Chapter 1_ Force Systems _part A

8/3/2019 Chapter 1_ Force Systems _part A

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CHAPTER 1

FORCE SYSTEMS

(Part A)

Topics

Basic Concepts

Force

Moment

Couples Resultant of force systems

Problems in 2-and-3 Dimensions



EME 1016 Applied Statics Chapter 1: Force Systems

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1.1 Basic Concepts (Fundamentals)

The following concepts and definitions are basic to the study of mechanics:

1.1.1 What is Mechanics?

Mechanics is the physical science that concerns the conditions of rest or motion of bodies

under the action of forces. In simple words it is the study of forces and their effects.

• Physical science that deals with the effects of forces on object.

• Study of the state of rest or motion of the bodies under the action of forces.

Note: Static can be regarded as a special case of dynamics in which the acceleration iszero (i.e. constant velocity)

Statics = Dynamics with zero acceleration (constant velocity)

The mechanics of rigid bodies is subdivided into two areas: Statics and Dynamics.

Statics deals with the equilibrium of bodies either at rest or move with constantvelocity; whereas

Dynamics deals with bodies in accelerated motion. Statics can be regarded as a

special case of dynamics, in which the acceleration is zero.




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1.1.2 Fundamental Concepts (Terms and Terminology)

A particle: *- It is a body whose size is very small (zero volume) that can be

analyzed as a mass concentrated at a point.For example, the size of the earth is insignificant compared to the size of its orbit, and

therefore the earth can be modeled as a particle when studying its orbit motion.

Rigid body: *- It is a combination of a large number of particles and these particlesremain at a fixed distance from one another both before and after applying a load, i.e.a body is considered rigid when the change in distance between two points is

negligible. In short, RIGID BODY CANNOT DEFORM.

In most cases, the actual deformation occurring in structures, machines, mechanisms,

and the like are relatively small, and the rigid-body assumption is suitable foranalysis.

Concentrated force

Concentrated force: Concentrated force describes the effect of a load on a small

area of a body (acts at a point on a body). We define a load as a concentrated force

when the area (over which the load is applied) is very small compared to the overallsize of the body.

Length: Is needed to locate the position of a point in space, thereby it describes the

size of a physical system. The length in SI unit is meter, ‘m’.

ParticleHook (represent

as a particle)

Railroad wheel

(rigid body)




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Time: It is measured by the intervals between the repeatable events such as the

swings of a clock pendulum or the vibrations of a quartz crystal in a watch. In SI unit,time is measured in second, (s), minute (min), hour (hr).

Mass: The concept of mass is a measure of body inertia (which is simply the

resistance of the body to change its velocity). It can also be thought of as the quantityof matter in a body. The earth will attract two bodies of the same mass in the same

manner and they offer the same resistance to a change in transnational motion. Themass of a body affects the gravitational attraction force between it and other bodies.

(Kg).

Force: It represents the action of one body on another. A force tends to move a body

in the direction of its action. It can be exerted by actual contact or at a distance (e.g.-gravitational and magnetic forces). The action of a force is characterized by: (1) its

magnitude, (2) its direction, and (3) its point of application. Thus the force is avector quantity. In SI unit, the force is measured in (Newton, N).

Velocity and acceleration: The rate of change of position of a point is called

velocity (m/s) and the rate of change of its velocity is called as the acceleration(m/s

2).

Space: It is a geometric region that occupied by bodies whose position is defined or

located by linear or angular measurements relative to a coordinate system. For three-

dimensional problems (3-D), three independent coordinates are needed from a certainreference point (O) known as origin. For two-dimensional problems, only two

coordinates are required.

Weight: Any two particles or bodies have a mutual attractive (gravitational) forceacting between them. This was illustrated on page 3 on Newton’s Law.

In the case of a particle located at or near the surface of the earth, the gravitationalforce between the earth (of mass Me) and the particle of (mass m) is termed as the

weight of the particle (W). If r is the distance between the earth’s center and theparticle, then we have the weight of the particle is given as:

2

.

r

M mGW e

Introducing a constant g = G.Me /r2

then the weight of the particle W may beexpressed as:

W = m.g Comparing with F = m.a, we term g the acceleration due to gravity. For a point on

the surface of the earth, the value g with a reasonable accuracy can be taken as 9.81

m/s2

in SI unit and 32.17ft/sec2 in U.S. unit




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1.1.3 Newton’s laws of Motion The entire subject of rigid-body mechanics is based on Newton’s three laws of

motion.

