Mechanics 3

Mechanics - 3
For more information about Newton’s
laws and applications search the
presentation- Mechanics-1 and
Mechanics-2 By Aditya Abeysinghe

1

Characteristics of forces


2

Requirements for two forces to be
equal
F

F

For two forces to be
equal, they should be:
1. Equal in magnitude
2. Of the same direction
3. Parallel (should display the
same inclination)
4. In the same line of action

3

Moment of a force
Moment of a force around a fixed distance is the
product of the force and the perpendicular
distance between the force and the fixed point.
That is if the force is F , the perpendicular
distance between the force and the fixed point
is d,
Then, the moment = F × d (Note:Moment is a vector
quantity – should also consider direction in calculations)
Moment around O,
F M=F×d
d
Direction- Anticlockwise
O
or
Counterclockwise 4

E.g.:
Find the total moment around A in the following
figure.
F1

D

F5

C

F2

F3

A

F4

B

Thus, the moment around
point A is,
GA = F2 d – F1d = (F2 - F1) d

First, since forces F3 , F4 and F5 go through A,
the distance between these forces and A is
zero. Thus, the moment of forces around A is
also zero.So, these can be ignored when
calculating the moment around A.

Second, we should categorize which forces create
a clockwise moment around A and which forces
create a counterclockwise or anticlockwise
moment around A.
Then, one type of moment should be taken as
positive and the other type as negative when
adding the moments.

In this example, I have taken the clockwise
direction as positive.
Consider ABCD to be a square with a side’s
length d.

5

If you are having trouble with the
direction, consider ABCD as follows:
F1

D

F5

F3

C

Counterclockwise or
anticlockwise direction
F1

F2
d

A

F4

F2

B

Thus, the moment around A,

A

d

GA = F2 d (Clockwise) + F1d (anticlockwise or
counterclockwise)

Clockwise
direction

Taking clockwise direction as positive,
GA = F2 d (Clockwise) - F1d (Clockwise)
Thus, the moment around A is,

GA = F2 d – F1d = (F2 - F1) d.


6

However, if you take the moment around D in
either direction, you will also have to consider the
inclined force, F3. In such a scenario, the moment
can be calculated using two ways.
1. By taking the perpendicular distance into
account and then calculating the moment.
F1

D

F5

F3

A

F4

C

F2

The perpendicular distance between F3
and D can be found to be d/√2.
Thus, the clockwise moment around D is,
GD = F2d – F4d – F3d/ √2

B

7

2. Dividing the inclined force into its vertical and
horizontal constituents and then taking the
moment around D.
F1

D

F5

F3

C

F2

F5

45°
A

F1

D

C

F2

F3 Sin 45°
F4

B

A F3 Cos 45°

F4

B

Since F3 Sin 45° goes through D, its effect can be minimized.

Thus, the clockwise moment around D is,
GD = F2d – F4d – F3 Cos45° d
Therefore, GD = F2d – F4d – F3d/ √2

8

The resultant force
A resultant force is any single force which can
replace a set of forces.
For example, consider the following two systems.
3N

3N

4N

8N

Since, all the forces are in the same
direction, their sum can be replaced by
their resultant(15 N) as below

4N

8N

When forces are not in the same
direction, their vector should be
considered.

1N
15N Mechanics-3 By Aditya Abeysinghe

9

If an inclined force is given, it should be divided
into its horizontal and vertical components and
then the resultant should be calculated.
8N

8 Sin 60° N
5N

5N
60°

8 Cos 60° N
6N

6N
Thus, the sum of the forces in the
horizontal direction ,
Rx = 5 + 6 + 8 Cos 60° = 15 N

The sum of the forces in the
vertical direction,
Ry = 8 Sin 60° = 4√3 N

Ry = 4√3 N
Rx = 15 N Mechanics-3 By Aditya Abeysinghe

10

Now we want to find the single force (resultant force) which can stand along
from the effect of these two forces.
It is clear that, if an inclined force is given, it should be divided into its
horizontal and vertical components and then the resultant should be
calculated.
Conversely, horizontal and vertical forces can be reduced to another inclined
force, which can stand out with the same effect produced by the two
individual and perpendicular forces.
Thus, the above problem can be simplified to,
Ry = 4√3 N
R

Ry = 4√3 N

α

Rx = 15 N

R2 = Rx2 + Ry2
R2 = 152 + (4√3)2
Therefore, R = 16.52 N
And, Tanα = Ry / Rx =
0.4619
Therefore, α = 24.79°

R

Rx = 15 N

Ry = 4√3 N
α
Rx = 15 N

11

Principle: The total moment exerted by the individual
forces around a point is equal to the the moment
exerted by the resultant of these forces around that
point.
Consider the following diagram:
F1

D

F5

F3

A
Consider point A

F4

C

C

D
R

F2

B

α

A

B
X

According to the principle,
From the above principle,
The moment of forces around A =
F2 d – F1d = R x Sinα.
The moment of the resultant of these
Thus, x can be calculated.
force around A

12

E.g.: Consider the following figureE

D
5N
8N

6N

F

C
2N
A

1N
4N

B

Taking the length of a side as 4 m,

Calculate the following:
i. Moment around points D and F.
ii. The resultant of the system
iii. The point where the resultant intersects the system.

