Analytical Toolbox

Vectors and Applications

By

Dr J.P.M. Whitty

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Learning objectives

• After the session you will be able to:• Explain two types of physical quantities• Create graphical representation of vector

quantities• Resolve vectors• Perform vector addition• Use math software (or otherwise) to solve systems

of vectors in order to answer examination type questions

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Scalar Quantities

• Definition• A scalar quantity is described by a single #

alone (i.e. magnitude); examples include:• Length• Volume • Mass• Time

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Vector Quantities

• Definition:• A vector quantity is described by a

magnitude AND a direction• Force• Velocity • Acceleration• Displacement

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Vector Quantities Cont…

• A vector quantity (such as force) can be depicted as an arrow at an angle to the horizontal, e.g. 10 Newtons acting at 30 degrees:

300

10N

For more generally a force F @ an angle

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Example

1. Which of the following are vector and which are scalar quantities:

a) Temperature at 373K

b) An acceleration downwards of 9.8ms-2

c) A weight of mass 7kg

d) £500

e) A north-westerly in of 20 knots

Scalar

Vector

Vector

Scalar

Vector

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Vector Representations

• A (position) vector may join two points in space (A and B say), then, we may say:

B

A

a

Bold face

ABAB aThey are usually written as:

ABABa aWith the magnitude written as:

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Equal vectors

• Two vectors are equal if they have the same magnitude and direction

CDAB ca

B

A

a

Here we say:

D

C

c

CDAB or

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Equal and opposite vectors

• Two vectors are equal and opposite then they have the same magnitude but act in opposite directions (sometimes referred to as negative vectors)

DCAB ca

B

A

a

Here we say:

D

C

c

DCAB or CDAB or CDAB or

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Addition of vectors

Any quantities can be added using the a parallelogram of triangular rules:

Parallelogram rule:Vectors drawn from a single point

Triangular rule: Vectors placed end to end

Resultant vector

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Example #1:

• Find the resultant force of two 5N@10o and

8N@70o

•

5Units@10o

8Units@70o

Measure R to give: 11.4N@42o

R

8Units@70o

5Units@10o

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Example #2:

• Find the resultant force of three 6N@5o

and 8N@40o and 10N@80o •

6Units@5o

10Units@70o

8Units@40o

Solution 1: Apply the Parallelogram rule twice

Measure R to give: 21.6@44o

R

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Example #2: Triangular rule

• Here we simply place the vectors end to end, thus:

•

6Units@5o

10Units@70o

8Units@40o

Measure R to give: 21.6@44o

R

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Sum of a number of vectors

In general the triangular rule takes less construction and it is also easier extended to account for a number of vectors, thus:

• Let a be a vector from A to B, b be from B to C and so on…

• Then a+b+c+d, can be evaluated by formation of vector chain

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Vector chains

• The sum a+b+c+d, is constructed thus:

rdcba or A

B

C

D

E

a b

c

d

r

Here we can say:

AE dcba

:Hence

AEDECDBCAB

Notice the pattern

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Example:

Given that P,Q,R and S are point in three dimensional space, find the vector sum of

STRSQRPQ Solution:

STRSQRPQ

This has the same pattern as previously, i.e. a connected path thus:

PTSTRSQRPQ Note: No need to draw the diagram the outside letters render the result so long as they are connected

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The null vector

Suppose we consider another case where the resultant vector r=-e, we have:

A

B

C

D

E

a b

c

d

e

Here we now have:

AA edcba

0dcba

i.e. the same position::gives , denoting 0AA

i.e. 0, the null vector, has no length & hence direction

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Class Examples Time

• Find the sum of the position vectors:

PQLPKLAK

AQPQLPKLAK

Solution:

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Vector components

The vector OP is defined by its magnitude r and its direction . It can also be defined in terms of the components a and b in the directions OX and OY, respectively. O

rb

a

P

x

y

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Unit vectors

• Hence OP=a (along OX) + b (along OY).• If a unit vectors i,j (i.e. vectors of length unity) are

introduced along OX and OY respectively then:• r=a i + b j = i a + j b • r= i rcos + j rsin • Where a and b are the lengths along OX and OY,

equal to the magnitudes of the original vectors

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Vector addition (analytic solution)

The use of unit vector allows the calculation of vector addition analytically. Returning to the previous example #2, viz:

