reaction Force Isolate the object and the
force exerted on that body is represented  FBD
Combining Force by parallelogram law and principle of
moment
Force Components along the specified coordinate
system to satisfy the parallelogram law (reverse step)
2.1 Overview of Forces
Ch. 2: Force Systems
31
2.1 Overview of Forces
Orthogonal Projection along the specified direction. The
components of a vector are not the same as the
orthogonal projections onto the same coord. system,
except the coordinate system is the orthogonal
(rectangular) coordinate system.
b
b'
Fb
F2
Fb’
R
R
F1
Fa
a
a'
Fa’
2.1 Overview of Forces
Ch. 2: Force Systems
32
2.1 Overview of Forces
Addition of Parallel Forces by graphic or algebraic
(principle of moment) approach
(1)
F1
F
F2
F1
R1
(2)
(3)
-F
R2
F2
R2
R
R1
2.1 Overview of Forces
Ch. 2: Force Systems
33
2.2 2-D Simple Rectangular Coordinate Systems
y
F = Fx + Fy
F = Fx ˆˆ
i + Fy j
Fy
F
F = Fx2 + Fy2
θ
Fx
Fx
x=
θ
Fcos
Fy Fsinθ
=
θ = arctan 2 ( Fy , Fx )
Magnitude is always positive
Scalar component includes sign information too!
2.2 2-D Rectangular Coord. Systems
Ch. 2: Force Systems
34
2.2 2-D Arbitrary Rectangular Coordinate Systems
by convenience, right-handed, geometry of the problem
F
y
β
F
y
β
β
x
y
x
Fx = Fsinβ
Fx = Fsin(π−β)
x
α
Fx = Fcos(α−β)
Fy = Fcosβ
Fy = -Fcos(π−β)
Fy = Fsin(α−β)
F
2.2 2-D Rectangular Coord. Systems
Ch. 2: Force Systems
P. 2/1 If the two equal tension T in the pulley cable
together produce a force of 5 kN on the
pulley bearing, calculate T.
5 kN
35
Ch. 2: Force Systems
36
P. 2/1
5 kN
5 kN
T
60゜
T
By cosine law:
52 = T 2 + T 2 + 2T ⋅ Tcos60°
T = 2.89 kN
Ch. 2: Force Systems
P. 2/2 While steadily pushing the machine up
an incline, a person exerts a 180 N force P
as shown. Determine the components of P
which are parallel and perpendicular to the
incline.
37
Ch. 2: Force Systems
P. 2/2
38
n
180 N
10゜
15゜
=
Pt 180 cos (10=
+ 15 ) 163.1 N
Pn =
−180sin (10 + 15 ) =
−76.1 N
t
15゜
Ch. 2: Force Systems
P. 2/3 Determine the resultant R of the two forces
applied to the bracket. Write R in terms of
unit vectors along the x- and y-axes shown.
39
Ch. 2: Force Systems
P. 2/3
40
clearly and carefully draw the picture!
y’
y
20゜
150 N
200 N
15゜
x’
20゜
x
200 N
10゜
R x = 200cos (15 + 20 ) − 150sin (10 + 20 ) =
88.8 N
=
=
R y 200sin (15 + 20 ) + 150 cos (10 +
20 ) 244.6 N
R x ' = 200 cos15 − 150sin10 = 167.1 N
R y' = 200sin15 + 150 cos10 = 199.5 N
15゜
110゜
R = 88.8i + 244.6
=
j N 167.1i' + 199.5 j' N
Non-Orthogonal Coordinate System x'-y
by law of sine and cosine
200 N --> 174.34i' + 55.1j N
150 N --> -79.8i' + 157.2j N
R = (174.34-79.8 ) i' + ( 55.1+157.2 ) j = 94.54i' + 212.3j N
80゜
150 N
70゜
30゜
Ch. 2: Force Systems
41
P. 2/4 It is desired to remove the spike from the
timber by applying force along its horizontal
axis. An obstruction A prevents direct access,
so that two forces, one 1.6 kN and the other P,
are applied by cables as shown. Compute
the magnitude of P necessary to ensure
axial tension T along the spike. Also find T.
Ch. 2: Force Systems
P. 2/4
y
x
No net force in y-direction


