ch. 2. statics of particles 질점의정역학isdl.cau.ac.kr/education.data/statics/ch2.pdf · ·...

2011-09-14

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School of Mechanical Engineering

Ch. 2. Statics of Particles(질점의 정역학)

중앙대학교기계공학부

최해진

Statics

2 - 1


Contents

• Introduction• Resultant of Two Forces• Vectors• Addition of Vectors• Resultant of Several Concurrent

Forces• Sample Problem 2.1• Rectangular Components of a Force:

Unit Vectors• Addition of Forces by Summing

Components

• Sample Problem 2.3• Equilibrium of a Particle• Free-Body Diagrams• Sample Problem 2.4• Sample Problem 2.6• Rectangular Components in Space

Statics

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Introduction 2 - 3

• The objective for the current chapter is to investigate the effects of forces on particles:

- replacing multiple forces acting on a particle with a single equivalent or resultant force, 주어진힘들을합력으로 표현가능.

- relations between forces acting on a particle that is in a state of equilibrium. 평형상태에서 힘들의 관계를 규명.

• Particles (질점) : 작은물체에국한되는것이아니라, 물체의크기나 모양이 해석에큰 영향을주지않을경우.

- 물체에작용한모든힘들이동일한점에작용한다는의미.

Statics


Resultant of Two Forces 2 - 4

• Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force.

• The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs.

• Force is a vector quantity.

• Force : action of one body on another ; characterized by its point of application, magnitude, line of action, and sense.작용점, 크기, 작용선, 방향(부호)

Statics

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Addition of Vectors 2 - 5

• Trapezoid rule for vector addition

• Triangle rule for vector addition

B

B

C

C

QPRBPQQPR

rrr+=

-+= cos2222

• Law of cosines,

• Law of sines,

PC

RB

QA sinsinsin

==

• Vector addition is commutative,

PQQPrrrr

+=+

• Vector subtraction

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Addition of Vectors 2 - 6

• Addition of three or more vectors through repeated application of the triangle rule

• The polygon rule for the addition of three or more vectors.

• Vector addition is associative,

( ) ( )SQPSQPSQPrrrrrrrrr

++=++=++

• Multiplication of a vector by a scalar

Statics

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Resultant of Several Concurrent Forces 2 - 7

• Concurrent forces : set of forces which all pass through the same point.

A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces.

• Vector force components : two or more force vectors which, together, have the same effect as a single force vector.

Statics


Sample Problem 2.1 2 - 8

The two forces act on a bolt at A. Determine their resultant.

SOLUTION:

• Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the diagonal.

• Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant.

Statics

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• 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 aR

• Graphical solution - A triangle is drawn with Pand 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 aR

Statics



• Trigonometric solution - Apply the triangle rule.From the Law of Cosines,

( ) ( ) ( )( ) °-+=-+=

155cosN60N402N60N40cos222

222 BPQQPR

AA

RQBA

RB

QA

+°=°=

°=

=

=

2004.15

N73.97N60155sin

sinsin

sinsin

a

N73.97=R

From the Law of Sines,

°= 04.35a

Statics

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

kN0.13kN5.18 21 == TT

• Trigonometric solution - Triangle Rule with Law of Sines

°=

°=

° 105sinkN25

30sin45sin21 TT

kN94.12kN30.18 21 == TT

A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 25 kN directed along the axis of the barge, determinea) the tension in each of the ropes for a = 45o,

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

°105

°45

°45

Statics



• The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in a.

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

( ) °= 30sinkN252T kN5.122 =T

( ) °= 30coskN251T kN65.211 =T

°-°= 3090a °= 60a

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Rectangular Components of a Force: Unit Vectors 2 - 13

• Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components.

Fx and Fy are referred to as the scalar components of

jFiFF yxrrr

+=

Fr

• May resolve a force vector into perpendicular components so that the resulting parallelogram is a rectangle. are referred to as rectangular vector components and

yx FFFrrr

+=

yx FFrr

and

• Define perpendicular unit vectors which are parallel to the x and y axes.

jirr

and

Statics


Addition of Forces by Summing Components 2 - 14

SQPRrrrr

++=

• Wish to find the resultant of 3 or more concurrent forces,

( ) ( ) jSQPiSQP

jSiSjQiQjPiPjRiR

yyyxxx

yxyxyxyxrr

rrrrrrrr

+++++=

+++++=+ )()()()(

• Resolve each force into rectangular components

å=++=

xxxxx

FSQPR

• The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.

å=++=

y

yyyyF

SQPR

x

yyx R

RRRR 122 tan-=+= q

• To find the resultant magnitude and direction,

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Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

SOLUTION:

• Resolve each force into rectangular components.

• Calculate the magnitude and direction of the resultant.

• Determine the components of the resultant by adding the corresponding force components.

