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Vector Mechanics for Engineers: Statics
Professor Nikolai V. Priezjev, Ph.D.
Tel: (937) 775-3214
Rm. 430 Russ Engineering Center
Email: nikolai.priezjev@wright.edu
Textbook: Vector Mechanics for Engineers: Dynamics,
Beer, Johnston, Mazurek and Cornwell, McGraw-Hill,
10th edition, 2012.
Moments of Forces
Brief Review: Moment of a Force About a Point
• A force vector F is defined by its magnitude and direction.
Its effect on the rigid body also depends on it point of
application.
• The moment of F about point O is defined as
FrMO 
• The moment vector MO is perpendicular to the
plane containing O and the force F.
• 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.
• Magnitude of MO 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.
FdrFMO  sin
Principle of Transmissibility!
3.8 Rectangular Components of the Moment of a Force
     kyFxFjxFzFizFyF
FFF
zyx
kji
kMjMiMM
xyzxyz
zyx
zyxO






The moment of F about O,
kFjFiFF
kzjyixrFrM
zyx
O



 ,
Remember the (–) sign for j.
 
xy
ZO
xyO
yFxF
MM
kyFxFM




0 yx MM
For 2D (z = 0 and Fz = 0)FdrFMO  sin
FFF
zyx
kji
zyx

MO


Review: Moment of a Force About a Given Axis
• Moment MO of a force F applied at the point A
about a point O,
FrMO


• Scalar moment MOL about an axis OL is the
projection of the moment vector MO onto the
axis:
 FrMM OOL

 
• Moments of F about the coordinate axes:
xyz
zxy
yzx
yFxFM
xFzFM
zFyFM



Unit vector:
kji zyx

 coscoscos 
FFF
zyx
zyx
zyx
),,( zyx  

)0,0,1(

In x-direction:
=
)0,1,0(

In y-direction:
)1,0,0(

In z-direction:
3.11 Moment of a Force About a Given Axis
• Moment MO of a force F applied at the point A
about a point O,
FrMO


• Scalar moment MOL about an axis OL is the
projection of the moment vector MO onto the
axis,
 FrMM OOL

 
The tendency to rotate the
body about the fixed axis.
Only the force component perpendicular
to the axis is important!
3.12 Moment of a Couple
• Two forces F and -F having
1. the same magnitude,
2. parallel lines of action, and
3. opposite sense are said to form a couple.
• Moment of the couple:
 
 
FdrFM
Fr
Frr
FrFrM
BA
BA




sin



• 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.
Example: Moment
of a Couple
Moment of a Couple (continued)
Two couples will have equal moments if
• 2211 dFdF 
• 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.
• Will be useful for drawing Free Body Diagram!
Addition of Couples
• Consider two intersecting planes P1 and
P2 with each containing a couple
222
111
planein
planein
PFrM
PFrM




• Resultants of the vectors also form a
couple
 21 FFrRrM


• By Varigon’s theorem
21
21
MM
FrFrM




• Sum of two couples is also a couple that is equal
to the vector sum of the two couples.
Couples Can Be Represented by 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., the point of application
is not significant.
• Couple vectors may be resolved into component vectors.
ATTENTION QUIZ
1. A couple is applied to the beam as shown. Its moment equals
_____ N·m.
A) 50 B) 60
C) 80 D) 100 3
4
51m 2m
50 N
2. What is the direction of the moment vector of the couple ?
A) pointing towards us B) parallel to the red vector
C) impossible to tell D) pointing away from us
APPLICATIONS
Free Body Diagram:
Several forces and a couple
moment are acting on this
vertical section of an I-beam.
Can you replace them with just
one force and one couple
moment at point O that will
have the same external effect?
If yes, how will you do that?
| | ??
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. Why?
• 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. Going backwards?
FrMO