- Newton’s first law: A particle remainsat rest or continues to move with

constant and uniform velocity (in astraight line with a constant speed) if

there is no unbalanced force acting on it.

- Newton’s second law: A particle actedupon by an unbalanced force F experiences an acceleration, (a), that has

the same direction as the force and amagnitude that is directly proportional to

the force.

Mathematically, if F is the applied force to a particle of mass m, then…………………………………………..(1.1)

- Newton’s third law: The forces of action and reaction between interacting

bodies are equal in magnitude,opposite in direction, and collinear

(they lie on same line of action).

- Newton’s law of gravitationattraction: It concerns the gravitational

attraction between any two particles. Itstates that “two particles of mass m1

and m2 are mutually attracted with

equal and opposite forces F and –F” and the magnitude of F is givenmathematically by:

2

21.

r

mmGF ……………………………………..(1.2)

Where:

F = the gravitation force between the two particles or bodies of masses m1and m2,G = the universal constant of gravitation (constant of gravity), according to

experimental evidence, G = 66.73(10-12

) m3 /(kg.s

2),

r = distance between the two particles or bodies

F = m.a




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SI units: Length in meter = m,Time in second = s,

Mass in (kilograms) = kg,Force in Newton = N,

The force unit N is derived from F = m.a. Thus 1 N is equal to a force required to

give 1kg of mass an acceleration of 1m/s2

(N=kg.m/s2). Therefore, a body of mass

1kg has a weight of 9.81N, a 2-kg body weight 19.62 N, and so on.

Prefix:

1.2- Working with vectors

1.2.1- Scalars and Vectors

Scalar: is a quantity that has magnitude only, such as mass, volume, density, speed (i.e.rpm), length and energy. A positive or negative number characterizes them. In other

words, scalars are quantities that associated with magnitude only.

Vector: A vector is a quantity that has both a magnitude and a direction. Or in otherwords, vector quantities possess a magnitude and direction. Examples of vector quantities

are force, displacement, velocity (i.e. m/s), acceleration, moment, and momentum.Speed is a scalar; it is the magnitude of velocity, which is a vector that is specified by a

direction as well as speed (magnitude).

Vectors are symbolized by boldface letter, such as V, A, B, … or

.,, B AV




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Conventions for Equations and diagrams:

A vector is represented graphically by an arrow to define its magnitude, direction,

and sense. The length of the arrow defines the magnitude of the vector, thedirection is defined by the angle between a reference axis, and the arrowhead

indicates the arrow’s line of action and the sense.

The negative of vector V is a vector -V as shown in the figure having the same magnitudeas V but in opposite direction and lies on the line of V action.

Example 1

1.2.2- Vector Operations

Multiplication and Division of a vector by a scalar:

If A is a vector and a is a scalar, then:

The product of the vector A by the scalar a = a A

Magnitude = |a A|

If a is positive, sense of aA is the same as sense of A

If a is negative sense of aA, it is opposite to the sense of A

Negative of a vector is found by multiplying the vector by ( -1 )

Law of multiplication applies: - Dividing the vector A by the scalar a = (1/a) A, a ≠ 0

Magnitude of Vector = 4 units

Direction of Vector = 20°measured

counterclockwise from the horizontal axis

Sense of Vector = Upward and to the right

The point O is called tail of the vector and

the point P is called the tip or head

Scalar multiplication and division

Vector A and its negative counterpart




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Parallelogram law and Triangle law

A and B are vectors that may be added to form a resultant vector R or vector sumR = A + B.

Using the Parallelogram law of combination,

It states that: two vectors A and B may be replaced by their equivalent vector R,

and R is a diagonal of the parallelogram that formed by A and B as its sides .

