RESULTANTS
Force – Couple Systems = Reduction of a Force to an Equivalent
Force and Moment (Moving a Force to Another Point)

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, and 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.

Let us consider for 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 Fare 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 M 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.
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 M . The magnitude and direction of
M
remains the same, but its new line of action will be d 
distance
F
away from point B.
1. A force F of magnitude 50 N is exerted on the automobile parkingbrake lever at the position x=250 mm. Replace the force by an
equivalent force-couple system at the pivot point O.
2. 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.
3. A 50 N horizontal force is applied to the handle of the industrial
water valve as shown. The force is perpendicular to the vertical plane
containing line OA of the handle. Determine the equivalent forcecouple system at point O.
Simplification of Force Systems – Resultants
If two force systems are creating the same external effect on the
rigid body they are exerted on, they are said to be “equivalent”.
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
380 mm
800 N
C
160 mm
150 mm
800 N
A
320 mm
B
4. The forces acting on the belts on a pulley system are equal with a
magnitude of 800 N. The pulley is secured to the steel column by means of
two bolts at A and B. Replace the two forces with a force-couple system, in
which the equivalent force will be at the midpoint of the bolts. Then,
determine the force each bolt will sustain by distributing this force and
couple to forces acting at points A and B.
5. Under nonuniform and slippery road conditions, the four forces shown are exerted
on the four drive wheels of the all-wheel-drive vehicle. Determine the resultant of
this system and the x- and y-intercepts of its line of action. Note that the front and
rear tracks are equal (i. e., AB  CD).
Three Dimensional Force Systems
The same principles can be enlarged to three dimensional force
  

systems. The resultant of forces F1, F2 , F3 , ...,Fn acting on a body
can be obtained by moving them to a desired point. In this way, the
given force system will be converted to
1) Three dimensional, concurrent forces comprising the same
magnitudes and directions as the original forces,
2) Three dimensional couples.
By calculating the resultants of these forces and couples, a single
resultant force and a single couple can be obtained.
The resultant force,
  


R  F1  F2  F3  ...  Fn 
Rx 
R


F
R  F
R  F
F
 F    F    F 
x
y
y
2
x
z
2
y
2
z
z
C
C
The resultant couple moment,

M 

 
rF 


C
The selection of point O is arbitrary, but the magnitude and direction
of

M
will depend on this point; whereas, the magnitude and

direction of R are the same no matter which point is selected.
6. Determine the force-couple
system at O which is equivalent
to the two forces applied to the
shaft AOB.


Is R perpendicular to M O ?
7. Represent the resultant of the force system acting on the pipe assembly by a
single force at A and a couple .
8. The special purpose milling-cutter is subjected to the force of 1200 N and a
couple of 240 N.m as shown. Replace this loading system by an equivalent
force-couple system at O.
9. The tension in cable AB is 450 N and the tension in cable CD is 270 N.
Suppose that you want to replace these two cables by a single cable EF so that the
force exerted on the wall at E is equivalent to the two forces exerted by cables AB
and CD on the walls at A and C. What is the tension in cable EF and what are the
coordinates of points E and F?
10. The threading dye is screwed onto the end of the fixed pipe which is bent
through an angle of 20°. Replace the two forces by an equivalent force at O and a
couple . Find and calculate the magnitude M of the moment which tends to screw
the pipe into the fixed block about its angled axis through O.





R  F  150 j  200 j  50 j









M o   0.15 sin 20 i  0.15 cos 20 k  0.2k  0.25 i   200 j





  0.15 sin 20 i  0.15 cos 20 k  0.2k  0.25 i  150 j



M o  17 i  85 k





eOC   sin 20 i  cos 20 k  0.34 i  0.94 k





M   M OC  M o  17 i  85 k   0.34 i  0.94 k  85 .68 Nm






y

1700 N
3400 N
8
tan 
15
34 cm
500 N
4
3
30 cm
50 cm
800
N. m

tan 
8
15
x
z
50 cm
50 cm
11. The pulleys and the gear are subjected to the loads shown. For these forces,
determine the equivalent force-couple system at point A.



3
4
F1  3400 i  3400 j  2040 i  2720 j
5
5


F2  500 i



15 
8 
F3  1700 i  1700
j  1500 i  800 j
17
17







R  F   2040 i  2720 j   500 i  1500 i  800 j



R  1040 i  3520 j

 
 



 
M A  r  F  C








  
 15  8  
r1  F1   0.5k  0.34  i 
j    2040 i  2720 j  0.3i  0.16 j  0.5k   2040 i  2720 j
17 
 17




 
r1  F1  1360 i  1020 j  489 .6k



 
r2  F2  0.3 j   500 i  150 k







 
r3  F3  0.5 j  0.5k  1500 i  800 j  400 i  750 j  750 k


C  800 k





 
M A  r  F  C  960 i  1770 j  1139 .6k






 


* Line MN lies in a plane parallel to the horizontal plane
* Line AD lies in the xz plane and makes a 37° angle with
the x axis
12. The direction cosines of robot arm AB are cos x=0.6, cos y (y <90°) and cos z=0’dır.
For arm BC the direction cosines are, cos x=7/9, cos y=4/9 and cos z=4/9. A force of
magnitude F=250 N and a couple of magnitude C=27 Nm along the axis BC are applied to
the end C of arm BC. Determine the moment about line AD. Replace the force and couple
acting on the robot assembly with an equivalent force-couple at point A.

