Mechanics 3

Mechanics-3 By Aditya Abeysinghe 1

Mechanics - 3

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

Abeysinghe


Characteristics of forces


Requirements for two forces to be equal

For two forces to be equal, they should be:1. Equal in magnitude2. Of the same direction3. Parallel (should display the

same inclination)4. In the same line of action

F F


Moment of a forceMoment 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)

dF

O

Moment around O,M = F × dDirection- Anticlockwise or Counterclockwise


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

F1

F2

F3

F4

F5

A B

CDFirst, 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.

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

6

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

F1

F2F3

F4

F5

A B

CD

Clockwise direction

A

F1

F2

d

d

Counterclockwise or anticlockwise direction

Thus, the moment around A,

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

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

GA = F2 d – F1d = (F2 - F1) d.Mechanics-3 By Aditya Abeysinghe


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

F2F3

F4

F5

A B

CDThe 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


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

F1

F2F3

F4

F5

A B

CD

45°

F1

F2

F3 Cos 45° F4

F5

A B

CD

F3 Sin 45°

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

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


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

4N8N

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

15N

3N

4N8N

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

1 N


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

5N

6N

60°

8N

5N

6N

8 Sin 60° N

8 Cos 60° N

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

Rx = 15 N

Ry = 4√3 N


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,

Rx = 15 N

Ry = 4√3 N

Rx = 15 N

Ry = 4√3 NR

Rx = 15 N

Ry = 4√3 NR

α

α

R2 = Rx2 + Ry

2 R2 = 152 + (4√3)2

Therefore, R = 16.52 NAnd, Tanα = Ry / Rx = 0.4619Therefore, α = 24.79°


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

F2F3

F4

F5

A B

CD

Consider point A

According to the principle, The moment of forces around A = The moment of the resultant of these force around A

R

α

X

From the above principle,F2 d – F1d = R x Sinα.Thus, x can be calculated.

D

AB

C


E.g.: Consider the following figure-

Taking the length of a side as 4 m,Calculate the following:i. Moment around points D and F.ii. The resultant of the systemiii. The point where the resultant intersects the system.

5 N

6 N

1 N

4 N

2 N

8 N

A B

C

DE

F


The system can be expanded as follows:5 N

4 N

6 Cos60°

6 Sin60°8 Sin60°

8 Cos60°

2 Sin60° 1 Sin60°

2 Cos60°1 Cos60°

A B

C

DE

F

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

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

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

= 14.72 N


Thus, the resultant is, R2 = RX

2 + RY2 = (3.5 N)2 + (14.72N)2

Therefore, R = 15.13 N and α = 77°To find where the resultant intersects the system, the system can be symplified as,

5 N

6 N

1 N

4 N

2 N

8 N

A B

C

DE

F

RX= 3.5 N

RY = 14.72NR

α

R

α

X

A B

C

DE

F

The moment of forces around A = The moment of the resultant of these force around A5.62 Nm = R Sinα × (4 – x) – R Cosα × 4√3 . Solving this, x = 2.03 m

Therefore, the resultant intersects 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.


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.

Weight line from step 1

Weight line from step 2The intersection point of these two lines indicates the center of mass


Centers of gravity of some basic shapes

GG

G

GG

2x

x

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


Finding the center of gravity of a combined object

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

0.1 m 0.05 m

G1G2

m1g m2gR

x1 x2

G1G2O

Taking clockwise moment about O,

m2g x2 - m1g x1 = 0

m1x1 = m2x2 x1 / x2 = m2/ m1If the mass of 1cm2 is m,Then, m1 = m × π × 102

And, m2 = m × π × 52

Therefore, m2 / m1 = 1/4

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 G1

If O is the center of gravity

And R the resultant of masses


Equilibrium of an objectBy two forces

By three forces By four or more forces

The two forces should be-1. Equal in magnitude2. Opposite in direction3. Linear or in the same

line of action

The resultant of any two forces should be –1. Equal in magnitude2. Opposite in direction3. Linear

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

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

F

To the third force

F1 F2R

F3

F1F2

F3

R


Some special definitions1. 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)


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

α

βθ

F1

F2F3

Lami’s theorem says that F1 = F2 = F3

Sinβ Sinθ Sinα


Stability of an objectUnstable 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 external unbalanced force, returns to the initial position, it is called stable equilibrium

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

EP = 0

EP = 0


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.E.g.:

mg mg

mg

mg

l

l /2

l / 4

l / 6

l /2

l

Stability of irregular shapesConsider the following object-

However, if you give a clockwise or counterclockwise moment to the

object,


R

mg

R

mg

R

mgR

mg

R

mg

Object rotated counterclockwise. Object rotates colckwise to gain initial stability

Object rotated clockwise. Object rotates counterclockwise to gain initial stability

Mechanics 3

Education

Transcript of Mechanics 3