Basics and statics of particles unit i - GE6253 PPT

ENGINEERING MECHANICSUnit – I

Basics and Statics of Particles

S.Thanga Kasi RajanAssistant ProfessorDepartment of Mechanical EngineeringKamaraj College of Engineering & Technology,Virudhunagar – 626001.Tamil Nadu, India

Email : stkrajan@gmail.com

S.ThangaKasiRajan, stkrajan@gmail.com 2

INTRODUCTION

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INTRODUCTION

To deal with the above situations, we need to know about

engineering mechanics

Mechanics is the foundation of most engineering

sciences and is an indispensable prerequisite to their study.

Mechanics is the science which describes and predicts the

conditions of rest or motion of bodies under the action of forces.

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Branches of Engineering Mechanics

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Mechanics: The actions and effects of forces on bodies.

Mechanics

Statics: Bodies at rest, or in equilibrium

Dynamics: Bodies in motion, or out of equilibrium

IN EQUILIBRIUM Will be static, OR move with constant velocity

OUT OF EQUILIBRIUM Will accelerate (velocity changing)

Velocity=0

Velocity=constant

Velocity changing with time

Engineering Mechanics

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Kinematics: how fast, how far, and how long it takes...

Kinetics: What forces were involved to produce this motion?

Dynamics: Kinematics: Study of motion without reference to forces producing motion: Relations between position, velocity, acceleration and time.

Kinetics: Relations between unbalanced forces and the changes in motion they produce.

E.g. Rollercoaster ride:

- Weight- Friction- Aerodynamic drag

What are the resulting accelerations?02/01/2017 S.ThangaKasiRajan, stkrajan@gmail.com 6

Six Fundamental Concept of Mechanics• Space - associated with the motion or the position of a point P given

in terms of three coordinates measured from a reference point or origin.

• Time - definition of an event requires specification of the time and position at which it occurred.

• Mass - used to characterize and compare bodies, e.g., response to earth’s gravitational attraction and resistance to change in translational motion.

A force is a push or pull. An object at rest needs a force to get it moving; a moving object needs a force to change its velocity.

The magnitude of a force can be measured using a spring scale.

Six Fundamental Concept of Mechanics

Six Fundamental Concept of MechanicsParticle

A particle has a mass but size is neglected.

When a body is idealised as a particle, the principles of mechanics reduces to a simplified form, since the geometry of the body will not be concerned in the analysis of the problem.

Six Fundamental Concept of MechanicsRigid Body

A combination of large number of Particles in which all the particles remain at a fixed distance from one another before and after application of load.

Here mass & size of the bodies are considered when analysing the forces.

Basic Laws of Mechanics

1. Newton’s Law

2. Newton’s Law of Gravitation

3. Triangle Law

4. Parallellogram Law

5. Lami’s Theorem

6. Principle of Transmissibility

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Basic Laws of Mechanics

• Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or continue to move in a straight line.

• Newton’s Third Law: The forces of action and reaction between two particles have the same magnitude and line of action with opposite sense.

• Newton’s Second Law: A particle will have an acceleration proportional to a nonzero resultant applied force.

• Newton’s Law of Gravitation: Two particles are attracted with equal and opposite forces,

GMgmgWr

MmGF 02/01/2017

Parallelogram Law

If two vectors acting at a point be represented in magnitude and

direction by the adjacent sides of a parallelogram, then their resultant is

represented in magnitude and direction by the diagonal of the

parallelogram passing through that point.

𝑅=√𝑃2+𝑄2+2𝑃𝑄𝑐𝑜𝑠Ө

Ө= tan−1( 𝑄𝑠𝑖𝑛 (𝛼 )𝑃+𝑄cos (𝛼 ) )

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Triangle LawIf two vectors acting at a point are represented by the two sides

of a triangle taken in order, then their resultant is represented by the

third side taken in opposite order

𝑃 𝑄

𝑅02/01/2017

sine and cosine rulesWhen adding forces it is often useful to

consider solving the problem using geometric rules, rather than considering components

The sine rule states that

sinsinsinXBA

The cosine rule states that

cos2222 BXXBA

cos2222 AXXAB

cos2222 ABBAX

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Principle of Transmissibility

According to this law the state of rest or motion of the rigid body is

unaltered if a force acting on the body is replaced by another force of

the same magnitude and direction but acting anywhere on the body

along the line of action of the replaced force.• Principle of Transmissibility -

Conditions of equilibrium or motion are not affected by transmitting a force along its line of action.NOTE: F and F’ are equivalent forces.