 100  
 150  
R y =Psin  atan 
1.6sin
atan
−


  =0

200
200






∴ P = 2.15 kN


 100  
 150  
T = R x Pcos  atan 
1.6
cos
atan
=
+




 200  
 200  


= 3.20 kN
42
Ch. 2: Force Systems
P. 2/5 As it inserts the small cylindrical part into
a close fitting circular hole, the robot arm
exerts a 90 N force P on the part parallel to
the axis of the hole as shown. Determine
the components of the force which the part
exerts on the robot along axes (a) parallel
and perpendicular to the arm AB, and (b)
parallel and perpendicular to the arm BC.
43
Ch. 2: Force Systems
44
P. 2/5
Quasi-Equilibrium
P is the force done by the robot on the part
-P is the force done by the part on the robot
may be part of strength analysis
of the robot arm
t2
t1
n2
−P = -90cos45n1 + 90sin 45t1 =
−63.6n1 + 63.6t1 N
−P = 90cos60n 2 + 90sin 60t 2 =
45n 2 + 77.9t 2 N
60゜
15゜
45゜
n1
90 N
Ch. 2: Force Systems
45
2.3 2-D Moment and Couple
Moment The measure of the attempt to rotate a body. It
is induced by force. The moment vector’s direction is
perpendicular to the plane established by the point and
the line of action of the force. It is a fixed vector. For
the rigid body problems or only the external effects of
the external moment onto the objects are of interested,
moment can be treated as a sliding vector. Hence, the
complete description must include magnitude, direction,
and line of action. And the problem makes use of the
principle of transmissibility.
2.3 2-D Moment and Couple
Ch. 2: Force Systems
MA
A
d
r
α
F
46
MA = r × F
MA =
F ⋅ r ( sinα ) =
Fd
Moment M A of F about point A
For 2-D (in-plane rotation) problem, moment vector always
points perpendicular to the plane. So it can be indicated by
the magnitude and the sense of rotation (CCW +, CW -)
about the point.
*** Sign’s consistency throughout the problem ***
2.3 2-D Moment and Couple
Ch. 2: Force Systems
47
Varignon’s theorem The moment of a force about any
point is equal to the sum of the moments of the
components of the force about the same point
(MA ) F
A
(MA ) P
(MA ) Q
r
F
( M A )F =
r×F
= r × (P + Q)
P
Q
= ( M A ) P + ( M A )Q
if F = P + Q
Usage Calculate the moment of the force from
its components. Moments of some components
may be trivial to calculate.
2.3 2-D Moment and Couple
Ch. 2: Force Systems
48
Couple The measure of the attempt to purely rotate a
body. It is produced by two equal, opposite, and noncollinear force. The couple vector’s direction is
perpendicular to the plane established by those two
lines of action of the force. It is a free vector and so no
moment center. Only the magnitude and direction are
enough to describe the couple.
arbitrary chosen
M
rB
d
rA
r
α
-F
F
M = rA × F + rB × ( -F )
= r×F
M = F ⋅ r ( sinα ) =
Fd
2.3 2-D Moment and Couple
M
d
Ch. 2: Force Systems
-F
F
M
M
-F
- d
F
F F
M
d
-F
49
-2F
2F
d/2
For rigid body, several pairs of equal & opposite forces
can determine the same couple. It is unique to
calculate the couple from the given pair of forces but
it is non-unique to determine the pair of forces which
produce that value of couple.
Usage Effect of the couple can be determined from
the equivalent pair of forces. Effect from some specific
pairs of forces may be trivial to calculate.
2.3 2-D Moment and Couple
Ch. 2: Force Systems
P. 2/6 Calculate the moment of the 250 N force on
