Eleventh Edition
3
CHAPTER
VECTOR MECHANICS FOR ENGINEERS:
STATICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
David F. Mazurek
Rigid Bodies:
Equivalent Systems
of Forces
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Eleventh
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Vector Mechanics for Engineers: Statics
Learning Outcomes
At the end of the chapter the students would be able to:
• Define a moment and moment of a couple (CO1-PO1)
• Use vector algebra, including vector product, scalar product and mixed triple
product (CO1-PO1)
• Apply vector product to solve moment about a point (CO1-PO1)
• Reduce a given system of forces with a simpler equivalent force-couple system
(CO1-PO1)
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Vector Mechanics for Engineers: Statics
Contents
Introduction
Moment of a Force About a Given Axis
External and Internal Forces
Sample Problem 3.5
Principle of Transmissibility: Equivalent Moment of a Couple
Forces
Addition of Couples
Vector Products
Couple Vectors
Moment of a Force About a Point
Varignon’s Theorem
Rectangular Components of the
Moment of a Force
Resolution of a Force Into a Force at O
and a Couple
Sample Problem 3.6
Sample Problem 3.1
Reducing a System of Forces to a ForceCouple System
Scalar Products
Further Reduction of a System of Forces
Applications of the Scalar Product
Sample Problem 3.8
Mixed Triple Products
Sample Problem 3.10
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Vector Mechanics for Engineers: Statics
Introduction
• Treatment of a body as a single particle is not always possible. In
general, the size of the body and the specific points of application of the
forces must be considered.
• Current chapter describes the effect of forces exerted on a rigid body and
how to replace a given system of forces with a simpler equivalent system.
• First, we need to learn some new statics concepts, including:
• moment of a force about a point
• moment of a force about an axis
• moment due to a couple
• Any system of forces acting on a rigid body can be replaced by an
equivalent system consisting of one force acting at a given point and one
couple.
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Vector Mechanics for Engineers: Statics
External and Internal Forces
• Forces acting on rigid bodies are
divided into two groups:
- External forces
- Internal forces
• External forces are shown in a free
body diagram.
• Internal forces, such as the force
between each wheel and the axel
it is mount on, are never shown
on a free body diagram.
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Vector Mechanics for Engineers: Statics
Principle of Transmissibility: Equivalent Forces
• Principle of Transmissibility Conditions of equilibrium or motion are
not affected by transmitting a force
along its line of action.
NOTE: F and F’ are 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.
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Vector Mechanics for Engineers: Statics
Vector Products
• Concept of the moment of a force about a point
requires the understanding 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.
• 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 
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Vector Mechanics for Engineers: Statics
Vector Products: Rectangular Components
• Vector products of Cartesian unit vectors,
   
 
 
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
• Vector products in terms of rectangular
coordinates

