chapter 2 force

7/30/2019 Chapter 2 Force

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Statics

react on rigid bodies which are at rest or not

in motion. This study is the basis for the

engineering principles, which guide the

design of all structures, since before we can

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eg n to es gn any structure we must rst

know the forces applied to it.

Forces

,

have both magnitude and direction.

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Resultant Forces

,

advantageous to replace all the forces,

which act on a body with a single force.

This single force, called a resultant force,

must produce the same movement and

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effects that all the original forces would

produce on the body.

Resolution of Forces

y

F

sinF

4

x

cosF


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Force Polygons

placing the two vectors (which act at the

same point) in a tip-to-tail fashion and

completing the triangle with a vector.

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a

b

a

b

c

The resultant vector (c) is the sum of the original two

vectors (a and b) and therefore replaces the original two

vectors.

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By using Cosine Law,

c2 = a2 +b22*a*b*cos C


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The resultant vectorc is the sum of the

original vectors a and b and thereforere laces the ori inal two vectors.

ef

f

r

7

dd

e

Moment of a force

Moment of a force is a measure of the turning effect of theforce.

The magnitude of a moment is dependent on two items --- themagnitude of the force and theperpendicular distancebetween the point or axis of rotation and the force involved.

Moment of a force about a point

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= force x perpendicular distance from the point

i.e. M = F x d


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Example

The frame is loaded as shown in the following

diagram. Calculate the moment caused about point

A at the base of the frame.

0.5

m

.

10

kN

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Moment , M = Force x Perpendicular Distance

= 10 x 0.25

= 2.5 kNm

Principle of moments

,

the clockwise moments about any point is

equal to the sum of anti-clockwise

moments about that point.

i.e.

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Equilibrium of Forces

A structure is considered to be in

equilibrium if it remains at rest when

su ec e o a sys em o orces an

moments.

If a structure is in equilibrium, then all its

members and parts are also in equilibrium.

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,

forces and moments (including support

reactions) acting on it must balance eachother.

For a plane structure subjected to forces in

its own plane, the conditions for equilibrium

can be expressed by the following

equations of equilibrium:

0

0

F

F

y

x

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The third equation above states that the sum

of moments of all forces about any point inthe lane of the structure is zero.

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ExampleExample

Find T1, T2 hanging 5kg by graphical method (using scale 1cm = 10N) ?Find T1, T2 hanging 5kg by graphical method (using scale 1cm = 10N) ?

SolutionSolution

W =W = (5kg)(10 m/s(5kg)(10 m/s2 )) == 5050N ; In the space diagram: mark region between W & T1, W & T2,N ; In the space diagram: mark region between W & T1, W & T2,

T2 & T1 as A, B, C respectivelyT2 & T1 as A, B, C respectively

==

representingrepresenting WW((5050N); Draw a line parallel to T2 from point b & draw another line parallel toN); Draw a line parallel to T2 from point b & draw another line parallel to

T1 from point a. These 2 lines intersect at point c. Measure bc & ca ?T1 from point a. These 2 lines intersect at point c. Measure bc & ca ?

bc =bc = 4cm4cm & ca =& ca = 3cm3cm T2 =T2 = 40N40N & T1 =& T1 = 30N30N >>>>

1430

900T1 T2

C a

c

900

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5kg

W

A B

b 1430

Space Diagram Force Diagram

Scale 1cm = 10NScale 1cm = 10N


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Newtons first law of motionLaw of inertia

Every object remains in a state of rest or of

uniform speed along a straight line unless

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.

Newtons second law of motion

proportional to, and in the same direction as,

the unbalanced force acting on it, and

inversely proportional to the mass of the

object:

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F = maUnit: F(N), m(kg), a(m/s2)


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Newtons third law of motion

opposite reaction.

Mass

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Mass (m) is a measure of the inertia of an

object and is an unchanging quantity.Unit: kg

Weight

object.