the handle of the monkey wrench about the
center of the bolt.
50
Ch. 2: Force Systems
P. 2/6
Varignon’s Theorem
MO =
−250 cos15 × 0.2 + 250sin15 × 0.03
y
= 46.4 Nm CW
x
51
Ch. 2: Force Systems
P. 2/7
Vector approach
r = 0.03i + 0.35 j m
F = 240cos10i - 240sin10 j N
M O = r × F = -84.0k Nm
y
x
52
Ch. 2: Force Systems
53
P. 2/8 The force exerted by the plunger of cylinder
AB on the door is 40 N directed along the line
AB, and this force tends to keep the door
closed. Compute the moment of this force
about the hinge O. What force Fc normal to
the plane of the door must the door stop at C
exert on the door so that the combined moment
about O of the two forces is zero?
Ch. 2: Force Systems
P. 2/8
MO
40 N
FC
θ = atan (100/400 ) = 0.245 rad
MO =
−40 cos θ × 0.075 − 40sin θ × 0.425 =
7.03 Nm CW
=
FC M
=
8.53 N
O / 0.825
54
Ch. 2: Force Systems
P. 2/9 While inserting a cylindrical part into the
circular hole, the robot exerts the 90 N force
on the part as shown. Determine the moment
about points A, B, and C of the force which
the part exerts on the robot.
55
Ch. 2: Force Systems
P. 2/9
-P
MC
rAC = ( 0.55cos60 + 0.45cos45 ) i + ( 0.55sin60 - 0.45sin45 ) j = 0.593i + 0.158 j m
F = -P = -90sin15i + 90cos15 j = -23.29i + 86.93 j N
90 × 0.15 =
13.5 Nm CCW
MC =
M -P at C about=
rAC × F = 55.23 Nm CCW
A
68.7 Nm CCW
MA =
M C + M -P at C about A =
56
Ch. 2: Force Systems
P. 2/10 As part of a test, the two aircraft engines are
revved up and the propeller pitches are
adjusted so as to result in the fore and aft
thrusts shown. What force F must be exerted
by the ground on each of the main braked
wheels at A and B to counteract the turning
effect of the two propeller thrusts? Necglect
any effects of the nose wheel C, which is
turned 90°and unbraked.
57
Ch. 2: Force Systems
P. 2/10
no rotation  resultant couple = 0
F
MC = 2 × 5 − F × 3 = 0
F = 3.33 kN
58
Ch. 2: Force Systems
P. 2/11 A lug wrench is used to tighten a square-head
bolt. If 250 N forces are applied to the wrench
as shown, determine the magnitude F of the
equal forces exerted on the four contact points
on the 25 mm bolt head so that their external
effect on the bolt is equivalent to that of the
two 250 N forces. Assume that the forces are
perpendicular to the flats of the bolt head.
59
Ch. 2: Force Systems
P. 2/11
F
Equivalent couple system at bolt head
250 × 0.7 = 2 ( F × 0.025 )
F = 3500 N
60
Ch. 2: Force Systems
61
2.4 2-D Resultants
Force – push / pull body in the direction of force
–- rotate the body about any axis except the
intersection line to the line of force
Dual effects : force and couple to separate the push / pull
and rotate effect while maintaining the
resultant force and moment (external effect)
B
A
F
-F
A
B
B
F
d
F
Force-Couple System
A
F
M=Fd
2.4 2-D Resultants
Ch. 2: Force Systems
62
Resultant is the simplest force combination which can
replace the original system of forces, moments, and
couples without altering the external effect of the system
on the rigid body. The resultant force determination will
be used in the Newton’s 2nd law :
∑ F = ma
2.4 2-D Resultants
Ch. 2: Force Systems
63
Resultant Determination
• Force Polygon : head to tail of force vectors
Note: only magnitude and direction are ensured
i.e., line of action may be incorrect!
For the specified rectangular coordinate system,
