# Engineering Mechanics CHAPTER 2

```ENGINEERING MECHANICS
CHAPTER 2
FORCES &amp; RESULTANTS
By Jeff Lamoon
www.jefflamoon.com
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 1 of 35
EE05042007
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
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 &amp; Resultants
Slide 2 of 35
EE21052005
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 &amp; Resultants
8 cm
9 cm
Slide 3 of 35
EE05042007
• 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.
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 4 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 5 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 6 of 35
EE05042007
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 &amp; Resultants
Slide 7 of 35
EE05042007
Lets draw the diagram
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.
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 8 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 9 of 35
EE05042007
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 &amp; Resultants
0
Slide 10 of 35
EE05042007
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.
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 11 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 12 of 35
EE05042007
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.
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 13 of 35
EE05042007
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.
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 14 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 15 of 35
EE05042007
a) Graphical Method
8 kN
4 kN
3 kN
400
200
6 kN
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 16 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 17 of 35
EE05042007
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.
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 18 of 35
EE05042007
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.
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 19 of 35
EE05042007
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 &amp; Resultants
x
Slide 20 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 21 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 22 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 23 of 35
EE05042007
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 &amp; Resultants
Fx = ( 4.6 kN)
A
Slide 24 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 25 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 26 of 35
EE05042007
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
•
Slide 27 of 35
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 &amp; Resultants
Fy
+ ve
Rx
Rx
Rx
Rx
Ry
R
Ry Ry
R
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 28 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 29 of 35
EE05042007
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 &amp; Resultants
4.10
Slide 30 of 35
R

Rx
Rx
EE05042007
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 &amp; Resultants
16 0
Slide 31 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 32 of 35
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 33 of 35
x
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 34 of 35
x
(1)
(2)
EE05042007
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
Engineering Mechanics – Chapter 2: Forces &amp; Resultants
Slide 35 of 35
EE05042007
x
```