# Lec 1 - Fundamental Concepts, Force Vectors

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• ES 11 Lecture 1

• the study of the relationship among

forces and their effects on bodies.

the science which describes and predicts the conditions for rest and motion of bodies under the action of forces.

• Mechanics

Rigid Bodies

Statics

Dynamics

Deformable Bodies

Fluids

Compressible

Incompressible

• represents the action of one body on another

may be exerted by actual contact or at a distance

characterized by its: Point of application

Magnitude

Line of action

represented by a vector.

• Development of other forces Reactions

Internal forces

Deformation of the body

Acceleration of the body

Applied Force

• Applied Force

Reaction

Development of force or forces at

points of contact with other

bodies (reactions).

• Development of forces within the

body itself (internal forces) A A

Applied Force

• Applied Force

Deformation of the body

• Applied Force

Acceleration of the body

• 10 N

30o

10 N 30o

Point of application (forces acting on the same particle have the same point of application)

Magnitude Direction Line of action (angle w.r.t. a fixed axis) Sense

• Treatment of bodies as particles - the shape and size of the object does not significantly affect the solution of the problems under consideration.

Rigid Bodies - the problems considered in this course are assumed to be non-deformable.

• Acceleration due to gravity

g = 9.81 m/s^2 or g = 32.2 ft/s^2

Quantity SI English

Length m (meter) ft (feet)

Mass kg (kilogram) slugs

Time s (seconds) s (seconds)

Force kg m/s2 OR N (newtons)

lbs (pounds)

• P

Q

A

R

A

Single equivalent force having the same effect as the original forces acting on the particle

• Parallelogram Law

The resultant is the diagonal of the parallelogram with the two forces as its sides

Triangle Law

Derived from the parallelogram law

If the two vectors are placed tip-to-tail, the resultant is the third side of the triangle

P

Q A R

P

A

Q A P

Q

• P

Q A P+Q = R

Q QPR

BPQQPR

cos2222

Law of cosines,

Law of sines,

P

C

R

B

Q

A sinsinsin

Vectors are defined as mathematical expressions possessing magnitude and direction, which add according to the parallelogram law.

Forces are sliding vectors

P

Q A

P+Q

Q

PQQP

P

Q A

Q+P

P

• Adding more than 2 vectors If the vectors are coplanar, the resultant may be obtained by using the polygon rule for the addition of vectors arrange the given vectors in a tip-to-tail fashion and connect the tail of the first vector with the tip of the last one

P

Q A

Q

S

R

P

Q A

Q

S

R

P+Q

P

A

Q

S

R

P+Q

P

A

Q

S

R

Q+S

(P+Q)+S

P+ (Q+S)

P+Q+S = (P+Q)+S = P+ (Q+S)

• PPP 2

Pn

- have the same direction as P with magnitude Pn

P 1.5P -2P

P

P

- Product nP of a positive integer n and a vector P

• 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.

• The two forces act on a bolt at

A. Determine their resultant.

SOLUTION:

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

• Trigonometric solution - Apply the triangle rule.

From the Law of Cosines,

155cosN60N402N60N40

cos222

222 BPQQPR

N73.97R

• From the Law of Sines,

A

A

R

QBA

R

B

Q

A

20

04.15N73.97

N60155sin

sinsin

sinsin

04.35

• 25

Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle if the resultant of the two forces applied to the hook support is to be horizontal, (b) the corresponding magnitude of R.

50 N

25O

P

• 26

50 N

25O

P

R

R

O sin

35

25sin

50

sin

(a) Determining &

OO

14.3735

25sin50sin 1

OO 25180

O86.117

• NR 22.73

27

50 N

25O

P

R

R

O sin

35

25sin

50

sin

(b) Determining R

sin

sin50R

• Two forces are applied as shown to a hook support. Knowing that the magnitude of the resultant of the two forces is a 50 N horizontal force, determine the value of for which the applied force 2 is minimum.

50 N

25O

P

Force 1

Force 2

• It has been shown that the resultant of forces acting at the same point (concurrent forces) can be found.

In the same way, a given force F can be resolved into components.

29

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

• How many components could a force be resolved into?

Infinite number of possible components

For a force resolved into two components,

• 32

15 . 0

40 . 0

45 . 0

a

b

P Given P is 800 N, determine the components of the force in a and b axes.

