# 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

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