 
V  Px i  Py j  Pz k  Qx i  Qy j  Qz k
  Py Qz  Pz Qy  i   Pz Qx  PxQz  j

  Px Qy  Py Qx  k
i
 Px
Qx
j
Py
Qy
k
Pz
Qz
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Vector Mechanics for Engineers: Statics
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 its point of application.
• The moment of F about O is defined as
MO  r  F
• The moment vector MO is perpendicular to the
plane containing O and the force F.
• Magnitude of MO, M O  rF sin   Fd
,
measures the tendency of the force to cause rotation of
the body about an axis along MO. The sense of the
moment may be determined by the right-hand rule.
• Any force F’ that has the same magnitude and
direction as F, is equivalent if it also has the same line
of action and therefore, produces the same moment.
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Vector Mechanics for Engineers: Statics
Moment of a Force About a Point
• Two-dimensional structures have length and breadth but
negligible depth and are subjected to forces contained
only 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 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 counterclockwise,
the sense of the moment vector is into the plane of the
structure and the magnitude of the moment is negative.
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Vector Mechanics for Engineers: Statics
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  
• Varignon’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.
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Vector Mechanics for Engineers: Statics
Rectangular Components of the Moment of a Force
The moment of F about O,
MO  r  F,
r  xi  yj  zk
F  Fx i  Fy j  Fz k
M O  M xi  M y j  M z k
i
 x
Fx
j
y
Fy
k
z
Fz
  yFz  zFy  i   zFx  xFz  j   xFy  yFx  k
The components of M o , Mx, My, and Mz, represent the
moments about the x-, y- and z-axis, respectively.
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Vector Mechanics for Engineers: Statics
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
M B   x A  xB 
Fx
j
 y A  yB 
Fy
k
 z A  zB 
Fz
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Vector Mechanics for Engineers: Statics
Rectangular Components of the Moment of a Force
For two-dimensional structures,
M O   xFy  yFz  k
MO  MZ
 xFy  yFz
M B    x A  xB  Fy   y A  yB  Fz  k
MB  MZ
  x A  xB  Fy   y A  yB  Fz
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Vector Mechanics for Engineers: Statics
Sample Problem 3.1
A 100-lb vertical force is applied to the end of a lever
which is attached to a shaft (not shown) at O.
Determine:
a) the moment about O,
b) the horizontal force at A which creates the same
moment,
c) the smallest force at A which produces the same
moment,
d) the location for a 240-lb vertical force to produce
the same moment,
e) whether any of the forces from b, c, and d is
equivalent to the original force.
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Vector Mechanics for Engineers: Statics
Sample Problem 3.1
STRATEGY: The calculations asked for all involve
variations on the basic defining equation of a
moment, MO = Fd.
MODELING and ANALYSIS:
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 which, by our
sign convention, would be negative or
counterclockwise.
MO  Fd
d  24 in.cos 60 12 in.
MO  100 lb12 in.
MO  1200 lb  in, or
 1200 lb in
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Vector Mechanics for Engineers: Statics
Sample Problem 3.1
b) Horizontal force at A that produces the same
moment,
d  24 in.sin 60  20.8 in.
M O  Fd
1200 lb  in.  F 20.8 in.
F

1200 lb  in.
20.8 in.
F  57.7 lb
Why must the direction of this F be to the right?
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Vector Mechanics for Engineers: Statics
Sample Problem 3.1
F(min) ?
What is the smallest force at A which produces the
same moment? Think about it and discuss with a
neighbor.
c) The smallest force at 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 lb  in.  F 24 in.
1200 lb  in.
F
24 in.
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F  50 lb
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Vector Mechanics for Engineers: Statics
Sample Problem 3.1
d) To determine the point of application of a 240 lb
force to produce the same moment,
M O  Fd
1200 lb  in.  240 lbd
1200 lb  in.
 5 in.
240 lb
OBcos60  5 in.
d
OB  10 in.

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Vector Mechanics for Engineers: Statics
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.
REFLECT and THINK: Various combinations of force and lever arm can
produce equivalent moments, but the system of force and moment produces
a different overall effect in each case.
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Vector Mechanics for Engineers: Statics
Sample Problem 3.4
STRATEGY: The solution requires
resolving the tension in the wire and
the position vector from A to C into
rectangular components. You will
need a unit vector approach to
determine the force components.
MODELING and ANALYSIS:
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.
The moment MA of the force F exerted
by the wire is obtained by evaluating
the vector product,



M A  rC A  F
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Vector Mechanics for Engineers: Statics
Sample Problem 3.4



M A  rC A  F
rC A  rC  rA   0.3 m  i   0.08 m  j
F  F    200 N 
  200 N 
rC D
rC D
  0.3 m  i   0.24 m  j   0.32 m  k
0.5 m
   120 N  i

i

M A  0.3
 120
  96 N  j   128 N  k


j
k
0 0.08
96  128
M A    7.68 N  m  i   28.8 N  m  j   28.8 N  m  k
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Vector Mechanics for Engineers: Statics
Sample Problem 3.4
REFLECT and THINK:
Two-dimensional problems often are solved easily
using a scalar approach, but the versatility of a
vector analysis is quite apparent in a threedimensional problem such as this.
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Vector Mechanics for Engineers: Statics
Scalar Products
• 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,
- are distributive,
- are not associative,
   
