COURSE CONTENT IN BRIEF 1.Introduction. 2.Resultant of concurrent and non-concurrent coplanar forces. 3.Equilibrium of concurrent and non-concurrent coplanar

COURSE CONTENT IN BRIEF1. Introduction.

2. Resultant of concurrent and non-concurrent coplanar forces.

3. Equilibrium of concurrent and non-concurrent coplanar forces.

4. Analysis of plane trusses.

5. Friction.

6. Centroid and Moment of Inertia.

7. Resultant and Equilibrium of concurrent non-coplanar forces.

8. Rectilinear and Projectile motion.

9. Newton’s second law, D’Alembert’s principle, banking and super elevation.

10. Work, Energy, and Power.

11. Impulse- Momentum principle.www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 1

Books for Reference

1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons.

2.Engineering Mechanics, by Irwing Shames, Prentice Hall of India.

3.Mechanics for Engineers, by Beer and Johnston, McGraw Hills Edition

4.Engineering Mechanics, by K.L. Kumar, Tata McGraw Hills Co.

www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 2

Axioms of Mechanics

(1) Parallelogram law of forces : It is stated as follows : ‘If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then the resultant of these two forces is represented in magnitude and direction by the diagonal of the parallelogram passing through the same point.’

B C

AO

P2

P1

R

10

Contd..www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 3

In the above figure, P1 and P2, represented by the sides OA and OB have

R as their resultant represented by the diagonal OC of the parallelogram OACB.

B C

AO

P2

P1

R

It can be shown that the magnitude of the resultant is given by:

R = P12 + P2

2 + 2P1P2Cos α

Inclination of the resultant w.r.t. the force P1 is given by:

= tan-1 [( P2 Sin ) / ( P1 + P2 Cos )]

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Contd..


(2) Principle of Transmissibility : It is stated as follows : ‘The external effect of a force on a rigid body is the same for all points of application along its line of action’.

PA B

P

For example, consider the above figure. The motion of the block will be the same if a force of magnitude P is applied as a push at A or as a pull at B.

The same is true when the force is applied at a point O.

P P

O

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(3) Newton’s Laws of motion:

(i) First Law :

If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant speed in a straight line (if originally in uniform motion). (ii) Second Law :

If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant and in the direction of this resultant i.e., F α a ,or F = m.a , where F, m, and a, respectively represent the resultant force, mass, and acceleration of the particle. (iii) Third law:

The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

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Note :

1. ‘Axioms’ are nothing but principles or postulates that are self – evident facts which cannot be proved mathematically but can only be verified experimentally and/or demonstrated to be true.

2. The three basic quantities of mechanics are length, time, and force. Throughout this Course we adopt SI units and therefore they are expressed in meters, seconds, and Newtons, written as m, s, and N respectively.

3. The ‘external effect’ of a force on a body is manifest in a change in the state of inertia of the body. While the ‘internal effect’ of a force on a body is in the form of deformation.


RESULTANT OF CONCURRENT COPLANAR FORCES

CHAPTER – 2

Y-Direction

X-DirectionF3

F1

R

Fx

Fy

F2

In the above diagram F1, F2, F3 form a system of concurrent

coplanar forces. If R is the resultant of the force system, then its magnitude and direction are given by:

Composition of forces and Resolution of force Resultant, R : It is defined as that single force which can replace a set of forces, in a force system, and cause the same external effect.


(i) Magnitude, R = (Fx)2 + (Fy)

2

(ii) Direction, θ = tan –1(Fy / Fx) , where:

ΣFx = Algebraic summation of x-components of all individual forces.

ΣFy = Algebraic summation of y-components of all individual forces.

θ = Angle measured to the resultant w.r.t. x-direction.

The process of obtaining the resultant of a given force system is called ‘Composition of forces’.

Note: The orientation of x-y frame of reference is arbitrary. It may be chosen to suit a particular problem.

Contd..