R = F1 + F2 + F3 =
∑F
=
Rx
Fx R y
∑=
∑ Fy
θ = arctan 2 ( R y , R x )
R=
( ∑ Fx ) + ( ∑ Fy )
2
2
2.4 2-D Resultants
Ch. 2: Force Systems
64
Resultant Determination
• Prin. Transmissibility & Parallelogram Law :
graphical method  quick and easy visualizable but
low accuracy
Note: magnitude, direction, and line of action are correct
2.4 2-D Resultants
Ch. 2: Force Systems
65
Resultant Determination
• Force-Couple Equivalent Method:
algebraic method  high accuracy
Note: magnitude, direction, and line of action are correct
1. Specify a convenient reference point
2. Move all forces so the new lines of action
pass through point O  Force-Couple Equivalence
2.4 2-D Resultants
Ch. 2: Force Systems
66
Resultant Determination
• Force-Couple Equivalent Method:
algebraic method  high accuracy
Note: magnitude, direction, and line of action are correct
3. Sum forces and couples to R and M O
4. Locate the correct line of action of R
 Force-Couple Equivalence
R = ∑F
=
MO
M ∑ (=
Fd )
∑=
i
i
i
i
i
Rd
Principle of Moment
2.4 2-D Resultants
Ch. 2: Force Systems
P. 2/12 In the design of the lifting hook the action of
the applied force F at the critical section of
the hook is a direct pull at B and a couple.
If the magnitude of the couple is 4000 Nm,
determine the magnitude of F.
67
Ch. 2: Force Systems
P. 2/12
Equivalent force-couple system
at the critical section
F × 0.1 = 4000 ∴ F=40 kN
68
Ch. 2: Force Systems
P. 2/13 Calculate the moment of the 1200 N force
about pin A of the bracket. Begin by replacing
the 1200 N force by a force couple system
at point C. Calculate the moment of the
1200 N force about the pin at B.
69
Ch. 2: Force Systems
P. 2/13
Force-Couple equivalent system
makes the moment calculation intuitive
M C = 1200 × 0.2 = 240 Nm CCW
1
× 0.6 = 562 Nm CCW
5
2
× 0.5 = 1099 Nm CCW
M B = M A + 1200 ×
5
M A = M C + 1200 ×
70
Ch. 2: Force Systems
P. 2/14 The combined drive wheels of a front-wheeldrive automobile are acted on by a 7000 N
normal reaction force and a friction force F,
both of which are exerted by the road surface.
If it is known that the resultant of these two
forces makes a 15°angle with the vertical,
determine the equivalent force-couple system
at the car mass center G. Treat this as a
2D problem.
71
Ch. 2: Force Systems
P. 2/14
R
Rcos15 = 7000 ∴ R = 7246.9 N
M=
7000 × 1 + 7246.9sin15 × 0.5
= 7937.8 Nm CW
G
72
Ch. 2: Force Systems
73
P. 2/15 Determine and locate the resultant R of the
two forces and one couple acting on the I-beam.
Ch. 2: Force Systems
typical step in strength analysis
P. 2/15
O
First, find the equivalent force-couple at point O
R = 8 - 5 = 3 kN downward
M O = 25 − 5 × 2 − 8 × 2 = 1 kNm CW
Then, locate the correct line of action by prin. of moment
1
3d = 1 ∴ d = 1/3 m & x = 4 m
3
74
Ch. 2: Force Systems
P. 2/16 If the resultant of the two forces and couple M
passes through point O, determine M.
75
Ch. 2: Force Systems
P. 2/16
Resultant passes through point O
means there is no moment at point O
MO =
M - 400 × 0.15cos30 - 320 × 0.3 = 0
M = 148 Nm CCW
76
Ch. 2: Force Systems
77
P. 2/17 The directions of the two thrust vectors of an
experimental aircraft can be independently
changed from the conventional forward direction
within limits. For the thrust configuration shown,
determine the equivalent force-couple system