PQ  Q P
  
   
P  Q1  Q2   P  Q1  P  Q2
  
P  Q  S  undefined
• Scalar products with Cartesian unit components,

 
P  Q  Px i  Py j  Pz k  Qx i  Qy 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
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Vector Mechanics for Engineers: Statics
Applications of the Scalar Product
• Angle between two vectors:
 
P  Q  PQ cos  Px Qx  Py Q y  Pz Qz
cos 
Px Qx  Py Q y  Pz Qz
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
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Vector Mechanics for Engineers: Statics
Mixed Triple Products
• 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
• Evaluating the mixed triple product,


S  P  Q  S x  Py Qz  Pz Qy   S y  Pz Qx  PxQz 
 S z  PxQy  Py Qx 
Sx
 Px
Qx
Sy
Py
Qy
Sz
Pz
Qz
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Vector Mechanics for Engineers: Statics
Moment of a Force About a Given Axis
• Moment MO of a force F applied at the point A
about a point O,

 
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 
• Moments of F about the coordinate axes,
M x  yFz  zFy
M y  zFx  xFz
M z  xFy  yFx
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Vector Mechanics for Engineers: Statics
Moment of a Force About a Given Axis
• Moment of a force about an arbitrary axis,
M BL    M B

   rA B  F

rA B  rA  rB
• The result is independent of the point B
along the given axis. That is, the same
result can be gotten using rA C , for
example.
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Vector Mechanics for Engineers: Statics
Sample Problem 3.5
A cube is acted on by a force P as
shown. Determine the moment of P
a)
b)
c)
d)
about A
about the edge AB and
about the diagonal AG of the cube.
Determine the perpendicular distance
between AG and FC.
STRATEGY: Use the equations presented in this
section to compute the moments asked for. You can
find the distance between AG and FC from
the expression for the moment MAG.
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Vector Mechanics for Engineers: Statics
Sample Problem 3.5
MODELING and
ANALYSIS:
a) Moment of P about A,
M A  rF A  P
rF A  ai  a j  a  i  j 
PP


2 i  2 j  P 2 i  j 
M A  a  i  j  P 2  i  j 
M A  aP 2

 i  j k 
• List an alternative to the position vector rF A ,
and discuss your answer with a neighbor.
b) Moment of P about AB,
M AB  i  M A

 i  aP 2
 i  j k 
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M AB  aP 2
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Vector Mechanics for Engineers: Statics
Sample Problem 3.5
c) Moment of P about the diagonal AG,
M AG    M A

MA 
M AG 
rG A
rG A
ai  aj  ak
1


i  j k
a 3
3





aP
i  j k
2

1
aP
i  j k 
i  j k
3
2


aP
M AG  
aP

1

1

1


6
6
• What if, for  , you had chosen rA G instead?
How would that change the answer?
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Vector Mechanics for Engineers: Statics
Sample Problem 3.5
d) Perpendicular distance between AG and FC,
P 
P
2
0


j k 


1
P
i  j k 
 0 11
3
6
 since by definition P   =P 1 cos 
Therefore, P is perpendicular to AG.
M AG 
aP
 Pd
6
d
a
6
REFLECT and THINK: In a problem like this,
it is important to visualize the forces and moments
in three dimensions so you can choose the
appropriate equations for finding them and also
recognize the geometric relationships between
them.
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Vector Mechanics for Engineers: Statics
Moment of a Couple
• Two forces F and -F having the same magnitude,
parallel lines of action, and opposite sense are said
to form a couple.
• Moment of the couple,
   

M  rA  F  rB   F 

 
 rA  rB   F
 
 r F
M  rF sin   Fd
• The moment vector of the couple is
independent of the choice of the origin of the
coordinate axes, i.e., it is a free vector that can
be applied at any point with the same effect.
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Vector Mechanics for Engineers: Statics
Moment of a Couple
Two couples will have equal moments if
• F1d1  F2 d 2
• the two couples lie in parallel planes, and
• the two couples have the same sense or
the tendency to cause rotation in the same
direction.
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Vector Mechanics for Engineers: Statics
Addition of Couples
• Consider two intersecting planes P1 and
P2 with each containing a couple
M 1  r  F1 in plane P1
M 2  r  F2 in plane P2
• Resultants of the vectors also form a
couple

