Definition of Force Definition of a Vector Vector Diagram Graphical Representation of a Vector Parallelogram of Forces Resolving and inclined force Resultant of two forces acting at a point Force Resultant of a system of forces acting at a point

1. Resultant of a System of forces
Engineering Science N3
Matric Rewrite and FET College
Registrations
Mechanical Engineering N1 – N6
Electrical Engineering N1 – N6
Civil Engineering N1 – N6
Business Qualifications N4 – N6
Email: topstudentz017@gmail.com
Phone:073 090 2954
Fax: 011 604 2771

2. Resultant of a System of forces
1) Definition of Force
2) Definition of a Vector
3) Vector Diagram
4) Graphical Representation of a Vector
5) Parallelogram of Forces
6) Resolving and inclined force
7) Resultant of two forces acting at a point
Force
8) Resultant of a system of forces acting at a
point

3. 1.Definition of a Force.
A force, F Newtons, can be defined in terms of
magnitude and direction as a pull or a push such
that it will displace an object from its resting point
from point A to another point B as illustrated in
the diagram.
Yverticaldistance
Point A Point B
(x,y = 0,0) (x, y = 5 meters, 0 meters) x horizontal distance
Fig. 1
F

4. This pull or push, F Newtons, can be also be
applied to a simple lever in order to do positive
work as shown.
In this case, the Force, F, applied will cause a
turning moment of magnitude M = F Newtons X
0.65 meters at the turning point, T.
Turning Point, T
Fig. 2
F

5. Generally, this formula is written thus:
Moment, M = Fd
where F = Force applied to the beam or lever in
newtons
d= perpendicular distance between the force
applied and the turning point as illustrated above.
2. Definition of a Vector
The forces as illustrated in Figure 1 and Figure 2
both have magnitude in Newtons and have
direction. They are thus referred to as vectors.
The direction of the force is usually measured in
terms of the angle in degrees that the force makes
with the positive x axis.

6. 3. Vector Diagram
Assuming that the force, F, applied to displace the block
in figure 3 is inclined at an angle 45 degrees to the
horizontal, this force would have to be resolved into
vertical and horizontal forces.
Figure 6 shows a vector diagram in which the
diagonal,DB, of the diagram ABCD is the given force, F,
which displaces the block from point A to point B.
Vector diagrams are drawn in a tail- head, tail – head
sequence.
In Fig 6, the vector FV is drawn vertically from the tail of
the given force F. Secondly, a horizontal line is drawn to
represent FH from the head of the inclined force F.
The intersection of FV and FH will give the point A.

7. 4. Graphical Representation of a Vector
Given an force of 600N, inclined to the negative x
axis at an angle 450 , FV and FH can be drawn
graphically by applying a scale of , say, 50N =
10mm.
The following diagrams illustrates the graphical
solution of resolving the inclined force of 600N.
Fig 1 a Step 1 Fig 1b Step 2
D
B x
600 600
D
B xC
y y

8. Fig 1 c Step 3 Fig 1 d Step 4
600
600
4.1 Graphical Solution of a vector.
Procedure (Fig 1a – d)
Step 1 Mark of the point B by drawing two lines such
that they intersect to mark point B. These two lines
are vertical line (y axis) and horizontal line (x axis)as
shown above.
Step 2 Using the negative x axis as reference,
measure an angle 45 degrees from this horizontal
line.
D
BC
D
BC

9. Step 3 Based on the scale of 50N = 10mm, mark
off point D by measuring 120mm along the 45
degree line to represent the magnitude of this
force. (600N /50N) X 10mm = 120mm
Step 4 Draw a vertical line downward from the
point D.
Step 5 Extend the horizontal negative x axis so
that it intersects the vertical line from point D.
mark this point A.
Step 6. Measure FH and FV in millimetres.
Step 7 Convert the distance of FH and FV in
millitres to force in Newtons by applying the
formula:
(distance in mm)/10mm X 50N

10. 5. Parallelogram of Forces
Parallelogram of Forces may also be called
Parallelogram of Vectors.
A parallelogram is a plane shape bounded by
four sides. The two pairs of opposite sides of the
rectangle in Fig 6, DA and CB, are parallel and
have the same magnitude and direction. This is
also true for the pair of opposite sides as shown,
namely, DC and AB.
Rectangles and squares are special types of
parallelograms because they have four equal
angles which are 90 degrees.
Most parallelograms look like the plane shape in
fig. 9

11. 6. Resolving an inclined force.
(analytically)
As stated in section 3 the inclined force, F, applied to
displace the block in figure 3 through a distance of 5
metres is inclined at an angle 45 degrees to the
horizontal. This force, F, has a vertical component, FV,
and horizontal component, FH.
The Forces, FV and FH may be calculated or resolved
by applying trigonometrical functions as shown on
page 14.
It will be very important to note here that the
resolved forces are of the same magnitude and
direction as FH and FV as obtained in the graphical
solution of section 6.

12. Yverticaldistanc
Point A Point B
(x,y = 0,0) (x, y = 5 meters, 0 meters) x horizontal distance
Fig. 3
F
450
Turning Point, T x
Fig. 4
F y
300 1500

13. F
450
450
D
FV
C
B
FH
Fig.6
A
300 1500
300
Z
300 W
Fig 5 Y
X
F
X
300 1500
300
300
300
Z
WY
Fig. 7
F
FV
FH

14. z
Fig 3 2 units Fig 4 2(2)1/2
units
600 600
300300
450 450
Trigonometric
Ratio
Angle 60 Angle 45 Angle 30
Sine Ө opp/hyp 30.5/2 20.5/2 1/2
Cos Ө Adj/hyp 1/2 20.5/2 30.5/2
Tan Ө Opp/adj 30.5 1 30.5/3
Opp. Opposite to the
angle
Adj. Adjacent to the angle
Hyp. The inclined side of a
right angle triangle
Table 1
31/2units

15. FV = the subscript V denotes vertical component of
the inclined force.
FH= the subscript H denotes horizontal The Fig 3 , Fig
4 and table 1 above illustrates clearly the fact that FV
and FH can be calculated by applying the sine, cosine
or tan of the angle the inclined force makes with the
positive or negative x axis.
We shall now proceed to calculate the values of FV
and FH as shown on Fig 6 based on the table on the
previous page.
Example 1. Calculate FV and FH for the inclined force F
in Fig. 3 given that F is 600N.