Find the resultant force of three Find the resultant force of three 6N@56N@5oo and 8N@40 and 8N@40oo and 10N@80 and 10N@80oo

Here the solution is to resolve the vectors into Here the solution is to resolve the vectors into components and add them to give the overall resultcomponents and add them to give the overall result

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Example #2: analytic solution

Letting the forces equal F1, F2 and F3 respectively

1) F1= i rcos + j rsin = i6cos5o + j6sin5o = 5.977i+0.526j (3dp)

2) F2= ircos + jrsin = i8cos40o+j8sin40o = 6.128i+5.142j (3dp)

3) F3=i rcos + j rsin=i10cos70o+j10sin70o = 3.420i+9.397j (3dp)

4) Therefore adding the individual components: 5) r=15.525i+15.065j (3dp)

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Example #2: analytic solution

The result is in Cartesian form, however we have been asked for the magnitude and direction of the resultant vector. To do this we must resort back to elementary trigonometry and Pythagoras, thus:

O

r

b=15.06515.065

a=15.52515.525

P

x

y

(dp)138.44

)525.15/065.15(tan &

(3dp)632.21

065.15525.15

1

22

22

o

bar

r

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Example #2: Alternative notation

The previous example can be evaluated using column or row vectors as follows:

70sin10

70cos10 &

40sin8

40cos8;

5sin6

5cos622

o

o

o

o

o

o

FFF1

397.9

420.3 &

142.5

128.6;

526.0

977.522

FFF1

065.15

525.15

397.9

420.3

142.5

128.6

526.0

977.5r

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Example #2: MatLab

This notation allows to solve such problems using math software such as MatLab

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Example # 3

Find the forces in the members of the structure; and evaluate the stress in each given that each bar is 50mm in diameter:

300N

60o

A B

C

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Example # 3: Solution

i(FAB)-i(FBCCos60o)+j(FBCSin60o)=j300

300N

60o

A B

C

j

i

i: i: FFABAB-F-FBCBCCos60Cos60oo=0=0

j: j: (F(FBCBCSin60Sin60oo)=300)=300

N173

31002

3200

060cos3200

AB

AB

oAB

F

F

F

34532003

3

3

2300

BCF

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Example # 4

Find the forces in the members of the structure; and evaluate the stress in each given that each bar is 25mm in diameter:

500N60o

A

B

C

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Example # 4: Solution

Apply vector eqn, thus: i(FABcos30o)+ j(FABsin30o)- i(FBCcos30o)+

j(FBCSin30o)=j500

500N60o

A

B

C

500

50030sin2

50030sin30sin:

AB

AB

BCAB

F

F

FFj

BCAB

BCAB

FF

FFi

030cos30cos;

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Examination type questions (Homework)

1. Find the value of the resultant force given that the following act on a specific point in a roof truss.

o20kN31 Fo10kN72 F

o32kN142 F

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Examination type questions (Homework)

2. Given that the following three forces act on a 12mm diameter bar:

o23kN21 Fo30kN62 F

o32kN132 F

Find the resultant force on the bar [3], and evaluate the maximum stress that bar can experience.

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Examination Type Question

Exploit symmetry conditions and find the stresses in each of red members the 20mm dia, steel members (E=200GPa). Hence or otherwise evaluate the resulting strains.

[20 marks]

250 250

1m1m

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Examination Type: Solution

Exploit symmetry thus:

250

250

A

B

C

0)60sin60cos(250 00 ACAB FF ijijNow @ A:

02

3

2250 AC

ABAB FFF

ijij

3

250

2 :Hence

2

3250:

2:

3500

3500

AC

ABAB

ACAB

F

FF

FF

j

i

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Examination Type Question: Strain value solutions

These can be evaluated from the elasticity definitions as well!

strainmili6.4GPa200

MPa46.0 :likewise

strainsmili6.4GPa200

MPa92.0

ACAC

E

EE

AB

Note: the units here are of utmost importance

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Examination Type Question: Stress value solutions

These can be evaluated from the elasticity definitions

MPa46.0

320

2504 :and

MPa92.0320

105004

2420

3250

AC

2

3

420

310500

2

2

3

A

F

A

F

AC

ABAB

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Summary:

• Have we met our learning objectives specifically are you now able to:• Explain two types of physical quantities• Create graphical representation of vector

quantities• Resolve vectors• Perform vector addition• Use math software (or otherwise) to solve systems

of vectors in order to answer examination type questions