# Trends in Project Management

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```					Chapter 3
Rigid Bodies :
Equivalent Systems of
Forces
Slide #: 1
Introduction

In previous lessons, the bodies were assumed to be particles.

Actually the bodies are a combination of a large number of
particles.

body

particle

Slide #: 2
Introduction

A force may cause a body to move in a straight path that is
called translation

Another force may cause the body to move differently, for
example, rotation
The external forces may cause a motion of translation or rotation
or both. The point of application of force determines the effect.
Slide #: 3
Principle of Transmissibility : Equivalent Forces
F

F’

According to the principal of transmissibility, the effect of an
external force on a rigid body remains unchanged if that force
is moved along its line of action. Two forces acting on the rigid
body at two different points have the same effect on that body
if they have the same magnitude, same direction, and same
line of action. Two such forces are said to be equivalent.
Slide #: 4
Principle of Transmissibility : Equivalent Forces

• Moving the point of application of
the force F to the rear bumper
does not affect the motion or the
other forces acting on the truck.

Slide #: 5
Vector Product of Two Vectors
• Concept of the moment of a force about a point is
more easily understood through applications of
the vector product or cross product.

• Vector product of two vectors P and Q is
defined as the vector V which satisfies the
following conditions:

1.Line of action of V is perpendicular to
plane containing P and Q.
2. Magnitude of V is V  PQ sin 

3. Direction of V is obtained from the
right-hand rule.                                   Slide #: 6
Vector Product of Two Vectors

• Vector products:
- are not commutative, Q  P   P  Q 
- are distributive,    P  Q1  Q2   P  Q1  P  Q2
- are not associative,  P  Q   S  P  Q  S 

Slide #: 7
Vector Product of Unit Vectors

It follows from the definition of the vector product of two
vectors that the vector products of unit vectors i, j, and k are:

ixi=jxj=kxk=0

                     
i i  0     j  i  k k  i  j
                           
i j k      j j 0    k  j  i
                   
i k   j   j k  i   k k  0

Slide #: 8
Vector Products in Terms of Rectangular Components

The rectangular components of the vector product V of two
vectors P and Q are determined as follows: Given

P = Px i + Py j + Pz k            Q = Qx i + Qy j + Qz k

V=PxQ
V = (Py Qz - Pz Qy ) i + (Pz Qx - Px Qz ) j + (Px Qy - Py Qx ) k

V = Vx i + Vy j + Vz k

Slide #: 9
Vector Products in Terms of Rectangular Components

V = P x Q = Vx i + Vy j + Vz k
where
Vx = Py Qz - Pz Qy
Vy = Pz Qx - Px Qz
Vz = Px Qy - Py Qx

i        j  k
Expansion of
V = P x Q = Px        Py Pz      Determinant
Qx        Qy Qz
Slide #: 10
Expansion of Determinant

P = Px i + Py j + Pz k
y                  Q = Qx i + Qy j + Qz k

i   j     k
V=PxQ=

Slide #: 11
Expansion of Determinant - Example

P=4i+3j+2k                     1
Q=6 i+1j+5k

i   j   k
V=PxQ=

Slide #: 12
Calculation of Determinant : Method -1

i j      k
V=PxQ=           4 3      2
6 1      5

Slide #: 13
Calculation of Determinant : Method -1

4 5 6

i   j    k     i    j
V=PxQ=           4   3    2     4    3
6   1    5     6    1

1 2 3
V=PxQ= 1+ 2 + 3 - 4 - 5 - 6
Slide #: 14
Calculation of Determinant : Method -1

18 k       2i   20 j

i   j    k          i        j
V=PxQ=           4   3    2          4        3
6   1    5          6        1

15 i   12 j         4k

V = P x Q = 13 i - 8 j -14 k
Slide #: 15
Calculation of Determinant : Method -2

a11 a12 a13
11
D=     a21 a22 a23
21
a31 a32 a33                       = D11
31

D11 : Minor of a11 in D
a22 a23
D11 =            = a22* a33 – a32* a23
a32 a33

Slide #: 16
Calculation of Determinant : Method -2

a11 a12 a13
11
D=     a21 a22 a23
21
a31 a32 a33                       = D12
31

D12 : Minor of a12 in D
a21 a23
D12 =    a31 a33 = a21* a33 – a31* a23

Slide #: 17
Calculation of Determinant : Method -2

a11 a12 a13
11
D=     a21 a22 a23
21
a31 a32 a33                         = D13
31