Component of a force, in simple terms, is the effect of a force in a certain direction. A force can be split into infinite number of components along infinite directions. Usually, a force is split into two mutually perpendicular components, one along the x-direction and the other along y-direction (generally horizontal and vertical, respectively). Such components that are mutually perpendicular are called ‘Rectangular Components’.

Component of a force :

Fy

Fig. 1Fx

FFy

F

FxFig. 2 Fig. 3

FFy

Fx

The process of obtaining the components of a force is called ‘Resolution of a force’.


The adjacent diagram gives the sign convention for force components, i.e., force components that are directed along positive x-direction are taken +ve for summation along the x-direction.

Sign Convention for force components:

+ve

+vex

xy

y

Also force components that are directed along +ve y-direction are taken +ve for summation along the y-direction.

When the components of a force are not mutually perpendicular they are called ‘Oblique Components’. Consider the following case.

Oblique Components of a force:


taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force F, the sense of the same is defined by its tail at the tail of the first force and its tip at the tip of the second force’.

F

F1

F2Let F1 and F2 be the oblique components of a

force F. The components F1 and F2 can be

found using the ‘triangle law of forces’, which states as follows: ‘If two forces acting at a point can be represented both in magnitude and direction, by the two sides of a triangle

F

F1

F2

F1 / Sin = F2 / Sin = F / Sin(180 - - )

Contd..


EXERCISE PROBLEMS1. A body of negligible weight, subjected to two forces F1=

1200N, and F2=400N acting along the vertical, and the horizontal respectively, is shown in Fig.1. Find the component of each force parallel, and perpendicular to the plane.

Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N

FIG. 1

= 1200 N

X

34

Y

F2

F1

= 400 N


2. Determine the X and Y components of each of the forces shown in

(Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N,

F3X = -306.42 N, F3Y= -257.12N )

30º40º

12

5

300 N

390 N

400 N

X

Y

F1 =

F2 =

F3 =

FIG. 2

FIG.2.

21


22

600N

200N

800N

20º 40º

30º

FIG. 3

3. Obtain the resultant of the concurrent coplanar forces shown in FIG.3

(Ans: R = 522.67 N, θ = 68.43º)www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 15

4. A disabled ship is pulled by means of two tug boats as shown in FIG. 4. If the resultant of the two forces T1 and T2 exerted by the ropes is a 300 N force acting parallel to the X – direction, find :

(a) Force exerted by each of the tug boats knowing α = 30º.

(b) The value of α such that the force of tugboat 2 is minimum,

while that of 1 acts in the same direction.

Find the corresponding force to be exerted by tugboat 2.

( Ans: a. T1= 195.81 N, T2 = 133.94 N

b. α = 70º, T1 = 281.91 N, T2(min) = 102.61 N )

T2

R = 300 N

T1

α

20º

FIG. 4

X - direction

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5. An automobile which is disabled is pulled by two ropes as shown in FIG. 5. Find the force P and resultant R, such that R is directed as shown in the figure.

P

Q = 5 kN

R20º

40º

FIG. 5

(Ans: P = 9.4 kN , R = 12.66 kN)

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6. A collar, which may slide on a vertical rod, is subjected to three forces

as shown in FIG.6. The direction of the force F may be varied .

Determine the direction of the force F, so that resultant of the three forces is horizontal, knowing that the magnitude of F is equal to

(a) 2400 N, (b)1400N

( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)

1200 N

800 N60º

θ

F

ROD

COLLAR

FIG.6

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7. Determine the angle α and the magnitude of the force Q such that the resultant of the three forces on the pole is vertically downwards and of magnitude 12 kN. Refer Fig. 7.

8kN5kN

Q30º

α

Fig. 7(Ans: α = 10.7 º, Q = 9.479 kN )

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Documents

COURSE CONTENT IN BRIEF 1.Introduction. 2.Resultant of concurrent and non-concurrent coplanar forces. 3.Equilibrium of concurrent and non-concurrent coplanar