M  r  R  r  F1  F2

• By Varignon’s theorem
M  r  F1  r  F2
 M1  M 2
• Sum of two couples is also a couple that is equal
to the vector sum of the two couples
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Vector Mechanics for Engineers: Statics
Couple Vectors
• A couple can be represented by a vector with magnitude
and direction equal to the moment of the couple.
• Couple vectors obey the law of addition of vectors.
• Couple vectors are free vectors, i.e., there is no point of
application – it simply acts on the body.
• Couple vectors may be resolved into component vectors.
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Vector Mechanics for Engineers: Statics
Resolution of a Force Into a Force at O and a Couple
• Force vector F can not be simply moved to O without modifying its
action on the body.
• Attaching equal and opposite force vectors at O produces no net
effect on the body.
• The three forces may be replaced by an equivalent force vector and
couple vector, i.e, a force-couple system.
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Vector Mechanics for Engineers: Statics
Resolution of a Force Into a Force at O and a Couple
• Moving F from A to a different point O’ requires the
addition of a different couple vector MO’

 
M O'  r   F
• The moments of F about O and O’ are related,

 
      
M O '  r 'F  r  s   F  r  F  s  F

 
 MO  s  F
• Moving the force-couple system from O to O’ requires the
addition of the moment of the force at O about O’.
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Vector Mechanics for Engineers: Statics
Sample Problem 3.6
STRATEGY:
Look for ways to add equal and opposite forces
to the diagram that, along with already known
perpendicular distances, will produce new
couples with moments along the coordinate
axes. These can be combined into a single
equivalent couple.
MODELING:
Determine the components of the •
single couple equivalent to the
couples shown.
Attach equal and opposite 20 lb forces in
the ±x direction at A, thereby producing 3
couples for which the moment components
are easily computed.
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Vector Mechanics for Engineers: Statics
Sample Problem 3.6
ANALYSIS:
You can represent these three couples by
three couple vectors Mx, My, and Mz
directed along the coordinate axes. The
corresponding moments are
M x  30 lb18 in.  540 lb  in.
M y  20 lb12 in.  240lb  in.
M z  20 lb9 in.  180 lb  in.

M    540 lb  in.  i   240lb  in.  j
  180 lb  in.  k
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Vector Mechanics for Engineers: Statics
Sample Problem 3.6
REFLECT and THINK:
You can also obtain the components of
the equivalent single couple M by
computing the sum of the moments of the
four given forces about an arbitrary point.
Selecting point D, the moment is
M  M D   18 in.  j   30 lb  k
   9 in.  j   12 in.  k    20 lb  i
M  540 lb  in.i   240lb  in. j
 180 lb  in.k
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Vector Mechanics for Engineers: Statics
Reducing System of Forces to a Force-Couple System
• A system of forces may be replaced by a collection of
force-couple systems acting at a given point O
• The force and couple vectors may be combined into a
resultant force vector and a resultant couple vector,
R F

M OR   r  F

• The force-couple system at O may be moved to O’
with the addition of the moment of R about O’ ,
M OR '  M OR  s  R
• Two systems of forces are equivalent if they can be
reduced to the same force-couple system.
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Vector Mechanics for Engineers: Statics
Further Reduction of a System of Forces
• If the resultant force and couple at O are mutually
perpendicular, they can be replaced by a single force acting
along a new line of action.
• The resultant force-couple system for a system of forces
will be mutually perpendicular if:
1) the forces are concurrent,
2) the forces are coplanar, or
3) the forces are parallel.
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Vector Mechanics for Engineers: Statics
Further Reduction of a System of Forces
• System of coplanar forces
isreduced to a