at point O. Then replace this force-couple
system by a single force and specify the point
on the x-axis through which the line of action
of this resultant passes.
Ch. 2: Force Systems
P. 2/17
R = ( T + Tcos15 ) i + ( Tsin15 ) j = 1.966Ti + 0.259Tj N
M=
Tcos15 × 3 - T × 3 - Tsin15 ×10 = 2.69T Nm CW
O
MO
R
MO
Rx
Ry
Since Rx of the new force system
does not contribute moment about O,
only Ry can be used in calculation.
0.259T × x = -2.69T ∴ x = -10.4 m
78
Ch. 2: Force Systems
P. 2/18 Two integral pulleys are subjected to the belt
tensions shown. If the resultant R of these
forces passes through the center O, determine
T and the magnitude of R and the CCW angle
θit makes with the x-axis.
79
Ch. 2: Force Systems
P. 2/18
Resultant force passes O  MO = 0
(160 − T ) ×100 + (150 − 200 ) × 200 = 0 ∴ T = 60 N
R = ( 200 +150 -160cos30 - 60cos30 ) i + (160sin30 + 60sin30 ) j
= 159.5i + 110 j N
R = 193.7 N θ = 34.6°
80
Ch. 2: Force Systems
P. 2/19
A rear-wheel-drive car is stuck in the snow between other
parked cars as shown. In an attempt to free the car, three
students exert forces on the car at points A, B, and C
while the driver’s actions result in a forward thrust of 200 N
acting parallel to the plane of rotation of each rear wheel.
Treating the problem as 2D, determine the equivalent
force-couple system at the car center of mass G and locate
the position x of the point on the car centerline through
which the resultant passes. Neglect all forces not shown.
81
Ch. 2: Force Systems
P. 2/19
• 400 N and y-direction of 250 N
cause no moment about O.
• Moments by thrust force 200 N
cancel each other.
R = ( 200 + 400 + 200 + 250sin30 ) i + ( 250 cos 30 + 350 ) j
= 925i + 566.5 j N
M G = 350 ×1.65 + 250sin 30 × 0.9 = 690 Nm CCW
Since Rx of the new force system
does not contribute moment about G,
only Ry can be used in calculation.
MO
Rx
Ry
566.5 × x = 690 ∴ x = 1.218 m
82
Ch. 2: Force Systems
P. 2/20
An exhaust system for a pickup truck is shown in the figure.
The weights Wh, Wm, and Wt of the headpipe, muffler, and
tailpipe are 10, 100, and 50 N, respectively, and act at the
indicated points. If the exhaust pipe hanger at point A is
adjusted so that its tension FA is 50 N, determine the required
forces in the hangers at points B, C, and D so that the
force-couple system at point O is zero. Why is a zero
force-couple system at O desirable.
83
Ch. 2: Force Systems
84
P. 2/20
E
Force-couple at point O is zero  force-couple at any point is zero too!
At point E,
Wh × ( 0.2 + 1.3 + 0.9 ) + Wm × ( 0.65 + 0.9 ) + Wt × 0.4 − FA × (1.3 + 0.9 ) − FB × 0.9 =
0
FB = 98.9 N
Force components in horizontal and vertical direction = 0
FA + FB + FC cos 30 + FD cos 30 − Wh − Wm − Wt =
0
FD sin 30 − FC sin 30 =
0
F=
F=
6.415 N
C
D
So the pipe is in equilibrium w/ no external reaction force at support O.
Therefore stress at O is zero  no breakage
Ch. 2: Force Systems
85
2.5 3-D Rectangular Coordinate Systems
=
Fxz Fsin
=
θ y Fxy Fsin
=
θ z Fyz Fsinθ x
F = Fx i + Fy j + Fz k
F = Fx2 + Fy2 + Fz2
=
Fx F=
cos θ x Fy F=
cos θ y Fz F cos θ z
directional unit vector n F = cos θ x i + cos θ y j + cos θ z k
F = Fn F
Force Vector description : directional unit vector
and magnitude
2.5 3-D Rectangular Coord. Systems
Ch. 2: Force Systems
86
Direction of Force Vector by Two Points