13

The system can be expanded as follows:
E

5N

D

8 Sin60°

6 Sin60°
8 Cos60°

6 Cos60°
C

F
2 Sin60° 1 Sin60°

2 Cos60°

1 Cos60°
A

4N

B

Rx = 4 + 1 Cos60° - 6 Cos60° -5 + 8 Cos60°
- 2 Cos60°
= 3.5 N

Thus, the clockwise moment about D =
-(8 Cos60° × 2√3 ) + (8 Sin60° × 2) +
(2 Sin60° × 4) + (2 Cos60° × 4√3) – (4 × 4)
– (1 Cos60° × 4√3)
= -5.62 Nm (counterclockwise direction)

Therefore, moment about D =
5.62 Nm
Thus, the counterclockwise moment about F =
( 5 × 2√3) + (4 × 2√3) + 6Sin60° × 8 +
(1 Sin60° × 6) + (1 Cos60° × 2√3)
= 79.67 Nm(counterclockwise direction)

Therefore, moment about F = 79.67
Nm

Ry = 2 Sin60° + 1 Sin60° + 6 Sin60° +
8 Sin60°
= 14.72 N

14

RY = 14.72N

R

Thus, the resultant is,
R2 = RX2 + RY2 = (3.5 N)2 + (14.72N)2
Therefore, R

α

RX= 3.5 N

= 15.13 N and α = 77°

To find where the resultant intersects the system, the system
can be symplified as,
D
E

E

D
5N
8N

6N

F
F

R

C
2N
A

1N
4N

C

α

A
B

The moment of forces around A =
The moment of the resultant of these force around A
5.62 Nm = R Sinα × (4 – x) – R CosαMechanics-3Solving Abeysinghe 2.03 m
× 4√3 . By Aditya this, x =

B

X

Therefore, the
resultant intersects
15
2.03m right of A.

Center of mass
The center of mass on any object is any point where
the total mass of the object is concentrated.
The center of mass of a simple laminar body can be
calculated as follows:
Step 1:
Tie the body to a fixed point so that the tesion of the string
is equal to the weight of the body.
Step 2:
Rotate the object by some degrees either in the clockwise
direction or in the counterclockwise/anticlockwise
direction and follow step 1.

16

Weight line from step 1

Weight line from step 2

The intersection point of these
two lines indicates the center
of mass

Thus, center of mass can also be described as the
equilibrium point in which a particular object can
be balanced in one or more ways.

17

Centers of gravity of some basic
shapes
G
G
G

x
G

G

2x
Unlike in other shapes, in a triangle the
center of gravity lies in a 2:1 ratio from its
weight/median line.

18

Finding the center of gravity of a
combined object
E.g.: What is the center of gravity of the following object.

0.1 m
G1

0.05 m

If O is the
center of
gravity

G2

x2

x1
G1

And R the
resultant
of masses

O

G2

Taking clockwise moment about O,

m2g x2 - m1g x1 = 0
m1x1 = m2x2
x1 / x2 = m2/ m1
Therefore, x1 / x2 = ¼. But, x1 + x2 = 15 cm.
Therefore, x1 = 3cm and x2 = 12 cm.
Thus, the center of gravity lies 3cm right of
G

m1 g

R

m2 g

If the mass of 1cm2 is m,
Then, m1 = m × π × 102
And, m2 = m × π × 52
Therefore, m2 / m1 = 1/4
19

Equilibrium of an object
By two
forces

By three forces

By four or
more forces

The two forces should be1. Equal in magnitude
2. Opposite in direction
3. Linear or in the same
line of action
F

The resultant of any two forces should
be –
1. Equal in magnitude
To the
2. Opposite in direction
third
3. Linear
force

Resultant of the all of the
forces should be zero.
(∑ R = 0)

F1

R

F2

F1

R

The moment around any
point should be zero.
F2 (∑ (F× d) = 0 )

F
F3

F3

20

Some special definitions
1. Coplanar forces
Coplanar forces are forces that lie in the same
plane.
2. Concurrent forces
Are three or more forces which meet at a
common point or insect each other at the same point.
3. Collinear forces
Are forces that occupy the same line, either
parallel or not parallel to each other.(line of action is
same)

21

Lami’s Theroem
Lami’s theorem explains how three
coplanar, concurrent and non-collinear forces
are kept in equilibrium.
F1

Lami’s theorem says that
F1 =
F2 = F3
Sinβ
Sinθ
Sinα

θα
β
F3

F2


22

Stability of an object
Unstable equilibrium

Stable equilibrium

Neutral equilibrium

If an object, when given a
external unbalanced force,
decreases its potential energy
and its stability, it is called
unstable equilibrium.

If an object, when given a
returns to the initial
position, it is called stable
equilibrium

If an object, when given an
does not change its potential
energy, but maintains the
initial potential energy, it is
called neutral equilibrium.

EP = 0

EP = 0


23

Stability of a set of objects at the
corner of another large object
The maximum distance an object can be kept
without falling in gravity is by keeping the object
at its center of gravity.
l
E.g.:
l

l /2
l/4
mg

l/6

l /2
mg

mg

mg

24

Stability of irregular shapes
Consider the following objectHowever, if you give a clockwise
R
or counterclockwise moment to
the object,
R
R

mg

Object
rotated
counterclock
wise. Object
rotates
colckwise to R
gain initial
stability


Object
rotated
clockwise.
Object
rotates
countercl
ockwise
to gain
initial
stability

mg

mg

mg

mg

R

25

Mechanics 3

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