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Coordinate System

1. An origin as the reference point

2. A set of coordinate axes with scales and labels

3. Choice of positive direction for each axis

4. Choice of unit vectors at each point in space

Coordinate system: used to describe the position of a point in space and consists of

Cartesian Coordinate System

Vector Representation of Forces

A vector is a quantity that has both direction and magnitude.

Application of Vectors(1) Vectors can exist at any point P in space.

(2) Vectors have direction and magnitude.

(3) Vector Equality: Any two vectors that have the same

direction and magnitude, are equal no matter where in

space they are located.

Resolution and Composition of Forces

Resolution of force is a reverse process in which a single force is expressed in terms of its components. These components are sometimes referred to as the resolved parts of the force.

Y OX = F cos

OY = F sin

Splitting of forces into their components30°

Examples

X = -15 Cos 42 = -11.1 N

Y = 15 Sin 42 = 10.0 N

X = 35 Cos 62 = 16.4 N

Y = -35 Sin 62 = - 30.9 N

32°10 N

X = -10 Sin 32 = - 5.30 N

Y = -10 Cos 32 = -8.48 N02/01/2017

Resolution and Composition of Forces

F cosӨ

F sinӨ

F cosӨF cosӨ

F sinӨ F sinӨ

F sinӨ

F cosӨӨ

1. It is convenient to have Fx = F cos ӨFy = F sin Ө

and Always measure angle from horizontal reference(acute angle).

2. Assume force pointing Right and Top as positive otherwise negative

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Procedure to find Resultant Force

Procedure to find the magnitude and direction of the resultant force1. Find ƩFx

2. Find ƩFy

3. Magnitude of the resultant force is given by

4. Plot ƩFx and ƩFy with its appropriate sign5. Direction of the resultant force is given by

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Problems

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Problem No 1

Two forces act on a bolt at A. Determine their resultant.

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Problem No 1 (contd…)• Graphical solution - A parallelogram with sides

equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured,

35N 98 R

• Graphical solution - A triangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured,

35N 98 R

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Problem No 1 (contd…)• Trigonometric solution - Apply the triangle rule.

From the Law of Cosines,

155cosN60N402N60N40cos222

222 BPQQPR

2004.15

N73.97N60155sin

sinsin

N73.97R

From the Law of Sines,

04.3502/01/2017

Problem No 1 (contd…)Sl. No. ƩFx ƩFy

1. + 40 cos20 + 40 sin202. + 60 cos 45 + 60 cos 45

80.01 56.10

R = 97.72 N

Magnitude of the resultant force Direction of resultant force

Ʃ Fx = 80.01

Ʃ F y =

tan Ө = ()

Ө = 35ᴼ

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Problem 2

a) the tension in each of the ropes for = 45o,

b) the value of for which the tension in rope 2 is minimum.

A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is a 25 kN directed along the axis of the barge, determine

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Problem 2

• Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides.

kN 12.8kN 18.5 21 TT

• Trigonometric solution - Triangle Rule with Law of Sines

105sinkN 25

30sin45sin21 TT

kN 12.94 kN 18.3 21 TT02/01/2017

Problem 2• The angle for minimum tension in rope 2 is

determined by applying the Triangle Rule and observing the effect of variations in .

• The minimum tension in rope 2 occurs when T1 and T2 are perpendicular.

30sinkN 252T kN 12.5 2 T

30coskN 251T kN 21.7 1 T

3090 60

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Problem 3

Knowing that the tension in cable BC is 725-N, determine the resultant of the three forces exerted at point B of beam AB.

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Problem 3SOLUTION:• Resolve each force into rectangular components.

• Calculate the magnitude and direction.

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Problems for Practice

A disabled automobile is pulled by means of two ropes shown in fig 1. Determine the Magnitude and direction of Resultant by (a) parallelogram law, (b) Triangle law (c) analytical method

Determine the magnitude and direction of resultant force shown in fig 2.

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S.ThangaKasiRajan, stkrajan@gmail.com

Equilibrium of a Particle

2 - 35

• When the resultant of all forces acting on a particle is zero, the particle is in equilibrium.

• Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line.