3.16 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’
FrMO

'
• The moments of F about O and O’ are related,
 
FsM
FsFrFsrFrM
O
O



 ''
• Moving the force-couple system from O to O’ requires the
addition of the moment of the force at O about O’.
CONCEPT QUIZ
1. F1 and F2 form a couple. The moment
of the couple is given by ____ .
A) r1  F1 B) r2  F1
C) F2  r1 D) r2  F2
2. If three couples act on a body, the overall result is that
A) the net force is not equal to 0.
B) the net force and net moment are equal to 0.
C) the net moment equals 0 but the net force is not
necessarily equal to 0.
D) the net force equals 0 but the net moment is not
necessarily equal to 0 .
F1
r1
F2
r2
Sample Problem 3.6
Determine the components of the
single couple equivalent to the
couples shown.
SOLUTION:
• 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.
• Alternatively, compute the sum of the
moments of the four forces about an
arbitrary single point. The point D is a
good choice as only two of the forces will
produce non-zero moment contributions..
x
z
  FrMD

Sample Problem 3.6
• Attach equal and opposite 20 lb forces in
the +x direction at A
• The three couples may be represented by
three couple vectors,
  
  
   in.lb180in.9lb20
in.lb240in.12lb20
in.lb540in.18lb30



z
y
x
M
M
M
   
 k
jiM


in.lb180
in.lb240in.lb540


Moment of the couple:
 
FdrFM
Fr
FrFrM BA



sin


z
Sample Problem 3.6
• Alternatively, compute the sum of the
moments of the four forces about D.
• Only the forces at C and E contribute to
the moment about D.
   
      ikj
kjMM D


lb20in.12in.9
lb30in.18


   
 k
jiM


in.lb180
in.lb240in.lb540


• 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.
PROBLEM
Given: Handle forces F1 and F2 are
applied to the electric drill.
Find: An equivalent resultant
force and couple moment at
point O.
Plan:
a) Find FRO =  Fi
b) Find MRO =  ( ri  Fi )
where,
Fi are the individual forces in Cartesian vector notation.
ri are the position vectors from the point O to any point on the line
of action of Fi .
SOLUTION
F1 = {6 i – 3 j – 10 k} N
F2 = {0 i + 2 j – 4 k} N
FRO = {6 i – 1 j – 14 k} N
r1 = {0.15 i + 0.3 k} m
r2 = {-0.25 j + 0.3 k} m
MRO = r1  F1 + r2  F2
= { 0.9 i + 3.3 j – 0.45 k + 0.4 i + 0 j + 0 k } N·m
= { 1.3 i + 3.3 j – 0.45 k } N·m
i j k
0.15 0 0.3
6 -3 -10
+
i j k
0 - 0.25 0.3
0 2 -4
MRO = { } N·m
System of Forces: Reduction to a Force and Couple
• A system of forces may be replaced by a collection of
force-couple systems acting a given point O
• The force and couple vectors may be combined into a
resultant force vector and a resultant couple vector,
   FrMFR R
O

• The force-couple system at O may be moved to O’
with the addition of the moment of R about O’ ,
RsMM R
O
R
O

'
• Two systems of forces are equivalent if they can be
reduced to the same force-couple system.
If the force system lies in the x-y plane (a 2-D case), then the
reduced equivalent system can be obtained using the following
three scalar equations.
SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
WR = W1 + W2
(MR)o = W1 d1 + W2 d2
ATTENTION QUIZ
1. For this force system, the equivalent system at P is
___________ .
A) FP = 40 N (along +x-dir.) and MP = +60 N·m
B) FP = 0 N and MP = +30 N·m
C) FP = 30 N (along +y-dir.) and MP = -30 N·m
D) FP = 40 N (along +x-dir.) and MP = +30 N·m
P
1m 1m
30 N
40 N
30 N
•
x
y
Further Reduction of a System of Forces (Special Cases)
• 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 (just add the forces)
2) the forces are coplanar (all components are  at O)
3) the forces are parallel (moment is in xz plane).
Further Reduction of a System of Forces
• System of coplanar forces is reduced to a
force-couple system that is
mutually perpendicular.
R
OMR

and
• System can be reduced to a single force
by moving the line of action of until
its moment about O becomes R
OM
R