Statics



SOLUTION:• Resolve each force into rectangular components.

°=== 1.4,N1199N314tan aa

..

RR

x

y

• Calculate the magnitude and direction.

N6.199sin

N3.14==

aR

°= 1.4a

• Determine the components of the resultant by adding the corresponding force components.

1.199+=xR 3.14+=yR9.256.961000.11001102.754.27800.759.129150

4

3

2

1

-+-+-++--

FFFF

compycompxmagforce

rrrr

Statics

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Equilibrium of a Particle 2 - 17

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

• 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

000

==

==

ååå

yx FFFRrr

• 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.

Statics


Free-Body Diagrams (FBD) 2 - 18

Space Diagram : A sketch showing the physical conditions of the problem.

Free-Body Diagram : A sketch showing only the forces on the selected particle.

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In a ship-unloading operation, a 3500N automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope? 20 300

SOLUTION:

• Construct a free-body diagram for the particle at the junction A of the rope and cable.

• Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the particle.

• Apply trigonometric relations to determine the unknown force magnitudes.

Statics



SOLUTION:

• Construct a free-body diagram for the particle at A.

• Apply the conditions for equilibrium.

• Solve for the unknown force magnitudes.

3500Nsin120 sin 2 sin 58

ACAB TT= =

° ° °

3570 kNABT =

144NACT =

Statics

3500N

3500N

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School of Mechanical Engineering 2 - 21

Sample Problem 2.6

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 40 N in cable AB and 60 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.

2 - 21


Sample Problem 2.6

SOLUTION:

• Choosing the hull as the free body, draw a free-body diagram.

°=

==

25.60

75.1m 4m 7tan

a

a

°=

==

56.20

375.0m 4m 1.5tan

b

b

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

0=+++= DAEACAB FTTTRrrrrr

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Sample Problem 2.6

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

( ) ( )( ) ( )

( )

( )( ) jT

iFTR

iFFiT

jTiTjTiTT

jijiT

AC

DAC

DD

ACAC

ACACAC

AB

rr

r

rrrr

rrrrr

rrrrr

609363.084.193512.073.34

0

N06

9363.03512.056.20cos56.20sin

N84.19N73.3426.60cosN4026.60sinN40

-++

++-=

=

=

-=

+=

°+°=

+-=

°+°-=

2 - 23


Sample Problem 2.6

( )( ) jT

iFTR

AC

DAC rr

r

609363.084.193512.073.34

0

-++

++-=

=

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

( )( ) 609363.084.1900

3512.073.3400-+==++-==

åå

ACy

DACxTF

FTF

N 66.19N 9.42

+=+=

D

AC

FT

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2.12 Rectangular Components in Space 2 - 25

• The vector is contained in the plane OBAC.

Fr

• Resolve into horizontal and vertical components.

Fr

yy FF qcos=

yh FF qsin=

• Resolve into rectangular components

hF

fq

f

fqf

sinsin

sin

cossincos

y

hy

y

hx

F

FF

FFF

=

=

==

Statics


2.12 Rectangular Components in Space 2 - 26

• With the angles between and the axes,Fr

( )

kjiF

kjiF

kFjFiFF

FFFFFF

zyx

zyx

zyx

zzyyxx

rrrrr

rrr

rrrr

qqql

l

qqq

qqq

coscoscos

coscoscos

coscoscos

++=

=

++=

++=

===

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

Fr

Fr

lr

zyx qqq cos and,cos,cos

λ는 힘 F 방향으로의 unit vector이다.

Statics

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Rectangular Components in Space 2 - 27

Direction of the force is defined by the location of two points,

( ) ( )222111 ,, and ,, zyxNzyxM

( )

dFdF

dFd

Fd

FdF

kdjdiddMN

MN

FF

zzdyydxxdkdjdid

NMd

zz

yy

xx

zyx

zyx

zyx

===

++=º

=

-=-=-=

++=

=

rrrr

rr

rrr

r

1

and joining vector

121212

l

l

kdjdidMNd zyx

rrrr++==

λ는 MN 방향으로의 unit vector 이다.

Statics



a. 직각성분과방향코사인

그림(a)와같이 x, y, z축방향의단위벡터는기저벡터라고부르며 i, j, k로표시된다.

kjiA zyx AAA ++=

zzyyxx θA AθA Aθ A A coscoscos ===

방향코사인(direction cosine) 혹 방향여현

zzyyxx qlqlql cos cos cos ===

222zyx AAAA ++=

kjiλ zyx λλλ ++=

)(

coscoscos

kjikjiA

zyx

zyx

λλλAAAA

++=

++= qqq

λ는 A방향으로의 단위벡터이다.