AB

=
n F 
=
AB
( x2 − x1 ) i + ( y2 − y1 ) j + ( z2 − z1 ) k
2
2
2
( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 )
F = Fn F
2.5 3-D Rectangular Coord. Systems
Ch. 2: Force Systems
87
Direction of Force Vector by Two Angles
n F = ( cos φ cos θ ) i + ( cosφ sin θ ) j + ( sinφ ) k
nF = 1
F = Fn F
2.5 3-D Rectangular Coord. Systems
Ch. 2: Force Systems
88
Orthogonal Projection of the Force Vector may not be
equal to its component. They are equal when the
rectangular coordinate system is used.
Orthogonal Projection of F in the n - direction
magnitude = dot product of F with n
nF
F = Fn F
Fn
n = Fn F n
Fn F=
cos θ
=
F
Fn = Fn n = ( Fn F n ) n
θ
2.5 3-D Rectangular Coord. Systems
Ch. 2: Force Systems
P. 2/21 In opening a door which is equipped with
a heavy duty return mechanism, a person
exerts a force P of magnitude 32 N as shown.
Force P and the normal n to the face of
the door lie in a vertical plane. Express P as
a vector and determine the angles θx θy θz
which the line of action P makes with the
positive x-, y-, and z-axes.
89
Ch. 2: Force Systems
P. 2/21
top view
20゜
Pxy=Pcos30
P = Pcos30cos20i + Pcos30sin20 j + Psin30k
= 26.0i + 9.48 j +16k N
P i
=
= 35.5°
θ x acos
P
P j
=
= 72.8°
θ y acos
P
angle description
P k
=
= 60°
θ z acos
P
90
Ch. 2: Force Systems
P. 2/22
91
The rectangular plate is supported by hinges
along its side BC and by the cable AE. If the
cable tension is 300 N, determine the projection
onto line BC of the force exerted on the plate
by the cable. Note that E is the midpoint of the
horizontal upper edge of the structural support.
Ch. 2: Force Systems
P. 2/22
( −0.4, 0,1.2sin 25)
B = ( 0, 0,1.2sin 25 )
C = ( 0,1.2 cos 25, 0 )
D = ( −0.4,1.2 cos 25, 0 )
E = ( 0, 0.6 cos 25, 0 )
A=

AE
T=T
= 142.1i + 193.2 j − 180.2k
AE

BC
=
n BC = 0.9063 j − 0.4226k
BC
2-point description
=
n BC 251.2 N
TBC T=
Orthogonal projection in a direction: magnitude = dot product
92
Ch. 2: Force Systems
P. 2/23
93
The power line is strung from the power-pole
arm at A to point B on the same horizontal
plane. Because of the sag of the cable in the
vertical plane, the cable makes an angle of 15°
with the horizontal where it attaches to A.
If the cable tension at A is 800 N, write T as
a vector and determine the magnitude of its
projection onto the x-z plane.
Ch. 2: Force Systems
P. 2/23
θ
 1.5 
°
 = 8.53
 10 
T = Tcos15cosθ i + Tcos15sinθ j − Tsin15k
= 764.2i +114.6 j − 207k
θ = atan 
Txz =
Tx2 + Tz2 = 792 N
 T j 
°
or θ y acos
=
=

 81.76
 T 
Txz Tsin
=
=
θ y 792 N
94
Ch. 2: Force Systems
95
2.6 3-D Moment and Couple
Scalar approach in 3-D is more difficult than vector approach
Moment M O of F about point O
MO = r × F
M O ⊥ r and M O ⊥ F
M O normal to the plane and through O
2.6 3-D Moment and Couple
Ch. 2: Force Systems
96
Vector Cross Product
r × F = ( ry Fz − rz Fy ) i + ( rz Fx − rx Fz ) j + ( rx Fy − ry Fx ) k
i
 rx
j
ry
k
rz
Fx
Fy
Fz
for remembrance
Proof check & Visualization
by Prin. of Moment
=
M
ry Fz − rz Fy
x
=
M
rz Fx − rx Fz
y
=
M
rx Fy − ry Fx
z
2.6 3-D Moment and Couple
Ch. 2: Force Systems
97
Moment M λ of F about axis λ through point
1. Find moment M O of F about point O
MO = r × F
2. Orthogonally project M O in the
n -direction along axis λ
=
Mλ
( M O n ) n = ( r × Fn ) n
rx
M λ = Fx
nx
ry
Fy
ny
rz
Fz
nz
* Point O can be any point on axis λ
2.6 3-D Moment and Couple
Ch. 2: Force Systems
98
3-D Couple
Couple as free vector
2.6 3-D Moment and Couple
Ch. 2: Force Systems
99
3-D Equivalent Force-Couple System
2.6 3-D Moment and Couple
Ch. 2: Force Systems
P. 2/24 The helicopter is drawn here with certain
3-D geometry given. During a ground test,
a 400 N aerodynamic force is applied to
the tail rotor at P as shown. Determine the
moment of this force about point O of
the airframe.
100
Ch. 2: Force Systems
P. 2/24
Force P does not cause moment in y-direction
For this simple force P, we can determine
the moment component-wise
MO =
( 400 ×1.2 ) i + ( 400 × 6 ) k = 480i + 2400k N
101
Ch. 2: Force Systems
P. 2/25
In picking up a load from position A, a cable
tension T of magnitude 21 kN is developed.
Calculate the moment that T produces about
the base O of the construction crane.
102
Ch. 2: Force Systems
P. 2/25
Vectorial approach
A = ( 0,18,30 ) B = ( 6,13, 0 )