• In terms of rectangular coordinates,
R
Oxy MyRxR 
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.
Note: Since the support reactions are
not included, the given system will
not maintain the beam in equilibrium.
SOLUTION:
a) Compute the resultant force for the
forces shown and the resultant
couple for the moments of the
forces about A.
b) Find an equivalent force-couple
system at B based on the force-
couple system at A.
Sample Problem 3.8
SOLUTION:
a) Compute the resultant force and the
resultant couple at A.
       jjjj
FR


N250N100N600N150 
 
 jR

N600
 
       
   ji
jiji
FrM R
A



2508.4
1008.26006.1


 
 kM R
A

mN1880 
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.
 jR

N600
The couple at B is equal to the moment about B
of the force-couple system found at A.
     
   kk
jik
RrMM BA
R
A
R
B



mN2880mN1880
N600m8.4mN1880



 kM R
B

mN1000 
CONCEPT QUIZ
2. Consider two couples acting on a body. The simplest possible
equivalent system at any arbitrary point on the body will have
A) One force and one couple moment.
B) One force.
C) One couple moment.
D) Two couple moments.
1. The forces on the pole can be reduced to
a single force and a single moment at
point ____ .
A) P B) Q C) R
D) S E) Any of these points.
R
Z
S
Q
P
X
Y•
•
•
•
Sample Problem 3.10
Three cables are attached to the
bracket as shown. Replace the
forces with an equivalent force-
couple system at A.
SOLUTION:
• 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,
 FR

• Compute the equivalent couple,
   FrM R
A

Sample Problem 3.10
SOLUTION:
• Determine the relative position
vectors with respect to A.
 
 
 m100.0100.0
m050.0075.0
m050.0075.0
jir
kir
kir
AD
AC
AB






• Resolve the forces into rectangular
components.
 
 N200600300
289.0857.0429.0
175
5015075
N700
kjiF
kji
kji
r
r
F
B
BE
BE
B












  
 N1039600
30cos60cosN1200
ji
jiFD




  
 N707707
45cos45cosN1000
ki
kiFC




Sample Problem 3.10
• Compute the equivalent force,
 
 
 k
j
i
FR




707200
1039600
600707300



 
 N5074391607 kjiR


• Compute the equivalent couple,
 
k
kji
Fr
j
kji
Fr
ki
kji
Fr
FrM
DAD
cAC
BAB
R
A










9.163
01039600
0100.0100.0
68.17
7070707
050.00075.0
4530
200600300
050.00075.0







 
kjiM R
A

9.11868.1730 

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Lecture Statics Moments of Forces