• 4045180

33

15 . 0

40 . 0

45 . 0

a

b

P 15.0O 25.0O P=800N

95O

40O

45O

AF

BF

A

O

B

OO

FF

40sin45sin

800

95sin

NFB 85.567

NFA 19.516

• A force vector may be resolved into perpendicular components

• j F F y y

j

i i F F x x

F

q

y

x

i j and - unit vectors of magnitude 1 directed along the

positive x and y axes, respectively.

xF yF - vector components of

FFx , Fy scalar components of

F

- may be positive or negative

depending upon the sense of xF

and yF

- the absolute values are equal

to the magnitude of the

component forces and xF yF

qcosFFx q sinFFy

F = Fxi +Fy j

• 0 35

0 145 q

N F 800

x F

y F FX = -655.3 N i

FY = 458.9 N j

• 2 - 37

SQPR

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

jSQPiSQPjSiSjQiQjPiPjRiR

yyyxxx

yxyxyxyx

Resolve each force into rectangular components

x

xxxx

F

SQPR

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

y

yyyy

F

SQPR

x

yyx

R

RRRR 122 tan q

To find the resultant magnitude and direction,

• 38

tan q = x

y

F

F

F = 22

yx FF q

F

x F

y F

x

y

qR

R

x R

y R

x

y

xx FR

yy FR xy

RR

Rqtan

• Determine the resultant of the three forces below

39

25o

45o 350 N

800 N

600 N

60o

y

x

• 40

25o

45o 350 N

800 N

600 N

60o

y

x

RX = F x = 350 cos 25O + 800 cos 70O -600 cos 60O

RX = 317.2 + 273.6 300 = 290.8 N

RY = F y = 350 sin 25 + 800 sin 70O +600 sin 60O

RY = 147.9 + 751 + 519.6 = 1419.3 N

F= 290.8 N i +1419.3 N j

Resultant, F

NF 14493.14198.290 22 O4.788.290

3.1419tan 1 q

F = 1449 N 78.4O

• The resultant of concurrent forces acting on a particle in space will also act at the same particle.

Only the magnitude and direction are to be determined.

y

z

x

o

F1

F2

F3

Note: In this illustration, the magnitude and direction of all of the forces are given.

• From the force polygon (warped), the resultant can be drawn from the tail of the first force to the head of the last force.

The magnitude and direction of the resultant can be computed using successive use of the triangle law.

y

z

x

o

F1

F2 F3

z q

x q

y q

R

Note: The sine and cosine laws are hard to implement because usually the given angles are absolute.

• The rectangular components of a force can be determined easily depending on the given characteristics of the force.

Given the Magnitude and Two Angles

y

z

x

o

F

zq

yq

y

z

x

o

F

zq

xyq

xyF

zF

zxy FF qsin

zz FF qcos

• xyzxyxyy FFF qqq cossincos

xyzxyxyx FFF qqq sinsinsin

Given the Magnitude and Two Angles

y

z

x

o

F

zq

xyq

xyF

zF

zxy FF qsin

y

z

x

o

xyq

xyF

zF

xF

yF

• zz FF qcos

Given the Magnitude and Two Angles

y

z

x

o

xyq

xyF

zF

xF

yF xyzxyxyy

xyzxyxyx

FFF

FFF

qqq

qqq

cossincos

sinsinsin

222

FzFyFxF

kFzjFyiFxF

In vector form,

• Given the Magnitude and Three Absolute Angles

Fx = Fcos qx Fy = F cos qy Fz = Fcos qz where cos qx, cos qy and cos qz are direction cosines F = Fxi + Fyj + Fzk

y

z

x

o

x q

F

y q

z q

• A force of 500N forms angles of 600, 450 and 1200, respectively, with the x,y and z axes. Find the components Fx, Fy and Fz of the force. Find also the vector representation of the force.

Fx = F cos qx = (500N)cos 600 Fx = +250N Fy = F cos qy = (500N) cos 450 Fy = +354N Fz = F cos qz = (500N) cos 1200 Fz = -250N

y

z

x

o

F

qx qy

qz

Fx

Fy

Fz

F = 250N i + 354N j 250N k

• y

z

x

o

F

qx qy

qz

Fx

Fy

Fz

F = 250N i + 354N j 250N k

Note: The angle a force F forms with an axis should be measured from the positive side of the axis and will always be between 0 and 1800.

• kzjyixFF coscoscos qqqlet = cos qx i + cos qy j+ cos qz k

= unit vector x= cos qx y= cos qy z= cos qz

x2 + y

2 + z2 = 1

cos2 qx + cos2 qy + cos

2 qz = 1

it follows that,

F = F The force vector is equal to the product of the magnitude of the force and the unit vector.