Thus the combination or resultant vector is given by the equation of vector sum as:

R = A + B

Note: R = A + B as vector but not as scalar, i.e. r a + b

Special case:

For Collinear vectors A and B which they both have the same line of action, the

parallelogram reduces to a simple algebraic or scalar addition, i.e. R = A + B

Concurrent – Vectors intersect at a particular point

Coplanar – Vectors lie on same plane




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- Vector Subtraction (a special case of addition)

The resultant difference R' between two vectors A and B may be given as:

R' = A + (-B) = A – B

- Resolution of a VectorA vector may be resolved into two components if their lines of action are known,

and this can be done by using the parallelogram, see the figure below;

1. R is a vector to be resolved into two components A and B acting along the lines a

and b,2. Extend lines from the head of R parallel to the action lines a and b until

intersection occurs with a and b, as shown in Fig.(a),3. Draw the two components A and B such that they extend from the tail of R to the

points of intersection.

Special case of vector addition:

– Resultant of two parallel noncollinear forces

– “Resultant of Resultants” – Given F1 and F2 – Draw F and – F (arbitrary magnitude and direction, but

equal magnitude, opposite direction and collinear) – Find resultants R1 and R2

– R is the resultant of F1 and F2




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Rectangular components in 2-D

It is usually convenient to deal with vector components, V1 and V2, which are mutually

perpendicular. These components are called rectangular components. Vx and Vy are

vector components in x and y direction and Vx' and Vy' are the x' and y' vector

components in x' and y' direction.The direction of the vector V with respect to x-axis is specified by the angle , where:

x

y

V

V 1tan

Example:

Fa and Fb are known two forces;their resultant Fc can be obtained using

the parallelogram as follows Draw lines parallel to the chains a and

b from the heads of Fa and Fb untilthe intersect.

If Fc along the chain c is given thenits two components Fa and Fb can be

determined in a similar manner byusing the parallelogram law. Here we

must start from the head of Fc.




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Procedure for analysis:

Addition of two forces can be solved as follows:

(1) Parallelogram LawSketching the vector addition by parallelogram law,

1. Adding two component forces (parallelogram) yielding a resultant force thatforms the diagonal of the parallelogram,

2. Resolving a force into its components along two known axes directed from the

tail of the force, construct lines from the head of the force parallel to the axes toform a parallelogram, then the sides of the parallelogram represent the

components.3. Label all the known and unknown force magnitudes and the angles.

(2) Trigonometry1. Redraw a half portion of the

parallelogram to illustrate the trianglehead-to-tail addition of the

components,2. The magnitude of the resultant force

can be obtained from the laws of

cosines, and its direction is determined

from the law of sines,3. The magnitudes of two force

components are determined from the

law of sines.

Example 2

The two forces act on a bolt at A.

Determine their resultant. SOLUTION:

• 1. Graphical solution - construct aparallelogram with sides in the samedirection as P and Q and lengths in

proportion. Graphically evaluate theresultant which is equivalent in direction and

proportional in magnitude to the thediagonal.

• 2. Trigonometric solution - use the trianglerule for vector addition in conjunction with

the law of cosines and law of sines to find

the resultant.




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1. GRAPHICAL SOLUTION:

2. TRIGONOMETRIC SOLUTION:

i. A parallelogram with sides equal to P and Q

is drawn to scale. The magnitude and

direction of the resultant or of the diagonal tothe parallelogram are measured,

R 98 N 35°

ii. A triangle is drawn with P and Q head-to-tailand to scale. The magnitude and direction of

the resultant or of the third side of thetriangle are measured,

R = 98 N = 35°

Apply the triangle rule. From the Law

of Cosines,

R = 97.73 N




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Example 3

SOLUTION:

1. Parallelogram Law

Unknown: magnitude of

FR and angle θ

2. Trigonometry

Law of Cosines

Law of Sines

DirectionΦ of FR measure from the horizontal

The screw eye is subjected to two

forces F1 and F2. Determine the

magnitude and direction of the resultantforce.




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Example 4

The ring is subjected to two forces F1 and F2. If it is

required that the resultant force have a magnitude of

1kN and be directed vertically downward, determine

a) Magnitude of F1 and F2 provided θ = 30°, andb) The magnitudes of F1 and F2 if F2 is to be a

minimum.

SOLUTION:

a) Parallelogram Law

Unknown: Forces F1 and F2

(a)

b) Minimum length of F2 occur when

its line of action is perpendicular to F1.

Hence when

F2 is a minimum

(c)

From the vector diagram

(d)

Law of Sines

b

Chapter 1_ Force Systems _part A

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