D13 : Minor of a13 in D

a21 a22
D13 = a a           = a21* a32 – a31* a22
31  32

Slide #: 18
Calculation of Determinant : Method -2

a11 a12 a13
11
D=      a21 a22 a23
21
a31 a32 a33
31

D = a11 D11 - a12 D12 + a13 D13

D= a11 (a22* a33 – a32* a23) - a12 (a21* a33 – a31* a23)
+ a13 (a21* a32 – a31* a22)
Slide #: 19
Calculation of Determinant : Method -2

i    j   k
V=PxQ=          4    3   2
6    1   5

V=PxQ= i            3 2 - j 4 2 +k 4 3
1 5     6 5    6 1

V = P x Q = i (15-2) – j (20-12) + k (4-18)

V = P x Q = 13 i – 8 j – 14 k
Slide #: 20
Moment of a Force About a Point

• A force vector is defined by its magnitude and
direction. Its effect on the rigid body also depends
on it point of application.

The moment of force F about point O
is defined as the vector product

MO = r x F
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 .
Slide #: 21
Moment of a Force About a Point

The magnitude of the moment of F
about O can be expressed as

MO = rF sin  = Fd
where d is the perpendicular distance
from O to the line of action of F.

• Magnitude of MO measures the tendency
of the force to cause rotation of the body

The sense of the moment may be
determined by the right-hand rule.            Slide #: 22
Moment of a Force About a Point
• Two-dimensional structures have length and breadth but
negligible depth and are subjected to forces contained in
the plane of the structure.

• The plane of the structure contains the point O and the
force F. MO, the moment of the force about O is
perpendicular to the plane.

• If the force tends to rotate the structure counter
clockwise, the sense of the moment vector is out of the
plane of the structure and the magnitude of the moment
is positive.
• If the force tends to rotate the structure clockwise, the
sense of the moment vector is into the plane of the
structure and the magnitude of the moment is negative.
Slide #: 23
Varignon’s Theorem

• The moment about a give point O of the
resultant of several concurrent forces is equal
to the sum of the moments of the various
moments about the same point O.
                    
r  F1  F2    r  F1  r  F2  

• Varigon’s Theorem makes it possible to
replace the direct determination of the
moment of a force F by the moments of two
or more component forces of F.

Slide #: 24
Rectangular Components of the Moment of a Force

y                                             A (x , y, z )
Fy j
yj

Fx i
r
Fz k
O                            xi
x
zk
z                      The moment of F about O,
                       
M O  r  F , r  xi  yj  zk
                  
F  Fx i  Fy j  Fz k
Slide #: 25
Rectangular Components of the Moment of a Force

The rectangular components of
the moment Mo of a force F are
determined by expanding the
determinant of r x F.

i j k
Mo = r x F =    x y z = Mx i + My j + Mzk
Fx Fy Fz
where
Mx = y Fz - z Fy   My = zFx - x Fz
Mz = x Fy - y Fx                       Slide #: 26
Rectangular Components of the Moment of a Force

The moment of F about B,
             
M B  rA / B  F
         
rA / B  rA  rB
                                 
 x A  xB i   y A  y B  j  z A  z B  k
                   
F  Fx i  Fy j  Fz k
                             
i                j             k

M B  x A  xB       y A  yB    z A  z B 
Fx               Fy            Fz

Slide #: 27
Moment of a Force About a Point
Sample Problem 3.4

SOLUTION:
The moment MA of the force F exerted
by the wire is obtained by evaluating
the vector product,
          
M A  rC A  F

The rectangular plate is supported by
the brackets at A and B and by a wire
CD. Knowing that the tension in the
wire is 200 N, determine the moment
about A of the force exerted by the
wire at C.
Slide #: 28
Moment of a Force About a Point
Sample Problem 3.4
SOLUTION:
         
M A  rC A  F
                                      
rC A  AC  rC  rA  0.3 m i  0.08 m k

     
F  F  200 N 
CD
CD
                         
 0.3 m i  0.24 m  j  0.32 m k
 200 N 
0.5 m
                       
 120 N  i  96 N  j  128 N k
        
i    j   k

M A  0.3   0 0.08
 120 96  128

Slide #: 29
Moment of a Force About a Point
Sample Problem 3.4
SOLUTION:

        
i    j   k

M A  0.3   0 0.08
 120 96  128

0     0.08      0.3 0.08      0.3 0
MA =     i 96         - j           + k -120 96
-128     -120 -128