Unit: N

W = mg

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where g: acceleration due to gravity, 9.81 m/s2


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Example

100N a

Calculate:

10kg

30N

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(a) the net force acted on 10 kg mass, and

(b) the acceleration of 10 kg mass.

Example

which is suspended from the ceiling of a lift.

If the balance reads 40 N, what is the

acceleration of the lift?

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Friction

another.

It always acts in the opposite direction to the

motion.

It is roduced b the interlockin of

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irregularities of the surfaces in contact.

There are two types of friction that can act onobjects namely static friction and kinetic

friction. Static friction refers to the frictional

force developed on the object prior to motion.

acts on an object while motion is occurring.

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It must be realized that static frictional forces develop in

response to any magnitude of push on the object. As a

push or sliding force is applied and increases in magnitude,the frictional force must likewise increase if the object is to

.

static frictional force at its maximum value, motion occurs.

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Coefficient of friction

)reaction(Rnormalx

r c onocoe c enr c on

R

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The normal force is the reaction which occurs

perpendicular to the moving object and the

surface over which it moves.

Normal Force = Weight of the object

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Example

A block weighing 50 kN rests on a level

surface as shown. If the coefficient of static

friction between the surfaces is 0.3,

e erm ne e orce nee e o s ar e oc

in motion.

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Solution

Since the force system is concurrent,

the equations of equilibrium that are

needed are

00 yx FF

0xF

0yF

P = F

N = 50 kN

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F = * N

Therefore, in this case,F = P = 0.3 * 50 kN

= 15 kN

Fluid Mechanics

A fluid is a material whose particles are freeto move their position. Liquids and gasesare both fluids.

Fluids can move and change the external

sha e to suit the container.

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Density

mass per unit volume.

volume

massDensity

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Unit: kg/m3

Density of water = 1000 kg/m3

Archimedes Principle

immersed in a fluid, the upthrust on the

object is equal to the weight of fluid

displaced.

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Example

.

m3 and a density of 2700 kg/m3. Calculate

(a) the weight of the aluminium block,

(b) the upthrust acting on the block when

immersed in water of density 1000 kg/m3.

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(c) the apparent weight of the block when

immersed in water.

Law of Flotation

of fluid in which it floats.

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Example

3 .

density of 700 kg/m3 floats in water of density

1000 kg/m3.

Calculate

(a) the weight of the block,

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(b) the weight of water displaced by the block,

(c) the volume of the block immersed in water.

Pressure Pressure on a surface is defined as the force

acting at right angle on that surface divided

b the area of the surface.

The force that acts on the area which is

submerged at a depth in a fluid is equal to

A

Fp

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the weight of the column above the area.

The pressure is then this weight divided by

the area


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ghp where

= pressure at a point in a fluid (N/m2 or Pa)

= density of fluid (kg/m3)

=

p

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the point (m)

Pressure increase with depth below theliquid surface.

Principle of Fluid Pressure

all direction.

Pressure always acts at right angles to the

containing surfaces.

Pressure is the same at oints of e ual de th

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irrespective of volume or shape.


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Force on Immersed Surface

from the pressure on the surface and the

area of the surface.

xApF average

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The centre of pressure is the point where the

line of action of the resultant force passes.

Water is contained by a vertical surface

produces a triangular pressure diagram.

The total force produced acts through the

centroid of the triangle, i.e. one-third of the

height.

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Example

.

retaining water to a depth of 0.8 m.

(a) calculate the lateral force on the gate

produced by the water;

b calculate its centre of ressure

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(c) calculate the moment acting about the

base.

Answer(a)

0.8 m

Pressure = gh

= 1000*9.81*0.8

= 7848 Pa

Gate

Centre of

Pressure

0.8*1/3 m

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Lateral force = Average Pressure *Area

= (7848/2)*(0.8*1.6)

= 5022.7 N


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b

Location of Centre of Pressure = 0.8*1/3

= 0.267 m

(c)

Moment = Force * distance

= 5022.7 * 0.267

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= 1341 Nm

chapter 2 force

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