• A force has the components Fx = 20N, Fy = -30N, Fz = 60N. Determine its magnitude F and the angles qx, qy, qz it forms with the coordinate axes.

y

z

x

o

Fx

Fy

Fz 222 FxFyFxF

NNF

NNNF

704900

)60()30()20( 222

• qx = cos-1 (Fx / F) = cos

-1 (20/70)

qx = 73.40

qy = cos-1 (Fy / F) = cos

-1 (-30/70)

qy = 115.40

qz = cos-1 (Fz / F) = cos

-1 (60/70)

qy = 31.00

y

z

x

o

Fx

Fy

Fz F

F = 70 N

• Given the Magnitude and Two Points where Force Passes

222 dzdydxd

zzd

yyd

xxd

oez

oey

oex

Fd

dxFx F

d

dyFy F

d

dzFz

kFzjFyiFxF

y

z

x

o

e e e z y x E , ,

o o o z y x O , ,

F

• kFzjFyiFxF

y

z

x

o

eee zyxE ,,

eee zyxO ,,

F

FxFy

Fz

d

dxx qcos

d

dyy qcos

d

dzz qcos

• Determine the vector representation of the given force.

z

x

O

2.4m E

1.5m

F=1.6kN

1.2m

1.5m

y

• EXAMPLE 1-8

z

x

O

2.4m E

1.5m

F=1.6kN

1.2m

1.5m

y

O(1.2, 1.5, 0.0) E(0.0, 2.4, 1.5)

dx = 0.0 -1.2 = -1.2 dy = 2.4 -1.5 = +0.9 dz = 1.5 - 0.0 = +1.5

222 5.19.02.1 d

d = 2.12m

• kNkNFx 905.0)6.1(12.2

2.1

kNkNFy 679.0)6.1(12.2

9.0

kNkNFz 131.1)6.1(12.2

5.1

kkNjkNikNF 131.1679.0905.0

Fx

Fz

z

x

O

E

y

Fy

• The resultant R of two or more forces in space will

be determined by summing their rectangular

components. Graphical or trigonometric methods are

generally not practical in the case of forces in space.

FR FzFyjFxiRzkRyiRxi kFzjFyiFx

FxRx FyRy FzRz

• 222 RzRyRxR

R

Rxx qcos

R

Ryy qcos

R

Rzz qcos

• Determine graphically, the magnitude and direction of the resultant of the two forces using (a) Parallelogram law and (b) the triangle rule.

900 N

600 N

30o

45o

• A parallelogram with sides equal to 900 N and 600 N is drawn to scale as shown.

From the scaled drawing of the forces, the resultant is

R 1400 N 900 N

600 N

30o

45o

R

q q 46o

Note: The triangle rule may also be used. Join the forces in a tip to tail fashion and measure the magnitude and direction of the resultant.

• For the magnitude of R, using the cosine law:

oR 135cos6009002600900 222

NR 13916.1390

For angle q, using sine law:

sin

600

135sin

O

R

900 N

600 N

30o

135o

R

q

OO 8.1730 qThe angle of the resultant:

OO

8.171391

135sin600sin 1

• Sample Problem 2.10 : (2.89) A frame ABC is supported in part by cable DBE which passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support D.

400mm

210mm 280mm

510mm

480mm 600mm

x z

A

B

C

D E

• O(0, 510, 280)

E(480, 0, 600)

dx = xE xO = 480-0=480

dy = yE yO = 0-510=-510

dz = zE zO = 600-280=320

d = 770 mm

400mm

210mm

280mm

510mm

480mm

600mm

x z

A

B

C

D E

TDB = 385N

TDBY

TDBX TDBZ

E(480, 0, 600)

O(0, 510, 280)

NTd

dT DB

x

DBX 240385770

480

NTd

dT DB

y

DBY 255385770

510

NTd

dT DB

zDBZ 160385

770

320

NkjikTjTiTT DBZDBYDBXDB 160255240

• Solution:

Position vector of BH = 0.6 m i + 1.2 m j - 1.2 m k

Magnitude, BH = 0 6 1 2 1 2 1 8 2 2 2 . . . . m

BH

BH BH BH BH

BH

x y z

BH

BH m i m j m k

T T T BH

BH

N

m m i m j m k

T N i N j N k

F N F N F N

| | . ( . . . )

| | . | | | | .

. . .

( ) (500 ) (500 )

1

1 8 0 6 1 2 1 2

750

1 8 0 6 1 2 1 2

250

250 500 500