MA = (0- (96)(0.08)) i – ((0.3)(-128) – (-120)(0.08)) j + (96)(0.3) k
MA = - 7.68 i + 28.8 j + 28.8 k

Slide #: 30
Moment of a Force About a Point
Sample Problem 3.4
SOLUTION:

                                                  
M A  7.68 N  m i  28.8 N  m j  28.8 N  mk

Slide #: 31
Rectangular Components of the Moment of a Force
For two-dimensional structures,                 
M O  rO A  F

MO  F d

        
i     j   k

MO  x     y   0
Fx   Fy   0

                 
M O  xFy  yFx k
MO  MZ
 xFy  yFx
Slide #: 32
Rectangular Components of the Moment of a Force
For two-dimensional structures,                     
M B  rB    A F

MB  F d

                       
i             j          k

M B  x A  xB      y A  yB       0
Fx            Fy          0

                                      

M B  x A  x B Fy   y A  y B Fx k 
MB  MZ
 x A  x B Fy   y A  y B Fx

Slide #: 33
Moment of a Force About a Point
Sample Problem 3.1

A 100-N vertical force is applied to the end of a
lever which is attached to a shaft at O.
Determine:
b) horizontal force at A which creates the same
moment,
c) smallest force at A which produces the same
moment,
d) location for a 240-N vertical force to produce
the same moment,
e) whether any of the forces from b, c, and d is
equivalent to the original force.

Slide #: 34
Moment of a Force About a Point
Sample Problem 3.1

a) Moment about O is equal to the product of the
force and the perpendicular distance between the
line of action of the force and O. Since the force
tends to rotate the lever clockwise, the moment
vector is into the plane of the paper.

M O  Fd
d  24 in. cos 60  12 in.
M O  100 lb 12 in.              M O  1200 lb  in

Slide #: 35
Moment of a Force About a Point
Sample Problem 3.1
b) Horizontal force at A that produces the same moment,

d  24 m sin 60  20.8 m
M O  Fd
1200 N m  F 20.8 m.
1200 N m
F
20.8 m                   F  57.7 N

Slide #: 36
Moment of a Force About a Point
Sample Problem 3.1

c) The smallest force A to produce the same moment
occurs when the perpendicular distance is a
maximum or when F is perpendicular to OA.

M O  Fd
1200 N m  F 24 m 
1200 Nm.
F
24 m
F  50 N

Slide #: 37
Moment of a Force About a Point
Sample Problem 3.1

d) To determine the point of application of vertical
240 N force to produce the same moment,

M O  Fd
1200 Nm  240 N d
1200 Nm.
d             5m
240 N
OB cos60  5 m                        OB  10 m

Slide #: 38
Moment of a Force About a Point
Sample Problem 3.1
e) Although each of the forces in parts b), c), and d)
produces the same moment as the 100 lb force, none
are of the same magnitude and sense, or on the same
line of action. None of the forces is equivalent to the
100 lb force.

Slide #: 39
Scalar Product of Two Vectors
• The scalar product or dot product between
two vectors P and Q is defined as
 
P  Q  PQ cos scalar result

• Scalar products:
   
- are commutative,       PQ  Q P
                
- are distributive,      P  Q1  Q2   P  Q1  P  Q2
  
- are not associative,   P  Q S  undefined
• Scalar products with Cartesian unit components,
                                              
P  Q  Px i  Py j  Pz k  Qx i  Q y j  Qz k 
                            

                                              
i i 1 j  j 1 k k 1 i  j  0            j k  0 k i  0
 
P  Q  Px Qx  Py Q y  Pz Qz
 
P  P  Px2  Py2  Pz2  P 2                                     Slide #: 40
Scalar Product of Two Vectors : Applications
• Angle between two vectors:
 
P  Q  PQ cos  Px Qx  Py Q y  Pz Qz
Px Qx  Py Q y  Pz Qz
cos 
PQ
• Projection of a vector on a given axis:
POL  P cos  projection of P along OL
 
P  Q  PQ cos
 
PQ
 P cos  POL
Q

• For an axis defined by a unit vector:
 
POL  P  
 Px cos x  Py cos y  Pz cos z

Slide #: 41
Mixed Triple Product of Three Vectors

• Mixed triple product of three vectors,
  
S  P  Q  scalar result

• The six mixed triple products formed from S, P, and
Q have equal magnitudes but not the same sign,
                          