AB
= 4.06i − 3.39 j − 20.32k kN
T=T
AB

r = OA = 18 j + 30k m
M O = r × T = -264.2i +121.9 j - 73.2k kNm
103
Ch. 2: Force Systems
P. 2/25
Algebraic approach
A = ( 0,18,30 ) B = ( 6,13, 0 )

AB
= 4.06i − 3.39 j − 20.32k kN
T=T
AB
translate force to B, moment at O by Tx
( M O )T
=
−4.06 × 13k
x
moment at O by Ty
( M O )T
=
−3.39 × 6k
y
moment at O by Tz
( M O )T
=
−20.32 × 13i + 20.32 × 6 j
z
∴ MO =
−264.16i + 121.92 j − 73.12k kNm
104
Ch. 2: Force Systems
105
P. 2/26 The special-purpose milling cutter is subjected
to the force of 1200 N and a couple of 240 Nm
as shown. Determine the moment of this
system about point O.
Ch. 2: Force Systems
P. 2/26
MO = moment induced by force + free vector couple
R = 1200cos30 j -1200sin30k = 1039 j - 600k N
r = 0.2i + 0.25k m
M O = r × R + 240cos30 j - 240sin30k = -259.8i + 327.8 j + 87.8k Nm
106
Ch. 2: Force Systems
P. 2/27
107
A 5 N vertical force is applied to the knob
of the window-opener mechanism when the
crank BC is horizontal. Determine the moment
of the force about point A and about line AB.
Ch. 2: Force Systems
P. 2/27
r = 75cos30i + 75 j + 75sin30k mm
MA =
r × ( -5k ) =
−375i + 325 j Nmm
=
n AB cos 30i + sin 30k
M AB =
−281i − 162.4k Nmm
( M A n AB ) n AB =
108
Ch. 2: Force Systems
109
2.7 3-D Resultants
Resultant is the simplest force combination which can
replace the original system of forces, moments, and
couples without altering the external effect of the
system on the rigid body.
Vectorial approach is more suitable in 3-D problems.
1. Define the suitable rectangular coord. System and
specify a convenient point O
2. Move all forces so the new lines of action pass
through point O  force-couple equivalence
3. Sum forces and couples to R and M
4. Locate the correct line of action of R
 force-couple equivalence solving piercing point
(2 unknowns: r × R = M rank-2 degenerated)
2.7 3-D Resultants
Ch. 2: Force Systems
R = ∑ F
go together to determine the resultant 
M = ∑ ( r × F )
110
Principle of Moment
The selected point O specifies the couple M
F = m
xG
∑
Dynamics:
 calculate the resultants

