Statics of a Particle

8/11/2019 Statics of a Particle

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Statics of a Particle


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Statics

Statics is the study of the effect of external forceson structures or

components which are in a state of static equilibrium.

Statics allows the calculation of internal forceswhich are acting on various

parts of a structure or on individual components of an assembly of parts.

For structural integritypurposes it is essential that the forces being carried

by individual parts of a structure or assembly are known

Once the forces are known, further calculations can be carried out to

determine the extent of any deformationand magnitude of any resultingstressesthat are generated in the parts

Decisionscan then be made as to the structural integrity of the structure

or part based on limiting valuesof deformation and/or stress


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Definitions

Particle: defined as a body possessing matter but of no significant

dimensions.

Rigid body: defined as an assembly of particles, the distance between

any two of which remains fixed (hence no deformation).

Force : defined as the 'action or cause which impresses or tends to

impress motion and/or deformation on matter'.

Coplanar forces: sets of forces having lines of action lying in one plane.


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Definitions

Forces encountered in practice are usually found to be applied to a 'solid

body' as a whole or over a finite area of the bodies boundary.

Initially, however, consider only the forces acting on a single particle.

The notion of a particle provides a convenient starting point for study, as

the results of the analysis of particle behaviourcan then be extended to a

finite rigid body.

In what follows, important concepts of resultant force, force componentsand equilibriumwill be defined.


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Addition of forces

Consider the following situations:

In all three cases there are two externally applied forces which, if added

according to the usual rules, would sum to 15 N. It is evident that in case

(a) the netforce is only 5 N (10-5) and that in case (b) the netforce is 15

N (10+5). Cases (a) and (b) are special in that the external forces have the

same line of action only in case (a) they are in opposite directions.

Case (c) is a more general situation where the lines of action of the two

forces are at an angle to each other. In this situation the forces must be

added using the parallelogram of forces.


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Parallelogram of forces

Consider a point A on which is acting two forces Pand Qin the directions

AB and AC respectively. The lengthof the lines AB and AC are inproportionto the magnitudes of the forces Pand Q. The two forces may

be replaced by a single force of magnitude R(known as the RESULTANT)

which is represented in bothmagnitude and direction by the diagonal AD

of the parallelogram ABCD as shown below.

R represents the sumof the forces Pand Q.

NOTE: ALLvector quantities may be added in this way.


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The parallelogram may be simplified to a triangle by taking the upper (or lower)

half. Note that the forces Pand Qgo 'tail-to-head' in any order and Ris the'closing vector' of the triangle.


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Resultant force

The resultantof any number of coplanar forces acting simultaneously on a

particle is the single force that is equivalentto that force set.

The three forces F1, F

2and F

3can be replaced by the resultant force F

R.


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Addition of more than two forces

The resultant of a coplanar force set can be found graphically by using the'polygon law'. By placing the force representations 'tail to head' in any

order, the resultant is then represented by the line joining the initial

point to the terminalpoint.

See example 2.2, p18, Ryder & Bennett.


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Force components

For ease of analysis forces may be resolved into two or more force

componentsand still have the same effect.

Coplanar component forces

Here the force is resolved into two components which are perpendicular

to each other and are coplanar with the actual force.

Fx = Fcos

Fy = Fsin


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From consideration of the figure below and the polygon law it follows that the

x-component of the resultant Rof any number of such forces is the sum of the

x-components of the individual forces; and similarly for the y-components.


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Thus, Rx = F1cos1 + F2cos

2

= F1x

+ F2x

Rx = F

x

and similarly,

Ry = F

y

Also, if the sums of the x-components and y-components are known for a

set for forces, the resultant can be determined in magnitude and

direction as follows:

)+ R(RR = yx 22

and

x

y

RR

R =tan


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Equilibrium of a Particle

If the resultant force on a particle is zerothen the particle is said to be in

EQUILIBRIUM.

On consideration of Newtons 1st Law, it follows that a particle in

equilibrium is one which is either at rest or moving with constant speed in

a straight line.

In practice, we may want to calculate unknown forces acting on a particle

which is known to be in equilibrium. This can be done knowing that the

equivalent resultant force acting on the particle is zero for equilibrium to

exist. To determine the unknown forces two conditions can be used:


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(i)

If the forces acting on a particle are summed by the polygon law, then if the

resultant is zero the polygon must close. The graphical condition: 'for a

particle to be in equilibrium the force polygon must close'.

(ii) Since the magnitude of the resultant |R| = [Fx)2+ (Fy)

2] then for R= 0

the summations Fxand Fymust both be zero. The analytical

condition: 'for a particle to be in equilibrium Fx= 0 and Fy= 0'.

'for a particle to be in equilibrium the force polygon must close'

'for a particle to be in equilibrium Fx = 0 and Fy = 0'


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Weight of a Particle

The weight of a particle is the vertical force experienced by the particle

from the attraction of the Earth. This gravitation force can be found by

considering Newton's 2nd Law, where, if a particle mass m(kg) is being

accelerated under the action of a force F(N), the acceleration a(m/s2) is

given by F= ma. Therefore, any particle falling freely under the action of

its weight alonedescends with acceleration magnitude g, (Galileo).

Therefore we can write,

W = mg(N)

The value of gmay vary slightly from one locality to the next, but normally

a value of 9.81 m/s2is adopted.


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Problem Solution

To obtain solutions to problems where the body considered is assumed to

be a particle, further assumptions can be made.

(1) If a force is applied by means of a massless cord then the direction of

the force and its line of action coincide with the cord.

(2) If a cord passes over a smooth pulley the two forces exerted by the

cord on the pulley are equal in magnitude, this magnitude being

referred to as the TENSION in the cord.

(3) If a body is in contact with a smooth surface the force of the surface

on the body is in the direction normalto the surface.


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Statics of a Particle