ENGINEERING MECHANICS
CHAPTER 2
FORCES & RESULTANTS
By Jeff Lamoon
www.jefflamoon.com
Engineering Mechanics – Chapter 2: Forces & Resultants
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2.1 Introduction
This chapter introduces the concept of vectors.
Upon completion of this chapter, the student will be able
1 Distinguish between vector and scalar quantities.
2 Determine the resultants of two force vectors by using
vector triangle, parallelogram law and vector polygon
addition.
3 Resolving a force vector into its components.
4 Determine the resultant of several force vectors using
the method of perpendicular components.
Engineering Mechanics – Chapter 2: Forces & Resultants
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2.2 Vector Quantities
• Scalar quantity has only magnitude.
Examples: 10 kg mass, 8 m long, 20 m2 area, etc.
• Vector quantity has magnitude + direction.
E.g. 1) 5 km south.
5 cm
E.g. 2) 8 kN horizontal force.
E.g. 3) 9 m/s north.
Engineering Mechanics – Chapter 2: Forces & Resultants
8 cm
9 cm
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• Vectors may be in any direction:
600
1500
3
4
• Scalar quantities can be added algebraically:
3 cm + 5 cm = 8 cm
6 kg + 8 kg = 14 kg
However, vector quantities cannot be added
or subtracted because vector has the extra
property of direction.
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2.3 Vector Triangle Method
y
8 kN
6 kN
60 0
40 0
x
a) Graphical Method (Tip To Tail)
i) Choose a scale like 1 cm to 1 kN, then
begin by drawing any force vector.
6 cm
400
x
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ii) Draw the other force taking care to join them
tip-to-tail.
y
8 kN
6 kN
600
40 0
x
600
Tip of first arrow
join to tail of next
arrow.
8 cm
6 cm
400
x
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iii) Complete the triangle by drawing the resultant R
from the start point of first arrow to end point of
last arrow (force vector).
y
600
End point
8 kN
6 kN
600
8 cm
40 0
x
R
6 cm
400
x
Start point
Engineering Mechanics – Chapter 2: Forces & Resultants
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Lets draw the diagram
another way : start with
the 8 kN force instead
Step 1
y
8 kN
60
40 •0
x
Step 2
Step 3
6 cm
6 cm
400
8 cm
6 kN
•0
400
R
1500
x
8 cm
1500
8 cm
x
x
The same resultant R is obtained and it makes no difference which
force you start to draw first, as long as you follow ‘Tip to Tail, join
start point to end point’ rule.
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b) Analytical method
Start by sketching the triangle ABC.
A
8 cm
R
Angle ABC = (300 + 400) = 700
600
As triangle is not a right-angle,
B
6 cm
400
C
Cosine Law is used here.
x
R 2  (8x103 ) 2  (6x103 ) 2 2(8x103 )(6x103 )cos700
R = 8.20 kN
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To find direction of R, use sine rule to find  ,
then  .
3
3
8 kN
8.2 kN


8x10 8.2x10

sin  sin 70 0
600
7.52
sin  
 0.917
8.2
B
6 kN
400
C
i.e. 
  66.50
x
 180  66.5  40
0
 73.5
0
8.2 kN
73.50
0
Engineering Mechanics – Chapter 2: Forces & Resultants
0
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2.4 Parallelogram Method
y
8 kN
6 kN
•0
60
40 •0
x
a) Graphical Method
i) Use scale of 1 cm to 1 kN, draw the forces
from the same point as in the diagram.
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ii) Construct a parallelogram by drawing
parallel lines as shown.
y
8 kN
6 kN
600
40 0
x
8 kN
600
6 kN
400
x
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iii) Draw the diagonal R from the point where
the tails of the two arrows meet.
y
8 kN
60•0
R
6 kN
40 •0
x
8 kN
6 kN

