1) The document discusses different types of force systems and how to find the resultant force for each type. It covers collinear forces, coplanar parallel forces, coplanar concurrent forces, non-coplanar parallel forces, and concurrent non-coplanar forces. 2) For each force system, it provides the method to resolve forces into components, sum the components, and determine the magnitude and direction of the resultant force. 3) Examples are included to demonstrate how to apply the methods to solve for the resultant force in specific problems involving different force system configurations.

1. 1

2. Resultant of Forces System
Resultant:
 it is a single force which can replace the original force system
without changing its external effect on rigid body.
 If R > 0 , the resultant will accelerate this body.
 If R = 0 , the body will remain in Equilibrium or balanced state.
F2
F3
F4
R
Collinear forces:
In this system, line of action of all the forces act
along the same line. For example, consider a rope
being pulled by two players. Then the resultant
force, in the direction of pull of first player is:
F1
F2
F1
R = (F1 - F2)
2

3. 3
Coplanar parallel force:
Resultant of Forces System
In this system, all forces are parallel to each other and lie in a
single plane. Consider a See-Saw. Two children are sitting in the
See-Saw and it is in equilibrium. If the weights of both the boys
are (W1 and W2), then for equilibrium, the fulcrum reaction
should be:
R = W1 + W2
Coplanar concurrent point:
In this system, line of action of all forces passes
through a single point and forces lie in the same plane.
An example is the weight suspended by two inclined
strings.
 Here, all the three forces pass through point A.
 The moment of the forces about the point A is zero.
W2
W1
R
2452 N
A
1) Resolve each force to ( Fy , Fx ),
with the direction of each force.
2) Find : Rx =  Fx , Ry =  Fy
3) Find :
4) Find :
To find the Resultant Force ( R ):
(direction of R )

4. 4
Lecture 4
Resultant of Forces System
Coplanar non-concurrent forces:
 All forces do not meet at a single
point, but lie in a single plane.
F3
F4
F1
F2
1) The forces ( F1 , F2 ) are moved along their lines of action to intersect by the principle
of transmissibility to fine the resultant ( R12 ).
2) The resultant ( R12 ) is then combined with (F3 ) to find the resultant (R123 ).
3) Finally (R1234 ) is determined by combining R123 ) with (F4 ).
 The Resultant Force ( R ) can be
obtained by means of parallelogram
law [(R ) is either a single force or a
couple] :
F3
F4
F1 F2
R12
R123
R1234
1) Resolve each force to ( Fy , Fx ), with the direction of each force.
2) Find : Rx =  Fx , Ry =  Fy
3) Find :
4) Find : (direction of R )
 Or the resultant it can be found by:
 To find the location of ( R ), [ R . d =  Mo ], where ( d ) is the perpendicular distance.
F
3
F
4
F
1
F
2

5. 5
Non coplanar parallel forces:
Lecture 4
Resultant of Forces System
All the forces are parallel to each other, but not in the same
plane. An example is forces acting on the table legs to lift it.
To find the Resultant Force ( R ) parallel to ( X or Y –
axis ) is determined by the following equations:
1) R =  Fx or R =  Fy
2) To find its location if the forces parallel to (x –
axis): R . y =  Mz , and R . z =  My
3) To find its location if the forces parallel to (y –
axis): R . x =  Mz , and R . z =  Mx
F1
F2
F3
X
Y
Z
F1
F2
F3
Y
Z
X
Parallel to ( X – axis ) Parallel to ( Y – axis )

6. 6
Concurrent - Non coplanar forces:
Lecture 4
Resultant of Forces System
 In this system, all forces do not lie in the same plane,
but their line of action passes through a single point ( O ).
 Any two forces of a concurrent force system
determine a plane, their resultant is determined by
parallelogram law.
 The resultant of any set of concurrent forces must be a
force ( not a couple or force and couple ).
1) Resolve each force to ( Fy , Fx and Fz ), with
the direction of each force.
2) Find : Rx =  Fx , Ry =  Fy and
Rz =  Fz
1) Find :
2) Find :
To find the Resultant Force ( R ):
F2
X
Z
Y
F3
F1
R
O x
y
z

