Namas ChandraIntroduction to Mechanical engineering Chapter 9-1

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CHAPTER 9

Statics, Dynamics, and Mechanical Engineering

Namas ChandraIntroduction to Mechanical engineering Chapter 9-2

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Objectives

Understand the concept of a VectorWrite a vector in component formUnderstand the concepts of a couple and momentConstruct a free-body diagram of a physical system

Sections

9.1 Introduction9.2 The Concept of a Vector9.3 Forces, Couples, and Moments9.4 Equilibrium and Free-body Diagrams9.5 Frictional Forces9.6 Motion of a Rigid Body

Namas ChandraIntroduction to Mechanical engineering Chapter 9-3

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9.1 Introduction

Statics deals with bodies (Solid, Liquid, Gases) that are in equilibrium with applied forces. At static equilibrium, a body is at rest or in constant motion (no acceleration).

Newton’s Law First: A body is at rest or in uniform motion (constant

velocity) along a straight-line unless acted upon by some external (unbalanced) forces.

Second: Every body accelerates in the direction of external forces. The acceleration is proportional to the magnitude of the force.

Third: For every action, there is an equal and opposite reaction.

Statics is critical in the study of bodies under forces in the design. Free body diagram isolates forces/bodies and is a critical tool.

Namas ChandraIntroduction to Mechanical engineering Chapter 9-4

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9.2 Concept of a Vector

While scalar has only magnitude, e.g.: mass, temperature -vector has magnitude and direction, e.g. velocity, force, moment

The above is a geometric representation of the magnitude of the vector is unity, it is called a unit vector

Head

Tail

Magnitude of the vector

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9.2.1Component of Vector

x y

x y

x y z

2 2 2x y z

F F F

ˆ ˆˆ =F i+F j (2-D)

ˆ ˆ ˆ =F i+F j+F k (3-D)

ˆ ˆ ˆi, j

Magnitude o

, k are unit vectors

F F F + F

f F

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9.2.2Component of Vector

x

y

If F=22 N is being pulled at 30 , then

F = F sin = 22 sin = 11N

F = F cos = 22 cos = 11 3N

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9.2.3Direction Cosines of Vector

x

y z

yx z

Let , , be the angle of vector F has with coordinate axis

projection of F on x-axiscos =

magnitude of F

F =

F

F F cos = ; cos =

F F

FF Fˆ ˆ ˆ ˆ ˆUnit vector c i + j + k = cos i+cos jF F F

2 2 2

ˆ+cos k

Note that cos cos cos 1

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9.2.4Addition/Subtraction of Vectors

Since vectors have magnitude and directions only components can be added

2

3 1 2

ˆ ˆFor example, adding 2i to 3j directly as (2+3) makes no sense

So always resolve components and then add

Example:

ˆ ˆ ˆ ˆ ˆ ˆˆ F 3i + 5j +8k F 2i + 4j + 5k

ˆ ˆˆ ˆ ˆ F F + F = i + 9j + 13

4 1 2

k̂

ˆ ˆ ˆˆ ˆ ˆ F F - F = 5i + j + 3 k

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9.3 Forces, Couples, and Moments

Magnitude of a couple: M = Fd Moment of a force M = Force x Lever arm The direction of the moment is normal to the plane containing the

force and the arm.

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9.4 Equilibrium and Free-Body Diagram

For static equilibrium, using Newton’s Second Law

Free body diagrams show all the forces and boundary conditions on a given body.Use the following steps

Isolate the body from surroundings and draw a simple sketch

Pick the right coordinate system (2D, 3D, Cartesian, Cylindrical)

Add all extreme forces (right direction) and internal forces Label all known quantities (magnitude/direction)

F 0 and M = 0

Namas ChandraIntroduction to Mechanical engineering Chapter 9-11

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9.5 Frictional forces

A body on a surface (horizontal or sloped) is acted upon by weight acting vertically down and a reaction normal to the surfaceUsing Coloumb’s Law for Static case

For bodies in motion,

s s

s

s

f N

f = frictional force

Static coefficient of friction

N = Normal force

k k

k s

s k

f N

Kinetic frictional coefficient, usually lower than

and do not depend on the force, only on the pair of surfaces

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9.5.2 Frictional Forces

s

x x

y y

0

Find: Force required to push the 15 kg crate up the hill

Solution:

F 0

M 0

Draw the free body diagram

For sliding f = N = .3 N

F m a P-m g sin 30 + f =0

F m a -m g cos 30 =0

Now M

N

0 f (0.1)+N (x) = 0; x= 0.03

Given 15 Kg crate up 30 degree slope with 0.3s

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9.5.2 Frictional Forces..2

2

m N=m g cos 30 = 15 kg 9.81 cos 30 = 127.4 N

s P=m g sin 30 + f P=111.8 N

Given 15 Kg crate up 30 degree slope with 0.3s

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9.6 Motion of Rigid Body

2 2

2

If F 0,

then there is unbalanced force leading the acceleration and hence motion

Now

F m a

M I

I - moment of inertia (kg m or slug ft )

rad - angular acceleration

s

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9.6.1 Crate Problem Revisited

s

s

y

x x

x

= 0.3

Let 0.3

F N - m g cos 0

N=m g cos

F -m g sin + f = m a

f= N = 0.3N = 0.3 (m g cos 30)

Thus, a = -g sin 30 + (0.3) g cos 30

f N N

x 2

2

= g (0.3 cos 30-sin 30)

m a = 9.81 (0.3 cos 30-sin 30)

secm

=2.36 sec

If the friction =0.3 compute the dynamics of motion?

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CHAPTER 9…concludes

Statics, Dynamics, and Mechanical Engineering