16. X
300
600
300
300
Z
WY
Fig. 8 Fig 9 Parallelogram
F
600
Solution: Based on the table 1
Sin 450 = FV/F Equation 1
Sin 450 = 0.70711 Equation 2
Therefore, FV/F = 0.70711
FV = 0.70711 X F
FV = 0.70711 X 600N = 424.2641N
Direction is in the negative y axis.

17. Similarly,
Sin 450 = FH/F Equation 1
Sin 450 = 0.70711 Equation 2
Therefore, FH/F = 0.70711
FH = 0.70711 X F
FH = 0.70711 X 600N = 424.2641N
Direction is in the positive x axis.
In order to verify our answer, we may apply
Pythagoras Theorem to the rectangle or system of
forces in Fig 6 as follows.
DB = √DA2 + AB2
DB = √FV
2 + FH
2
DB = √(424.2641N)2 + (424.2641N)2 =600N

18. Example 2. Calculate the FH and FV for the Fig 5
and Fig 7. The force, F, as applied on the lever is
750N.
Solution: Considering the vector diagram of fig.
7 and applying the trigonometric ratios:
cos 600 = FH /F Equation 1
cos 600 = 0.5 Equation 2
FH /F = 0.5
FH = 0.5F
FH = 0.5 X 750N = 375N
Likewise, FV = 750N X sin 60 = 649.5191N
Confirming our answer: (3752 + 649.51912)1/2 =
750N

19. 7. The Resultant of two forces acting at a point.
Given two forces FV, 250N, acting vertically
upwards and FH, 620N, acting in the positive x
direction as shown, the resultant of the two
forces is the diagonal of the corresponding
parallelogram of forces.
The resultant R = √F2
H + F2
V
y
x
FV = 250N
Fig 10 FH = 620N

20. y
x
FV = 250N
Fig 11 FH = 620N
The magnitude and direction of the resultant is given by the
Pythagoras theorem formula and the arctan formula as
shown above will give the direction of the resultant.
The resultant of two forces acting at a point is that single
force that will represent the two forces in magnitude and
direction.
There is, thus, a relationship between the resolution of an
inclined force and the resultant of a force.
tan -1 (FV/FH)

21. The Example 3. By applying the Pythagoras
Theorem and the arctan formula, calculate
the resultant of the force in Fig 11 in terms
of magnitude and direction.
Solution:
The resultant R = √F2
H + F2
V
And Ө = tan-1 (FV/FH)
Substituting the values of FV = 250N and FH
= 620N into the equations above, we have:
R = √ 2502 + 6202 =
and
Ө = tan-1 (250N/620N)= 21.9610

22. 8. The Resultant of a system of forces
acting at a point.
y
x
FA = 250N
Fig 12 Calculating the resultant of a system of forces
FE = 250N
FB = 250N
FC = 250N
FD = 250N

23. Each of the forces, FA, FB, FC, FD and FE has to be
resolved into the vertical and horizontal
components as illustrated in the next page.
All vertical forces acting downward in the
negative y direction are configured to be
negative. Likewise, vertical upward forces in
the positive y direction are positive.
This configuration is applied to the horizontal
components of the inclined forces.
The table 2 summarizes the resolution of these
five forces acting at a point.
Example 4. Calculate the resultant of the
system of forces as shown in Fig 12.

24. y
x
FA = 85N
Fig 13 FH = 620N
FE = 350N
FB = 150N
FC = 250N
FD = 350N
600
200300
350

25. Vertical
Component
FV
F orce
(Newtons)
Horizontal
Component
FH
F orce
(Newtons)
FA
+85N sin 60 +73.6122N +85N cos 60 +42.5N
FB
+150Nsin 35 +86.0365N -150N cos 35 -122.8728N
FC
350N sin 0 0 N -250N cos 0 -250N
FD
-350N sin 30 -175N -350N cos 30 -303.1089N
FE
-350N sin 20 -119.7071N +350N cos 60 +175N
∑FV -135.0584N ∑FH -155.3728N
Table 2

26. y
x
Fig 13 Resultant in terms of magnitude and direction
FH = -155.3728N 240.9980
40.9980 FV = -135.0584N
R= √ (155.3728)2 + (135.0584)2
R = 205.8676N and Ө = tan-1 (FV/FH)= 40.9980
Based on the configurations as explained above, the
sums of FV and FH respectively are -135.0584N and -
155.3728N.
The resultant, R, has a direction and magnitude as
illustrated in Fig 13 below.

27. The direction of the resultant, Ө, may be stated based
on two perspectives. A) the
positive x axis
In this case, the direction is stated thus: (1800 + 40.9980)
= 240.9980
B)The negative x axis.
Ө = 40.9980
The two answers above answers are correct. However,
it is important to understand the fact that A) has been
stated in terms of a positive angle. On the other hand
B) has been stated in terms of a reference angle. All
positive angles are measured with respect to the
positive x axis in the counter clockwise direction.