  • 1. Vector Mechanics for Engineers: Statics Professor Nikolai V. Priezjev, Ph.D. Tel: (937) 775-3214 Rm. 430 Russ Engineering Center Email: nikolai.priezjev@wright.edu Textbook: Vector Mechanics for Engineers: Dynamics, Beer, Johnston, Mazurek and Cornwell, McGraw-Hill, 10th edition, 2012. Moments of Forces
  • 2. Brief Review: Moment of a Force About a Point • A force vector F is defined by its magnitude and direction. Its effect on the rigid body also depends on it point of application. • The moment of F about point O is defined as FrMO  • The moment vector MO is perpendicular to the plane containing O and the force F. • 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. • Magnitude of MO 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. FdrFMO  sin Principle of Transmissibility!
  • 3. 3.8 Rectangular Components of the Moment of a Force      kyFxFjxFzFizFyF FFF zyx kji kMjMiMM xyzxyz zyx zyxO       The moment of F about O, kFjFiFF kzjyixrFrM zyx O     , Remember the (–) sign for j.   xy ZO xyO yFxF MM kyFxFM     0 yx MM For 2D (z = 0 and Fz = 0)FdrFMO  sin FFF zyx kji zyx  MO  
  • 4. Review: Moment of a Force About a Given Axis • Moment MO of a force F applied at the point A about a point O, FrMO   • Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis:  FrMM OOL    • Moments of F about the coordinate axes: xyz zxy yzx yFxFM xFzFM zFyFM    Unit vector: kji zyx   coscoscos  FFF zyx zyx zyx ),,( zyx    )0,0,1(  In x-direction: = )0,1,0(  In y-direction: )1,0,0(  In z-direction:
  • 5. 3.11 Moment of a Force About a Given Axis • Moment MO of a force F applied at the point A about a point O, FrMO   • Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis,  FrMM OOL    The tendency to rotate the body about the fixed axis. Only the force component perpendicular to the axis is important!
  • 6. 3.12 Moment of a Couple • Two forces F and -F having 1. the same magnitude, 2. parallel lines of action, and 3. opposite sense are said to form a couple. • Moment of the couple:     FdrFM Fr Frr FrFrM BA BA     sin    • 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.
  • 8. Moment of a Couple (continued) Two couples will have equal moments if • 2211 dFdF  • 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. • Will be useful for drawing Free Body Diagram!
  • 9. Addition of Couples • Consider two intersecting planes P1 and P2 with each containing a couple 222 111 planein planein PFrM PFrM     • Resultants of the vectors also form a couple  21 FFrRrM   • By Varigon’s theorem 21 21 MM FrFrM     • Sum of two couples is also a couple that is equal to the vector sum of the two couples.
  • 10. Couples Can Be Represented by 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., the point of application is not significant. • Couple vectors may be resolved into component vectors.
  • 11. ATTENTION QUIZ 1. A couple is applied to the beam as shown. Its moment equals _____ N·m. A) 50 B) 60 C) 80 D) 100 3 4 51m 2m 50 N 2. What is the direction of the moment vector of the couple ? A) pointing towards us B) parallel to the red vector C) impossible to tell D) pointing away from us
  • 12. APPLICATIONS Free Body Diagram: Several forces and a couple moment are acting on this vertical section of an I-beam. Can you replace them with just one force and one couple moment at point O that will have the same external effect? If yes, how will you do that? | | ??
  • 13. 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. Why? • 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. Going backwards? FrMO  
  • 14. 3.16 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’ FrMO  ' • The moments of F about O and O’ are related,   FsM FsFrFsrFrM O O     '' • Moving the force-couple system from O to O’ requires the addition of the moment of the force at O about O’.
  • 15. CONCEPT QUIZ 1. F1 and F2 form a couple. The moment of the couple is given by ____ . A) r1  F1 B) r2  F1 C) F2  r1 D) r2  F2 2. If three couples act on a body, the overall result is that A) the net force is not equal to 0. B) the net force and net moment are equal to 0. C) the net moment equals 0 but the net force is not necessarily equal to 0. D) the net force equals 0 but the net moment is not necessarily equal to 0 . F1 r1 F2 r2
  • 16. Sample Problem 3.6 Determine the components of the single couple equivalent to the couples shown. SOLUTION: • 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. • Alternatively, compute the sum of the moments of the four forces about an arbitrary single point. The point D is a good choice as only two of the forces will produce non-zero moment contributions.. x z   FrMD 
  • 17. Sample Problem 3.6 • Attach equal and opposite 20 lb forces in the +x direction at A • The three couples may be represented by three couple vectors,          in.lb180in.9lb20 in.lb240in.12lb20 in.lb540in.18lb30    z y x M M M      k jiM   in.lb180 in.lb240in.lb540   Moment of the couple:   FdrFM Fr FrFrM BA    sin   z