7. 7
Lecture 4
Resultant of Forces System
 All forces do not lie in the same plane and their lines of
action do not pass through a single point. This is the most
general force system.
 The Resultant of Non-Concurrent , Non coplanar & Non-
parallel force system can be a single force or a couple, but in
general, it is a force and a couple.
Non-Concurrent , Non coplanar & Non-parallel forces:
 The Resultant of a parallel force system, either
coplanar or noncoplanar, and of a nonconcurrent coplanar
force system is either a force or a couple.
 It is a force when the sum of the force components in
any direction is different from zero (not equal to zero).
 And zero or a couple when the sum of the force
components is equal to zero in every direction.
 The Resultant force can be located by applying the
principle of moments with respect to one or more axes.
 When the Resultant is a couple, its moment can be
determined by summing moments with respect to one or
more axes.
In general:

8. 8
Obtain the resultant of the concurrent coplanar forces acting as shown in Figure ( A ) below.
Lecture 4
Example ( 1 ): (Concurrent coplanar force system)
15 kN
15
0
105 kN
75 kN
45 kN
40
0
60 kN
35
0
Fig. A
∑ Fx = + 15 Cos 15 – 75 – 45 Sin 35 + 60 Cos 40
∑ Fy = + 15 Sin 15 + 105 – 45 Cos 35 – 60 Sin 40
= + 33.453 kN
= - 40.359 kN
= 40.359 kN
Solution:
+ ve
+ ve
∑Fx
f
∑Fy
R

9. 9
Lecture 4
Example ( 2 ): (Concurrent coplanar force system)
F2 = 200 N
F3 = 260 N
F1
30o
f
12
5
u
y
x
13
If the magnitude of the resultant force acting on the bracket is to be (450 N) directed along
the positive (u axis), determine the magnitude of (F1) and its direction ( ϕ. )?
Solution:
F1(x) = F1 sin f F1(y) = F1 cos f
F2(x) = 200 N F2(y) = 0
F3(x) = F3 sin 𝛼 = 260 (5/13) = 100 N
F3(y) = F3 cis 𝛼 = 260 (12/13) = 240 N

10. 10
Lecture 4
Example ( 2 ): (Concurrent coplanar force system)
FR(x) = FR cos 30 = 450 cos 30 = 389.7 N
FR(y) = FR sin 30 = 450 sin 30 = 225 N
(  + ) FR(x) =  Fx ; 389.7 = F1 sin f + 200 +100
F1 sin f = 89.7 ……… ( 1 )
(  + ) FR(y) =  Fy ; 225 = F1 cos f - 240
F1 cos f = 465 ……... ( 2 )
Solving equations ( 1 & 2 ), it yields to:
f = 10.9o F1 = 474 N
Example ( 3 ): (Non-Concurrent coplanar force system)
The force (100 lb) is the resultant of the couple
and three forces, two of which are shown in the
diagram. Determine the third force and locate it
with respect to point (A).
Solution:
The location and the magnitude of the third force
are unknown, so we will suppose the location and
its direction as shown in figure below:

11. 11
Lecture 4
Example ( 3 ): (Non-Concurrent coplanar force system)

12. 12
Lecture 4
Example ( 4 ): (Non-Concurrent Non-coplanar force system)
The ( 50 lb ) force ( R ) in ( Figure below) is the resultant of the four forces, three of which
are shown. Determine the fourth force and show it on the sketch.
Solution:
Let the fourth force to be ( F ),
then:
50 = F – 45
Taking the moment about the Y –
axis :
Taking the moment about the X – axis :

13. 13
Lecture 4
Example ( 5 ): (Concurrent Non-coplanar force system)
Determine the Resultant force of the tensions of the guy wire shown in the below figure.
Solution:
It is convenient to solve such problems to find the
length of the wire, then the force multiplier for each
force, from which the component for each wire
tension is calculated as follows:
The length of wire of tension ( 17 lb ) is:
The length of wire of tension ( 18 lb ) is:
The force multiplier is:
For the force of ( 12 lb ) its components are:
The length of wire of tension ( 12 lb ) is:

14. 14
Lecture 4
Example ( 5 ): (Concurrent Non-coplanar force system)
For the force of ( 18 lb ) its components are:
For the force of ( 17 lb ) its components are:
Rx = - 1.95 - 4.31 + 4.07 = - 2.19 lb Ry = 11.68 + 17.24 + 16.28 = 45.2 lb
Rz = - 1.95 + 2.87 + 2.71 = 3.63 lb