400
x
Note that half of the parallelogram is
exactly the same triangle as that obtained by
the Vector Triangle Method in 2.3.
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b) Analytical Method
8 kN
R
6 kN
400
x
We can select any half of the
parallelogram to solve for
the resultant. The right half
here is exactly the same as
the one in 2.3 b) above and
the analytical method use is
also exactly the same.
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2.5 Vector Polygon Method
This method is used to find the resultant of
three or more forces. Example below illustrates
this principles
8 kN
4 kN
3 kN
400
200
6 kN
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a) Graphical Method
8 kN
4 kN
3 kN
400
200
6 kN
Start with any force drawn to scale, and follow
tip-to-tail rule , draw the other 3 forces.
8 kN
40 0
end point
R
4 kN
start point
3 kN
20 0
6 kN
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b) Analytical Method
For 3 or more forces, this method involves more calculation steps.
The polygon is divided into several triangles and cosine and sine
rules are then applied. Using the example above:
4 kN
40 0
3
R
R2
R1
2 8 kN
1
3 kN
20 0
6 kN
The polygon is divided into three
triangles 1, 2 and 3 as shown. Starting
with triangle 1, using cosine and sine
rules, calculate R1. Then the process is
repeated for triangle 2 to find R2, and
finally triangle 3 to find R, the resultant
of the four forces. This is very tedious
and not the preferred method.
* Check sections 2.6 and 2.7 for an easier method
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2.6 Resolution of A Force Into Components
We know that resultant is the sum of 2 or more forces and
it produces the same effect on a body.
Hence a system of forces can be replaced by a single
resultant force which have the same effect on the body.
Conversely, a single force F (like a resultant), can be
replaced by 2 forces or components of F, which produce
the same effect on the body.
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The method of resolving a single force vector into two
components is the reverse of vector addition.
Any single force can be resolved into 2 components in any
direction, but in order for us to develop an easier
analytical method, we would choose to resolve a single
force into its perpendicular components based on the xand y-axis.
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2.7 Resolution of A Force Into Perpendicular
Components
One of the most important and useful concept is the
resolution of a force vector into two perpendicular
components, in the x and y axis as denoted by the Fx (x)
and Fy (y) components.
y
Fx = F cos 
F
Fy
Fy = F sin 

Fx
Engineering Mechanics – Chapter 2: Forces & Resultants
x
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Example 2.1
Resolve the 300 N force into components along the x and
y axes.
300 N
300
Sketch the parallelogram with Fy perpendicular to Fx.
y
300 N
300
Fy
x
Fx
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Select any half of the triangle, a vector triangle, for
analysis.
300
Fy
Fx
x
As this is a right-angle triangle,
cos 300 = Fx / 300
i.e
Fx = 300 x cos 300
= 259.81 N
Also since
i.e
sin 300 = Fy / 300
Fy = 300 sin 300
= 150 N
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2.8 Resolution of A Force By Fx And Fy
Components
For a number of forces, this section introduces an easier
analytical method of finding the resultant.
Students should master this method as it would form the
basis of all your calculations for resultant forces.
Let us use the same simple example in section 2.3 above
to demonstrate the method:
y
8 kN
6 kN
60
0
40 0
x
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1. We first resolved the 2 forces into perpendicular i.e.
Fx and Fy components.
Fy = ( 8 x 103 sin 30o)
Fy = ( 6 x103 sin 40o)
8 kN
6 kN
30 0
40 0
Fx = ( 6 x 103 cos 40o)
Fx = ( 8 x 103 cos 30o)
2. The 2 forces can now be replaced by the above 4
components as shown below.
Fy = ( 4.0 kN)
Fy = ( 3.86 kN)
Fx = ( 6.93 kN)
Engineering Mechanics – Chapter 2: Forces & Resultants
Fx = ( 4.6 kN)
A
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3. Now we can add the two Fx forces to form a single Rx, as they
are both horizontal forces but only opposite in direction, if we
adopt a sign convention. Similarly, we can also add the two Fy
forces to form a single Ry.
Fy = ( 4.0 kN)
Fy = ( 3.86 kN)
Fx = ( 6.93 kN)
Fx = ( 4.6 kN)
A
 Fx = Rx, is the sum of all the Fx forces,
 Fy = Ry , is the sum of all Fy forces.
 Fx = (4.6x103) - (6.93x103) {Fx pointing left is –ve}
Rx = - 2.33 kN
 Fy = (3.86x103) + (4.0x103) {Fy pointing up is +ve}
Ry = + 7.86 kN
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4. These two forces, Rx and Ry can now replace the previous four
component forces in (2) above, and we can use the vector
triangle method to add them back to form a single force, which
is the resultant R, as shown by the vector triangle below.
Ry = 7.86 kN
R

From the right-angle triangle, we
then use the Pythagoras Theorem to
find the magnitude of R.
Rx = 2.33 kN
The angle  of R is:
The magnitude of R is:
 = tan -1 (Ry/Rx)
R =  Rx2 + Ry2 
= tan -1 ((7.86x103)/(2.33x103))
= (2.33x103)2 + (7.86x103)2
0