∑ M G = IGθ at C.M.
F=0
∑
Statics:
 calculate the resultants at any point
∑M = 0
2.7 3-D Resultants
Ch. 2: Force Systems
111
Resultants of Special Force Systems
Concurrent Forces No moment about the point
of concurrency R = ∑ F M = ∑ ( r × F ) =
0
Parallel Forces Magnitude of R = magnitude of
algebraic sum of the given forces
R =
∑F
Wrench Resultant
of screwdriver
MO =
∑ M=
r×R
R  M as the resultant
2.7 3-D Resultants
Ch. 2: Force Systems
112
Wrench Resultant – Force-Couple Equivalence
a) Determine the force-couple resultant R and M
at convenient point O
b) Orthogonally project M along and perp. to n R
nR
=
R
M1
=
R
M2
( M n R ) n=
R
M - M1
c) Transform couple M 2 into equiv. pair of R and -R
with -R applied at O to cancel R
d) Resultant R with correct line of action and
M1  R remains  wrench resultant
Wrench resultant is the simplest form to visualize
the effect of general force system on to the object :
translate and rotate about the unique axis – screw axis
2.7 3-D Resultants
Ch. 2: Force Systems
113
axis of the wrench, which is  R , lies in a plane
through O and ⊥ plane defined by R and M
2.7 3-D Resultants
Ch. 2: Force Systems
P. 2/28
The pulley and gear are subjected to the loads
shown. For these forces, determine the
equivalent force-couple system at point O.
114
Ch. 2: Force Systems
P. 2/28
typical problem in shaft analysis
R = ( 800 + 200 -1200sin10 ) i + 1200 cos10 j
= 792i +1182 j N
−800 × 0.55 j - 800 × 0.1k
800 N : M1 =
−200 × 0.55 j + 200 × 0.1k
200 N : M 2 =
1200 N : M 3 =
1200sin10 × 0.22 j + 1200cos10 × 0.075k + 1200cos10 × 0.22i
M O = M1 + M 2 + M 3 = 260i − 504 j + 28.6k Nm
115
Ch. 2: Force Systems
P. 2/29
Two upward loads are exerted on the small 3D
truss. Reduce these two loads to a single
force-couple system at point O. Show that R
is perpendicular to Mo. Then determine the
point in the x-z plane through which the
resultant passes.
116
Ch. 2: Force Systems
P. 2/29
R = 2400 j N
800 × 2.4k + 1600 × 2.4k + 1600 × 0.9i
MO =
= 1440i + 5760k Nm
determine line of action of R
R must be x m far from yz plane to produce 5760k Nm
5760 = 2400 × x ∴ x = 2.4 m
R must be - z m far from xy plane to produce 1440i Nm
1440 = 2400 × ( -z ) ∴ z = -0.6 m
117
Ch. 2: Force Systems
P. 2/30
Replace the two forces acting on the block
by a wrench. Write the moment M associated
with the wrench as a vector and specify the
coordinates of the point P in the x-y plane
through which the line of action of the wrench
passes.
118
Ch. 2: Force Systems
P. 2/30
a) Determine force-couple resultant at O
R = Fi - Fk
M O = F ( a + c ) j - Fbk
b) Project MO || and ┴ nR
nR =
1
1
ik
2
2
Fb Fb
M
=
nR
i− k
( M O n R )=

2
2
Fb
Fb
M⊥ =
MO − M =
− i + F (a + c) j − k
2
2
119
Ch. 2: Force Systems
P. 2/30
c) Transform couple M┴ into pair of force R and –R
If r points to the piercing point of the xy plane,
r = xi + yj
M⊥ = r × R
b
x=a+c, y=
2
d) Wrench consisting of R and M  acts
b
through xy plane at x = a+c, y =
2
120
Ch. 2: Force Systems
121
P. 2/31 The resultant of the two forces and couple
may be represented by a wrench. Determine
the vector expression for the moment M of
the wrench and find the coordinates of the point
P in the x-z plane through which the resultant
force of the wrench passes.
Ch. 2: Force Systems
P. 2/31
R = 100i + 100 j N
Let point P in xz plane, where the wrench passes, has the coordinate ( x, 0, z ) .
Moment about point P = 100 × zi + 100 × ( 0.4 − x ) k + 100 × ( 0.4 − z ) j - 100 × 0.3k - 20j
M P = 100zi + ( 20-100z ) j + (10-100x ) k Nm
This moment at point P must equal to the couple of the wrench passing through point P.
And since it is the wrench, M P  R.
∴ x = 0.1 m, z = 0.1 m
M=
10i + 10 j Nm
P
122