1222 =++ zyx lll

직각성분을사용한벡터의표현

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b. 직각성분을이용한벡터의덧셈

벡터A와 B의합인C는다음과같이표현한다.

kjikjiC

kjikjiBAC

)()()(

)()(

zzyyxx

zyx

zyxzyx

BABABACCC

BBBAAA

+++++=

++=

+++++=+=

zzzyyyxxx B A C B A C B A C +=+=+=

각단위벡터의성분을같다고두면C의직각성분은

Statics



위치벡터

c. 위치벡터, 상대위치벡터, 그리고단위벡터

OB좌표계의원점O로부터점 B로그은벡터 는 B의위치벡터이다.

상대위치벡터 AB점 A로부터점 B로그은벡터 는 B의A에대한상대위치벡터이다.

상대위치벡터 AB

dzzyyxx

ABAB ABABAB kjiλ )()()( -+-+-

==

크 기222 )()()( ABABAB zzyyxxdAB -+-+-==

점A에서점 B로향하는단위벡터는 λ로표시

kji )()()( ABABAB zzyyxxAB -+-+-=

AB의방향코사인(방향여현) 또는 λ의성분은

dxz

dyy

dxx

ABzz

AByy

ABxx

-==

-==

-==

q

q

q

cos

cos

cos

λ

λ

λ

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d.벡터를직각성분형태로나타내는방법

그림에서벡터 F는선분 AB를따라가는방향이다.

F를직각성분으로표현하는두가지방법

방법1 : 방향코사인이용

그림에서 각도θx , θy, 그리고 θz를알고있다면

)kji(

k cosjcosicos

zyx

zyx

λλλFFFFF

++=

++== qqqλF

방법2 : 상대위치벡터이용

단계 1 : 상대위치벡터 를직각성분형태로표현한다. 이단계는점A와 B의좌표를이용하여구한다.

단계 2 : 단위벡터 / 를계산한다.

단계 3 : F=F λ로쓴다.

AB

AB AB

Statics



예제

그림(a)에서크기가 500 N인힘 F에대한직각성분들을표현하라.

4 6 3 mAB = - + + -i j kuuur

kji

kjiλ

3841.07682.05122.0

)3(6)4(364

222

-+-=

-++-

-++-==

ABAB

F=F λ로쓰면

500( 0.5122i 0.7682j 0.3841k) 256i 384j 192k N= - + -= - + -

F

그림(b)에서

Unit vector)3,0,4(,)0,6,0( == AB

Statics

3m

6m

3m

6m6 m

192 N 384 N

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예제 그림(a)에서 (1) 위치벡터A의직각성분을결정하고(2) A와 각양의축과이루는각도를구하라.

m 392.10857.3596.4 m 857.340sin640sin m 596.440cos640cos

m 630sin1230sin m 392.1030cos1230cos )1문

kjikjiA ++=++=

=°=°=

=°=°=

=°=°==°=°=

zyx

xyy

xyx

xy

z

AAAAA AA AA AA

°===

°===

°===

--

--

--

0.3012392.10coscos

3.7112857.3coscos

5.6712596.4coscos 2)문

11

11

11

AAAAAA

zz

yy

xx

q

q

q

Statics



예제 직각성분을사용하여그림(a)에있는벡터 P와Q의합력인R을구하라.

120cos30 120sin 30 103.9 60.0 m100cos 70 100sin 70 34.2 94.0 m

= ° + ° = += - ° - ° = - -

P i i jjQ i j i j

(103.9 34.2) (60.0 94.0) 69.7 34.0 m= + = - + -= -

R P Q i ji j

69.7m

34.0m

R

θ

2 2 1 34.034.0 69.7 77.6 m tan 26.069.7

- R q= + = = = °

합력R 의크기와방향은

Statics

N

N

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2 - 35Sample Problem 2.7

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 qx, qy, qz 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.



SOLUTION:• Determine the unit vector pointing from A

towards B.

( ) ( ) ( )

( ) ( ) ( )m 3.94

m30m80m40

m30m80m40222

=

++-=

++-=

AB

kjiABrrr

• Determine the components of the force.

( )( )( ) ( ) ( )kji

kjiFF

rrr

rrr

rr

N 795N 2120N1060

318.0848.0424.0N 2500

++-=

++-=

= l

kji

kjirrr

rrrr

318.0848.0424.03.94

303.94

803.94

40

++-=

÷øö

çèæ+÷

øö

çèæ+÷

øö

çèæ -=l

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

kji

kji zyxrrr

rrrr

318.0848.0424.0

coscoscos

++-=

++= qqql

o

o

o

5.71

0.32

1.115

=

=

=

z

y

x

q

q

q


Practice Problems

q 2.4, 2.19, 2.30, 2.45, 2.63, 2.87, 2.125

2 - 38

ch. 2. statics of particles 질점의정역학isdl.cau.ac.kr/education.data/statics/ch2.pdf · ·...

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