force-couple system R and M OR that is
mutually perpendicular.
• System can be reduced to a single force
by moving the line of action of R until
its moment about O becomes M OR
• In terms of rectangular coordinates,
xR y  yRx  M OR
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Vector Mechanics for Engineers: Statics
Sample Problem 3.8
For the beam, reduce the system of
forces shown to (a) an equivalent
force-couple system at A, (b) an
equivalent force couple system at B,
and (c) a single force or resultant.
STRATEGY:
The force part of an equivalent
force-couple system is simply the
sum of the forces involved. The
couple part is the sum of the
moments caused by each force
relative to the point of interest.
Once you find the equivalent
force-couple at one point, you can
transfer it to any other point by a
moment calculation.
Note: Since the support reactions are
not included, the given system will
not maintain the beam in equilibrium.
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Vector Mechanics for Engineers: Statics
Sample Problem 3.8
MODELING and ANALYSIS:
a) Compute the resultant force and the
resultant couple at A.
R F
  150 N  j   600 N  j   100 N  j   250 N  j
R    600 N  j

M AR   r  F

  1.6 i    600 j    2.8 i    100 j 
  4.8 i    250 j 
M AR    1880 N  m  k
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Vector Mechanics for Engineers: Statics
Sample Problem 3.8
b) Find an equivalent force-couple system at B
based on the force-couple system at A.
The force is unchanged by the movement of the
force-couple system from A to B.


R  600 N  j
The couple at B is equal to the moment about B
of the force-couple system found at A.
M BR  M AR  rB A  R
   1880 N  m  k   4.8 m  i   600 N  j
   1880 N  m  k   2880 N  m  k
M BR    1000 N  m  k
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Vector Mechanics for Engineers: Statics
Sample Problem 3.8
c) The resultant of the given system of
forces is equal to R, and its point of
application must be such that the
moment of R about A is equal to M RA. This
equality of moments leads to
r  R  M RA
xi   600 N j  1880 N  mk
 x600 N k  1880 N  mk
Solving for x, you get x = 3.13 m.
Thus, the single force equivalent
to the given system is defined as
R  600 N , x  3.13 m
REFLECT and THINK:
This reduction of a given system of forces to a single equivalent force uses the
same principles that you will use later for finding centers of gravity and centers
of mass, which are important parameters in engineering mechanics.
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Vector Mechanics for Engineers: Statics
Sample Problem 3.10
STRATEGY:
• Determine the relative position vectors
for the points of application of the
cable forces with respect to A.
• Resolve the forces into rectangular
components.
• Compute the equivalent force,
R F
Three cables are attached to the
bracket as shown. Replace the
forces with an equivalent forcecouple system at A.
• Compute the equivalent couple,

M AR   r  F
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
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Eleventh
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Vector Mechanics for Engineers: Statics
Sample Problem 3.10
• Resolve the forces into rectangular
components.
FB   700 N  

rE B
rE B

75 i  150 j  50k
175
 0.429 i  0.857 j  0.289k
MODELING and ANALYSIS:
• Determine the relative position
vectors with respect to A.



rB A  0.075 i  0.050k m 



rC A  0.075 i  0.050k m 



rD A  0.100 i  0.100 j m 
FB  300 i  600 j  200k
N
FC   1000 N   cos 45 i  cos 45 j 
 707 i  707 j
N
 600 i  1039 j
N
FD   1200 N   cos 60 i  cos 30 j 
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Vector Mechanics for Engineers: Statics
Sample Problem 3.10
• Compute the equivalent force,


R  F

 300  707  600  i

  600  1039  j

 200  707 k




R  1607i  439 j  507k N 
• Compute the equivalent couple,
R
 
M A   r  F 



i
j
k




rB A F B  0.075
0
0.050  30i  45k
300  600 200



i
j
k



rC A F c  0.075 0  0.050  17.68 j
707 0  707



i
j
k



rD A F D  0.100  0.100 0  163.9k
600
1039
0

R


M A  30 i  17.68 j  118.9k
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Vector Mechanics for Engineers: Statics
Sample Problem 3.10
REFLECT and THINK:
The determinant approach to calculating
moments shows its advantages in a general
three-dimensional problem such as this.
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