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Chapter 2Force Vectors

Scalars and VectorsVectorsA mathematical quantity possessing magnitude and direction.

Scalar

A mathematical quantity possessing magnitude only.

Name some vectors: forces, velocity, displacement

Name some scalars: Area, volume, mass energy

Representation of vector

Bold RWord Processors Book uses this.

Arrow R

Long Hand, Word Processors

Underline RLong Hand, Typewriter, Word Processors

Magnitude of a Vector

R Book uses italics for all scalars

Types of Vectors

1). Fixed (or bound) vectorsa vector for which a unique point of

application is specified and thus cannot be moved without modifyingthe conditions of the problem.

2). Free vectora vector whose action is not confined to or associatedwith a unique line in space. (couple)

3). Sliding vectora vector for which a unique line in space (line of

action) must be maintained.

For 2 vectors to be equal they must have the same:

1). Magnitude P P

2). Direction

They do not need to have the same point of application.

A negative vector of a given vector has same magnitude but opposite direction.P

-P

P andP are equal and opposite P + (-P) = 0

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Polygon RuleSuccessive applications of triangle rule.

Q

Q P

S

P R

S

Note: P +Q + S = (P + Q) + S = P + (Q + S) vector addition is associative

Vector Subtractionthe addition of the corresponding negative vector

PQ = P + (-Q)

Resolution of vector into components

A single vector can be represented by 2 or more vectors. These vectorsare components of the original vector. Finding these is called resolving

the vector into its components.

There is an infinite number of ways to resolve one vector.

etc.

2 cases of particular interest are:

1). One of the 2 components is known. Easy (see above)2). The line of action of both components is know.

When would #2 happen? When you are given a coordinate system!

What are the x and y components ofP if P = 1000 lbs, = 30o

Px = P cos 30o

= 866 lbs Py = P sin 30o

= 500 lbs

Note: Given Px and Py, what is P?

P2

= Px2

+ Py2=866

2+ 500

2= 1000 lbs

P

R-Q

P

P1

P2

P1

P2

P

P

x

y

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1). Given: The fixed structure shown below.

P = 500 N

T = 200 N

Find: Combine P and T into a single force R

4.48

75cos53

75sin5tan

AD

BD

Law of cosines:

NR

R

cabbac

5.396

)4.48cos()500)(200(2500200

)cos(2

222

222

Law of sines:

4.48sin

5.396

sin

200

2.22

NR 5.396

2.22

3m

T

P

A

B

C D

5m

75o

R

P = 500

T = 200

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30o

Dir of

Dir of

Dir of

2.) Given: A barge is pulled by 2 tugboats. The resultant of the forces exerted by thetugboats is a 5000 pound force directed along the center axis of the barge.

A

1

30

B

C

2

Find: a). tension in each rope if =45 degrees

b). value of such that the tension in rope 2 is minimum.

5000

a).

45 30

T2 1T

105sin

5000

30sin45sin

21

TT

lbsT

lbsT

2590

3660

2

1

b). 5000

dir of1

T

lbsT

lbsT

250030sin5000

433030cos5000

60

2

1

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3). Given: The vertical force F of 350 lbs acts downward at A on the two-membered

frame.

F

A

B

C

30o

45o

Find: The magnitudes of the two components ofF directed along AB and AC.

75sin

350

45sin60sin

ACABFF

lbs256

lbs314

AC

AB

F

F

75

45

60lbs350