S  P  Q   P  Q  S   Q  S  P 
                              
  S  Q  P    P  S  Q   Q  P  S 

Slide #: 42
Mixed Triple Product of Three Vectors

• Evaluating the mixed triple product,
  
S  P  Q   S x Py Qz  Pz Q y   S y  Pz Qx  Px Qz 

 S z Px Q y  Py Qx   
Sx     Sy     Sz
 Px     Py     Pz
Qx     Qy     Qz

Slide #: 43
Moment of a Force About a Given Axis
• Moment MO of a force F applied at the point A
      
MO  r  F

• Scalar moment MOL about an axis OL is the
projection of the moment vector MO onto the
axis,
         
M OL    M O    r  F 

x   y    z
M OL  x       y     z
where
Fx    Fy   Fz      Fx , Fy , Fz  componentsof forceF
x, y , z  coordinates of point of application of F
x ,  y , z  componentsof force ( direction cosines of axis OL)

Slide #: 44
Moment of a Force About a Given Axis

• Moment MOL of the force F about OL
measures the tendency of F to make the
rigid body rotates about the fixed axis
OL

• Moments of F about the coordinate axes,
M x  yFz  zFy
M y  zFx  xFz
M z  xFy  yFx

Slide #: 45
Moment of a Force About a Given Axis

If the axis does not pass through the origin

• Moment of a force about an arbitrary axis,
 
M BL    M B
         
   rA B  F 
       
rA B  rA  rB

x          y          z
M BL  x A  x B     y A  yB    z A  zB
Fx          Fy          Fz

Slide #: 46
Sample Problem 3.5

A cube is acted on by a force P as
shown. Determine the moment of P
b)   about the edge AB and
c)   about the diagonal AG of the cube.
d)   Determine the perpendicular distance
between AG and FC.

Slide #: 47
Sample Problem 3.5

• Moment of P about A,
            
M A  rF A  P
        
                   
rF A  a i  a j  a i  j
                   1            1
P  P λ FC  P                 k
j-
       2      2 

      
                    1      1 
MA  a i  j  P           j-    k
       2      2 

MA          
 aP    
i  j k     
 2
• Moment of P about AB,
 
M AB   i MA

 i      
  aP    
i  j k          
 2                                aP
M AB 
2
Slide #: 48
Sample Problem 3.5

     
M AG    M A
               
 rA G ai  aj  ak

rA G
                
1   
i  j k   
a 3       3

MA       
aP   

i  j k
2
M AG             
1    aP   
i  j k    i  j k 
3               2

aP
1  1  1
6
aP
M AG  
6

Slide #: 49
Sample Problem 3.5

Alternative method using determinant

x         y          z
M AG  x F  x A   yF  y A    zF  z A
Fx         Fy          Fz

1/ 3  1/ 3  1/ 3
M AG    a     a      0    aP / 6
0  P/ 2  P/ 2

Slide #: 50
Problem 3.54 on page 105
The frame ACD is hinged at A
and D.

It is supported by a cable that
passes through a ring at B and is
attached to hooks at G and H

Knowing that the tension in the
cable is 1125 N, determine the
the force exerted on the frame by
portion BG of the cable

Slide #: 51
Problem 3.54 on page 105

G

D

A
B

Slide #: 52
Problem 3.54 on page 105

G



A                              
B             rB   A

            

M AD  x BA      y BA      z BA
Fx         Fy        Fz
Slide #: 53
Problem 3.54 on page 105
     
F  F BG
                     
        BG  400i  740 j  320k
 BG       
BG   400 2  740 2  320 2

                   
        BG  400i  740 j  320k
 BG       
BG          900


        BG                           
 BG        0.444i  0.822 j  0.3555k
BG

                                         
FF    BG  1125 (0.444i  0.822 j  0.3555k )
                       
M AD  x BA      y BA      z BA        FF    BG  500 i  925 j  400 k
 500     925       400
Slide #: 54
Problem 3.54 on page 105

G



A
B
                    
       AD   800 i  0 j  600 k
AD    800 2  0 2  600 2

                 
0.8     0      0.6             AD 800 i  0 j  600 k
M AD  x BA    y BA    z BA
 500    925     400     

             
 AD        0 .8 i  0 j  0 .6 k
Slide #: 55
Problem 3.54 on page 105

G

D

A                        
B          rB   A

x BA  400 mm  0.4m    y BA  0          z BA  0

0.8    0      0.6
M AD  0.4    0      0        222 N m
 500   925    400
Slide #: 56

```
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