As a special case, if the resultant couple  M is perpendicular to

the resultant force R , these two vectors can further be simplified to


obtain a single resultant force R . The force R can be slided a

distance d to form a moment   M , which is equal in magnitude

and opposite in direction  M , so that they will cancel each other
out. The distance d will be equal to d=M/R.
Wrench Resultants

When the resultant couple vector M is parallel to the resultant

force R , the resultant is called a “wrench”.
The wrench is the simplest form in which the resultant of a general
force system may be expressed.
By definition, a wrench is positive if the couple and force vectors
point in the same direction, and negative it they point in opposite
directions.
Wrench Resultants
A common example example of a wrench is found with the
application of a screw driver.
All force systems can be reduced to a wrench acting at a particular
line of action.
M
R
Equivalent force-couple system at point O
M is resolved into components M1 along the
direction of R and M2 normal to R.




  
M 1  M R //  M  eR eR


 
M 2  M R   M  M R //
Positive wrench
d
M2
R
Positive
Wrench
Negative
Wrench
13. In tightening a bolt whose center is at point O, a person exerts a 180 N force on the
ratchet handle with his right hand. In addition, with his left hand he exerts a 90-N force as
shown in order to secure the socket onto the bolt head. Determine the equivalent force-
couple system at O. The find the point in the x-y plane through which the line of action of
the resultant force of the wrench passes.
13.
Z
F1 = 30 N
F2 = 75 N
F3 = 40 N
C1 = 60 Nm
C2 = 100 Nm (in yz plane)

C2

C1
60°
C3 = 80 Nm (in plane ABCD)
y < 90o
y

F45°
1
G
6m
30°
A
X
37°
B
4m
O

C3

F2
C
53°
E

F3
O (0, 0, 0) m
A (12, 0, 0) m
Y
B (in xz plane)
C (12, 8, 0) m
E (6, 10, -3) m
G (10, 4, 4) m
D
14. Replace the system comprising two forces, two couples and a positive wrench with an
equivalent force-couple acting at point O. Then, reduce the system further into a wrench and
determine the coordinates of point P, of which the line of action of the wrench cuts through
the yz plane.
Force:
cos 2 45  cos 2 60  cos 2 y  1

 y  60 o







F1  30 cos45 i  cos60 j  cos60k  21 .21 i  15 j  15k








12  18 i  (8  0) j  (0  8)k
F2  75
 35 .15 i  46 .88 j  46 .88 k
4 2  82  82







F3  40 cos 60 cos 53 i  40 cos 60 sin 53 j  40 cos 30 k  12 i  16 j  34 .64 k





R  F  1.95 i  77 .88 j  96 .52 k








R  1.95 i  77 .88 j  96 .52 k
eR   
 0.015 i  0.63 j  0.78 k
2
2
2
R
1
.
95

77
.
88

96
.
52



124.04
Moment:


 
M o  r  F  C




 


 
r1  F1  10 i  4 j  4k  21 .21 i  15 j  15k  65 .16 j  65 .16 k
 






 
r2  F2  12 i  8 j   35 .15 i  46 .88 j  46 .88 k  375 .04 i  562.56 j  843 .84 k





 


 
r3  F3  6 i  10 j  3k  12 i  16 j  34 .64 k  394 .4 i  243 .84 j  24 k







C1  60 cos45 i  cos 60 j  cos 60 k  42.42 i  30 j  30 k


C 2  100 i
 



BA  BD   6 i  8k  4 j  32 i  24 k





 32 i  24 k
C 3  80
 64 i  32k
2
2
32

24














 
40

 










M o   65 .16 j  65 .16 k  375 .04 i  562.56 j  843 .84 k  394 .4 i  243 .84 j  24 k






 42.42 i  30 j  30 k  100 i   64 i  32k




M o  847 .86 i  841 .56 j  867 k



Equivalent force-couple system at point O




R  1.95 i  77.88 j  96 .52 k




M o  847 .86 i  841 .56 j  867 k
Reduction to a wrench in yz plane







 


M //  M o  eR  847.86 i  841 .56 j  867 k   0.015 i  0.63 j  0.78 k  133 .35 Nm








M //  M // eR  133 .35  0.015 i  0.63 j  0.78 k  2 i  84 .02 j  104 .02 k



Mo
z
Positive wrench




R  1.95 i  77.88 j  96 .52 k




M //  2 i  84.02 j  104 .02 k
Positive wrench




R  1.95i  77.88j  96.52k

M




M //  2 i  84.02 j  104.02k
O
y
x
The coordinates of point P, of which the line of action of the wrench cuts through the yz plane:

 









M   M o  M //  847.86 i  841.56 j  867 k   2 i  84.02 j  104 .02 k




M   849.86 i  925 .58 j  762 .98 k
 






y j  zk   1.95 i  77 .8 j  96 .52 k  849.86 i  925 .58 j  762 .98 k



 
 








r
R
M







1.95 yk  96 .52 yi  1.45 zj  77 .8 zi  849.86 i  925 .58 j  762 .98 k

j

1.95 z  925 .58

z  474 .66 m

k

1.95 y  762 .98

y  391 .27 m
z




R  1.95i  77.88j  96.52k
Positive wrench




M //  2 i  84.02 j  104.02k

r
x
P(0;391.27;474.66)
O
y