  • 18. Sample Problem 3.6 • Alternatively, compute the sum of the moments of the four forces about D. • Only the forces at C and E contribute to the moment about D.           ikj kjMM D   lb20in.12in.9 lb30in.18        k jiM   in.lb180 in.lb240in.lb540   • 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.
  • 19. PROBLEM Given: Handle forces F1 and F2 are applied to the electric drill. Find: An equivalent resultant force and couple moment at point O. Plan: a) Find FRO =  Fi b) Find MRO =  ( ri  Fi ) where, Fi are the individual forces in Cartesian vector notation. ri are the position vectors from the point O to any point on the line of action of Fi .
  • 20. SOLUTION F1 = {6 i – 3 j – 10 k} N F2 = {0 i + 2 j – 4 k} N FRO = {6 i – 1 j – 14 k} N r1 = {0.15 i + 0.3 k} m r2 = {-0.25 j + 0.3 k} m MRO = r1  F1 + r2  F2 = { 0.9 i + 3.3 j – 0.45 k + 0.4 i + 0 j + 0 k } N·m = { 1.3 i + 3.3 j – 0.45 k } N·m i j k 0.15 0 0.3 6 -3 -10 + i j k 0 - 0.25 0.3 0 2 -4 MRO = { } N·m
  • 21. System of Forces: Reduction to a Force and Couple • A system of forces may be replaced by a collection of force-couple systems acting a given point O • The force and couple vectors may be combined into a resultant force vector and a resultant couple vector,    FrMFR R O  • The force-couple system at O may be moved to O’ with the addition of the moment of R about O’ , RsMM R O R O  ' • Two systems of forces are equivalent if they can be reduced to the same force-couple system.
  • 22. If the force system lies in the x-y plane (a 2-D case), then the reduced equivalent system can be obtained using the following three scalar equations. SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM WR = W1 + W2 (MR)o = W1 d1 + W2 d2
  • 23. ATTENTION QUIZ 1. For this force system, the equivalent system at P is ___________ . A) FP = 40 N (along +x-dir.) and MP = +60 N·m B) FP = 0 N and MP = +30 N·m C) FP = 30 N (along +y-dir.) and MP = -30 N·m D) FP = 40 N (along +x-dir.) and MP = +30 N·m P 1m 1m 30 N 40 N 30 N • x y
  • 24. Further Reduction of a System of Forces (Special Cases) • 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 (just add the forces) 2) the forces are coplanar (all components are  at O) 3) the forces are parallel (moment is in xz plane).
  • 25. Further Reduction of a System of Forces • System of coplanar forces is reduced to a force-couple system that is mutually perpendicular. R OMR  and • System can be reduced to a single force by moving the line of action of until its moment about O becomes R OM R  • In terms of rectangular coordinates, R Oxy MyRxR 
  • 26. 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. Note: Since the support reactions are not included, the given system will not maintain the beam in equilibrium. SOLUTION: a) Compute the resultant force for the forces shown and the resultant couple for the moments of the forces about A. b) Find an equivalent force-couple system at B based on the force- couple system at A.
  • 27. Sample Problem 3.8 SOLUTION: a) Compute the resultant force and the resultant couple at A.        jjjj FR   N250N100N600N150     jR  N600              ji jiji FrM R A    2508.4 1008.26006.1      kM R A  mN1880 
  • 28. 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.  jR  N600 The couple at B is equal to the moment about B of the force-couple system found at A.          kk jik RrMM BA R A R B    mN2880mN1880 N600m8.4mN1880     kM R B  mN1000 
  • 29. CONCEPT QUIZ 2. Consider two couples acting on a body. The simplest possible equivalent system at any arbitrary point on the body will have A) One force and one couple moment. B) One force. C) One couple moment. D) Two couple moments. 1. The forces on the pole can be reduced to a single force and a single moment at point ____ . A) P B) Q C) R D) S E) Any of these points. R Z S Q P X Y• • • •
  • 30. Sample Problem 3.10 Three cables are attached to the bracket as shown. Replace the forces with an equivalent force- couple system at A. SOLUTION: • 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,  FR  • Compute the equivalent couple,    FrM R A 
  • 31. Sample Problem 3.10 SOLUTION: • Determine the relative position vectors with respect to A.      m100.0100.0 m050.0075.0 m050.0075.0 jir kir kir AD AC AB       • Resolve the forces into rectangular components.    N200600300 289.0857.0429.0 175 5015075 N700 kjiF kji kji r r F B BE BE B                 N1039600 30cos60cosN1200 ji jiFD         N707707 45cos45cosN1000 ki kiFC    
  • 32. Sample Problem 3.10 • Compute the equivalent force,      k j i FR     707200 1039600 600707300       N5074391607 kjiR   • Compute the equivalent couple,   k kji Fr j kji Fr ki kji Fr FrM DAD cAC BAB R A           9.163 01039600 0100.0100.0 68.17 7070707 050.00075.0 4530 200600300 050.00075.0          kjiM R A  9.11868.1730 