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The force 𝐹 acting on a body has two effects:
the first one is the tendency to push or pull the body in the direction
of the force,
the second one is to rotate the body about any fixed axis which does
not intersect nor is parallel to the line of the force.
This dual effect can more easily be represented by replacing the
given force by an equal parallel force and a couple to compensate for
the change in the moment of the force.
Consider 𝐹 acting at point A in a rigid body. It is
possible to slide force 𝐹 along its line of action, but
it is not possible to directly move it to point B
without changing the external effect on the rigid
body.
In order to do this, two equal and opposite forces
𝐹
and −𝐹
are added to point B without
introducing any net external effects on the body.
It is seen that, the original force at A and and the
equal and opposite one at B constitute the couple
M=Fd, which is counterclockwise for this case.
Therefore, we have replaced the original force at A by the same
force acting at a different point B and a couple, without altering
the external effects of the original force on the body. Since 𝐶
is a free vector, its location is of no concern. The combination of
the force and couple is referred to as a force-couple system.
C = Fd
By reversing this process, we can combine a given couple and a
force which lies in the plane of the couple (normal to the couple
vector) to produce a single, equivalent force. Force 𝐹 can be moved
to a point by constructing a moment equal in magnitude and
opposite in direction 𝐶 . The magnitude and direction of
C
remains the same, but its new line of action will be d 
distance
F
away from point B.
C = Fd
1. A force F of magnitude 50 N is exerted
on the automobile parking-brake lever at
the position x=250 mm. Replace the
force by an equivalent force–couple
system at the pivot point O.





F  50 sin 20i  50 cos 20 j  17.1i  46.98 j
+
M o  50
200.1 cos 25  0.25 cos10   50
200.1sin 25  0.25 sin 10 

cos



sin


Fy
M o  17.29 N  m
Fx

F
EQUIVALENT FORCE–COUPLE SYSTEM
20o
Mo=C=17.29 N.m
If two force systems are creating the same external effect on the
rigid body they are exerted on, they are said to be “
”.
The resultant of a force system is the simplest combination that
they can be reduced without altering the external effects they
produce on the body.
Coplanar Force Systems
  

If the resultant of all forces F1 , F2 , F3 , ...,Fn lying in a single plane

such as xy is R , this resultant is calculated by the vector sum of these
forces.
  


R  F1  F2  F3  ... Fn
Rx  Fx
R
R y  Fy
 F    F 
  tan
2
x
1
Ry
Rx
2
y

The location of the line of action of the resultant force R to an
arbitrary point (such as point O is the origin of the xy coordinate
system) can be determined by using the Varignon’s theorem. The

moment of R about point O will be equal the sum of the couple
moments constructed by moving its components to point O.

R
Mo 


F
 M  Fd 
Mo
d
R
2. Determine the x- and y-axis intercepts of the line of action of the resultant of
the three loads applied to the gearset.
3. The device shown is part of an automobile seat-back-release mechanism. The part
is subjected to the 4 N force exerted at A and a 300 Nmm restoring moment exerted
by a hidden torsional spring. Determine the y-intercept of the line of action of the
single equivalent force.
4. A machine component is subjected to the forces and couples shown. The
component is to be held in place by a single rivet that can resist a force but not a
couple. Determine the location of the rivet hole if it is to be located (a) on line FG,
(b) on line GH.
120 N
200 N
42 N.m
80 N
60 N
40 N.m
Determine the location of the rivet hole if it is to be
located (a) on line FG, (b) on line GH.
Resultant Force:


R  F


 200 cos15i  200 sin 15 j 


  120 cos 70i  120 sin 70 j 


  80 j   60i 




R   F  212.14i  244.53 j
Total Moment about G (resultant couple):
M G  42  40  800.59   600.13  120 cos 700.47   120 sin 700.19 
 200 sin 150.05  200 cos150.47 
M G  63.34 N  m
Determine the location of the rivet hole if it is to be
located (a) on line FG, (b) on line GH.
Equivalent Force-Couple System at point G:



R  212.14i  244.53 j
M G  63.34 N  m
Equivalent Force-Couple System
at G:
F
(a)
(b)



R  212.14i  244.53 j
G
MG


y
y



R  212.14i  244.53 j
G
G
63.34
 0.298 m
212.14
H
x



R  212.14i  244.53 j
x
63.34
 0.259 m
244.53