=
73.5
= 8.2 kN
Same answers as before.
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Resultant By Method of Perpendicular Components:
Use the Sign convention for
the force components:
1. Resolve all (but not horizontal
or vertical) forces into Fx and
Fy components.
2. Use
 Fx = Rx,
 Fy = Ry
Fx
- ve
•
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Fx
+ ve
-ve
Fy
4. Direction of R:
R
3. Magnitude: R =  Rx2 + Ry2 
Angle:
 = tan -1 (Ry/Rx)
(from x axis)
Engineering Mechanics – Chapter 2: Forces & Resultants
Fy
+ ve
Rx
Rx
Rx
Rx
Ry
R
Ry Ry
R
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R
Ry
Example 2.2
Four forces act on a bolt as shown. Determine the
resultant force.
Solution:
y
Fx components in the x direction
80 N
200
150 N
30 0
15 0
x
100 N
110 N
F1x = +150 cos 30 0 = 129.9 N
F2x = +100 cos 15 0 = 96.6 N
F3x = - 80 cos 70 0 = - 27.4 N
F4x = 0 N, (for 110 N force)
i.e. Fx = Rx = 150 cos 30 0 + 100 cos 15 0 - 80 cos 70 0
= 129.9 + 96.6 – 27.4 + 0
Rx = 199.1N
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80 N
0
y
20
150 N
30 0
15 0
x
100 N
110 N
Fy components in the y direction
F1y = +150 sin 30 0 = 75 N
F2y = - 100 sin 15 0 = - 25.9 N
F3y = + 80 sin 70 0 = 75.2 N
F4y = - 110 N (it is a Fy or
vertical force)
i.e. Fy = Ry = 150 sin 30 0 - 100 sin 15 0 + 80 sin 70 0
= 75 – 25.9 +75.2 - 110
Ry = 14.3 N
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80 N
0
y
20
150 N
30 0
15 0
The magnitude of R is
x
R =  ( Rx2 + Ry2)
R =  ( 199.12 + 14.32)
= 199.6 N
100 N
110 N
The angle , R makes with the horizontal is
 = tan-1 Ry / Rx
Since Rx
= tan-1 14.3 / 199.1
= 4.10
and Ry
are positive values,
R = 199.6 N and  =
Engineering Mechanics – Chapter 2: Forces & Resultants
4.10
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R

Rx
Rx
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Example 2.3
The rocket’s main engine exerts a total thrust of 900,000 N parallel to the y
axis. Each of its two small side engines exerts a thrust of 22,250 N in the
direction as shown.
Determine the magnitude and direction of the resultant force exerted on
the rocket by the engines.
y
900,000 N
22250 N
22250 N
31 0
Engineering Mechanics – Chapter 2: Forces & Resultants
16 0
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Solution:
Fx = Rx = 22,250 sin 16 – 22,250 sin 31
= - 5326.67 N
Fy = Ry = -22,250 cos 16 – 22,250 cos 31 – 900,000
= - 21388 – 19072 – 900,000
= - 940460 N
R = (Rx2 + Ry2) = (5326.672 + 9404602) = 940475 N
tan  = Ry / Rx = 940460 / 5326.67 = 176.55
 = 89.68 0
R
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Special situations:
1. Resultant’s direction is along the x-axis (Horizontal)
Fy = Ry = 0
Fx = Rx = R
2. Resultant’s direction is along the y-axis (Vertical)
Fx = Rx = 0
Fy = Ry = R
Example 2.4
A bracket is being pulled by 3 forces as shown. Determine the
angle  and the magnitude of force P if the resultant’s direction is
along the y-axis and its magnitude is 20 kN.
y
P kN
10 kN
60
0

8 kN
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Since the resultant’s direction is along the y-axis.
i.e Fx = Rx = 0
(No Rx component)
and Fy = Ry = 20x103 N ( R must be = Ry)
y
P kN
10 kN
60

0
8 kN
Fx = Rx = 0
(10x103)(cos 60o) + (8x103) – (Px103)(cos ) = 0
and Fy = Ry = 20x103 N
(10x103)(sin 60o) + (Px103)(sin ) = 20
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x
(1)
(2)
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From equation (1),
P cos  = 13
(3)
From equation (2),
P sin  = 11.34
(4)
Equation (4)/(3),
P sin / P cos  = (11.34 / 13)
i.e. tan  = 11.34 / 13 = 0.8723
 = tan -1 (0.8723)
y
P kN
 = 41.1 0
10 kN
60

0
8 kN
To find magnitude of P, substitute  = 41.1 0 into
equations (1) or (2).
P cos  = 13
P = 13 / cos  = 17.25 (P kN = 17.25 kN)
End of Chapter 2
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x