Chapter 3
Rigid Bodies :
Equivalent Systems of
Forces
Part -2
Slide #: 1
Couple
• Two forces F and -F having the same magnitude,
parallel lines of action, and opposite sense are said
to form a couple.
Slide #: 2
Moment of A Couple
B
• Moment of the force F,
-F
M O= r A x F
r
rB
F
A
• Moment of the force -F,
rA
MO= rB x (-F)
O
• Moment of the couple,
MO= rA x F + rB x (-F)
MO= (rA - rB) x F
M O= r x F
Slide #: 3
Moment of A Couple
• Moment of the couple,
   

M  rA  F  rB   F 

 
 rA  rB   F
 
 r F
M  rF sin   Fd
Where d is the perpendicular distance between
the lines of action of F and -F
• 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.
Slide #: 4
Problem 3.69 on page 116
A piece of plywood in which
several holes are being drilled
has been secured to a
workbench by means of two
nails. Knowing that the drill
applies a 12 Nm couple of
moment on the plywood,
determine the magnitude of
resulting forces applied to the
nails if they are located to
(a)At A and B
(b)At A and C
Slide #: 5
Problem 3.69 on page 116
(a)Nails are at A and B
12 Nm = (0.45m) * F
---> F = 26.67 N
F=26.67 N
-F= 26.67 N
Slide #: 6
Equivalent Couples
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.
Slide #: 7
Equivalent Couples
M
M
1m
120 N
20 N
6m
120 N
20 N
M
120 N
120 N
1m
Slide #: 8
Non- Equivalent Couples
M
1m
120 N
20 N
6m
20 N
M
120 N
Slide #: 9
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 Varigon’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
Slide #: 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.
Slide #: 11
Sample Problem 3.6 on page 113
Determine the components of the
single couple equivalent to the
couples shown.
Slide #: 12
Sample Problem 3.6 on page 113
SOLUTION:
compute the sum of the moments of the
four forces about D.






M  M D  M D1  M D 2  (r1 x F1 )  (r2 x F2 )






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
Slide #: 13
Sample Problem 3.6 on page 113
ALTERNATIVE 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.
Slide #: 14
Sample Problem 3.6 on page 113
• Attach equal and opposite 20 lb forces
in the +x direction at A
• The three couples may be represented by
three couple vectors,
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
Slide #: 15
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.
Slide #: 16
Resolution of a Force Into a Force at O and a
Couple
F
r
O
A
F
MO
A
O
Any force F acting at a point A of a rigid body can be
replaced by a force-couple system at an arbitrary point O.
It consists of the force F applied at O and a couple of
moment MO equal to the moment about point O of the force
F in its original position.
The force vector F and the couple vector MO are always
perpendicular to each other.
Slide #: 17
Sample Problem 3.7 on page 114
Replace the couple and force shown
by an equivalent single force applied
to the lever
Slide #: 18
Sample Problem 3.7 on page 114
First calculate the moment
created by the couple

 
M o  r x F  0.12m j x - 200Ni

M o  (24 Nm)k
Secondly resolve the given force into a force at O and a couple
We move the force F to O and add a couple of moment about O

 
 
M o  r x F  OB x F  (0.150m)i  (0.260m) j  x - 400N j

M o  (60 Nm)k
Slide #: 19
Sample Problem 3.7 on page 114


 

M o  (84 Nm)k  OC x F  (OC) cos600 i  (OC) sin 600 j x (400N ) j

M o  (84 Nm)k  (OC) cos600 (400)k
 (OC) cos600  0.210m  210m m
 (OC)  420m m
Slide #: 20
Problem 3.80 on page 118
A 135 N vertical force P is applied at A to
the bracket shown. The bracket is held
by two screws at B and C.
(a) Replace P with an equivalent forcecouple system at B
(b) Find the two horizontal forces at B and
C that are equivalent to the couple of
moment obtained in part a
Slide #: 21
Problem 3.79 on page 118



M B  rAB x F  (0.125 i  0.05 j  x (135N ) j

M B  16.875Nm k
Slide #: 22
Problem 3.79 on page 118
Find the two horizontal forces at B and C
that are equivalent to the couple of
moment obtained in part a
-F= 225 N

M B  16.875Nm k  (0.075) * F
F= 225 N
 F  225N
Slide #: 23
Reduction of a System of Forces to One Force and
One Couple
Any system of forces can be reduced to a force-couple
system at a given point O.
First, each of the forces of the system is replaced by an equivalent
force-couple system at O.
Then all of the forces are added to obtain a resultant force R,
and all of couples are added to obtain a resultant couple vector MO.
In general, the resultant force R and the couple vector MO
will not be perpendicular to each other.
Slide #: 24
Reduction of a System of Forces to One Force and
One Couple
First, each of the forces of the system is replaced by an equivalent
force-couple system at O.
Then all of the forces are added to obtain a resultant force R,
and all of couples are added to obtain a resultant couple vector MO.


R  F

R
 
MO   r  F

Slide #: 25
Sample Problem 3.8 on page
SOLUTION:
a) Compute the resultant force for the
forces shown and the resultant
couple for the moments of the
forces about A.
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.
Note: Since the support reactions are
not included, the given system will
not maintain the beam in equilibrium.
b) Find an equivalent force-couple
system at B based on the forcecouple system at A.
c) Determine the point of application
for the resultant force such that its
moment about A is equal to the
resultant couple at A.
Slide #: 26
Sample Problem 3.8 on page
SOLUTION:
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
R
 
M A   r  F 




 1.6 i   600 j   2.8 i  100 j 


 4.8 i   250 j 

R
M A  1880 N  mk
Slide #: 27
Sample Problem 3.8 on page
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.
R
R 

M B  M A  rB A  R



 1880 N  m k   4.8 m i   600 N  j


 1880 N  m k  2880 N  m k

R
M B  1000 N  mk
Slide #: 28