• Particle acted upon by two forces:- equal magnitude- same line of action- opposite sense

• Particle acted upon by three or more forces:- graphical solution yields a closed polygon- algebraic solution

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Examples for Equilibrium

Cables AB and AC carries the spool of weight

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Problem 4Given: Sack A weighs 20 N. and

geometry is as shown.

Find: Forces in the cables and weight of sack B.

1. Apply Equilibrium condition at Point E and solve for the unknowns (TEG & TEC).

2. Repeat this process at C.

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Problem 4

+ Fx = TEG sin 30º – TEC cos 45º = 0

+ Fy = TEG cos 30º – TEC sin 45º – 20 N = 0

Solving these two simultaneous equations for the two unknowns, we get

TEC = 38.6 N

TEG = 54.6 N

Note that the assumed directions for the forces in the two cables EG and EC are tensile in nature.

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Problem 4

Fx = 38.64 cos 45 – (4/5) TCD = 0

Fy = (3/5) TCD + 38.64 sin 45 – WB = 0

Solving the first equation and then the second we get

TCD = 34.2 N and WB = 47.8 N .

Apply Equilibrium Condition

Now move on to the point C and consider equilibrium at C

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Problem 5

It is desired to determine the drag force at a given speed on a prototype sailboat hull. A model is placed in a test channel and three cables are used to align its bow on the channel centerline. For a given speed, the tension is 200-N in cable AB and 300-N in cable AE.

Determine the drag force exerted on the hull and the tension in cable AC.

SOLUTION:• Choosing the hull as the free body,

draw a free-body diagram.

• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.

• Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.

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Problem 5

• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.

0 DAEACAB FTTTR

SOLUTION:• Choosing the hull as the free body, draw a

free-body diagram.

75.1m 4m 7

375.0m 4m 1.5

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Problem 5• Resolve the vector equilibrium equation into

two component equations. Solve for the two unknown cable tensions.r

ACACAC

3009363.021.99

3512.066.173

9363.03512.0

56.20cos56.20sin

N 21.99N 66.173

26.60cosN 20026.60sinN 200

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Problem 5

3009363.021.9900

3512.066.17300

N 98.35N 214.45

3009363.021.99

3512.066.173

This equation is satisfied only if each component of the resultant is equal to zero

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Problem 6A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that a = 25ᴼ and b = 15ᴼ and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB.

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Problem 6

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Forces in Space

• The vector is contained in the plane OBAC.

F • Resolve into

horizontal and vertical components.

yh FF sin

yy FF cos

• Resolve into rectangular components

sinsin

cossincos

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Forces in Space

• With the angles between and the axes,F

kFjFiFF

FFFFFF

zzyyxx

coscoscos

• is a unit vector along the line of action ofand are the direction cosines for

zyx cos and,cos,cos

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Problem 6

The tension in the guy wire is 2500 N. Determine:

a) components Fx, Fy, Fz of the force acting on the bolt at A,

b) the angles x, y, z defining the direction of the force

SOLUTION:• Based on the relative locations of the

points A and B, determine the unit vector pointing from A towards B.

• Apply the unit vector to determine the components of the force acting on A.

• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

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Problem 6SOLUTION:• Determine the unit vector pointing from A

towards B.

m 3.94

m30m80m40

m30m80m40222

• Determine the components of the force.

N 795N 2120N1060

318.0848.0424.0N 2500

318.0848.0424.03.94

303.94

803.94

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Problem 6• Noting that the components of the unit vector are

the direction cosines for the vector, calculate the corresponding angles.

kji zyx

318.0848.0424.0

coscoscos

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Problem 7

A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 3750 N.

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Problem 7Net Force acting at point A is

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Problem 7

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Problems for practice

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Problems for practice

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References1. Ferdinand P Beer & E.Russell Johnston “VECTOR MECHANICS FOR ENGINEERS STATICS & Dynamics”, (Ninth Edition) Tata McGraw Hill Education Private Limited, New Delhi.2. Engineering Mechanics – Statics & Dynamics by S.Nagan, M.S.Palanichamy, Tata McGraw-Hill (2001).

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Thank you

Any Queries contact

S.Thanga Kasi RajanAssistant ProfessorDepartment of Mechanical EngineeringKamaraj College of Engineering & Technology,Virudhunagar – 626001.Tamil Nadu, India

Email : stkrajan@gmail.com

02/01/2017

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