Statics Math

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Statics (ENGR 2214)Prof. S. Nasseri

What you need to know from Math!GEN 240




Part 1

Preliminary

(Algebra, Geometry, Trigonometry)




sin(0) 0, cos(0) 1, tan(0) 0

sin(90) 1, cos(90) 0, tan(90)

sin(180) 0, cos(180) 1, tan(180) 0sin(270) 1, cos(270) 0, tan(270)

1 3 3sin(30) cos(60) , cos(30) sin(60) , tan(30) , tan(60

2 2 3

! ! !

! ! ! g

! ! !! ! ! g

! ! ! ! ! ) 3

2

sin(45) cos(45) , tan(45) 12

!

! ! !

P ythagorean Theorem2 2 2

sin bsin , cos , tan =

cos a

a b c

b a

c c

UU U U

U

!

! ! !

a

bc

Usin

cos

tan

0

1

0

0

-1

�

1

0

�

-1

0

-�

sin

cos




P ythagorean Theoremsin(0) 0, cos(0) 1, tan(0) 0

sin(90) 1, cos(90) 0, tan(90)

sin(180) 0, cos(180) 1, tan(180) 0

sin(270) 1, cos(270) 0, tan(270)1 3 3

sin(30) cos(60) , cos(30) sin(60) , tan(30) , tan(602 2 3

! ! !

! ! ! g

! ! !

! ! ! g

! ! ! ! ! ) 3

2sin(45) cos(45) , tan(45) 1

2

!

! ! !

hypotenuse




P ythagorean Theorem

2 2

Law of sines:sin sin sin

Law of cosines: 2 cos

a b c

c a b ab

E F K

K

! !

! a

b

c

F

E

K

a

b

c

K




The unit circle andtrigonometric functions

sin( ) sin , cos( ) cos , tan( ) tan

sin(180 ) sin , cos(180 ) cos , tan(180 ) tan

sin(180 )

(4th quadrant)

(2nd quadrant)

(3rd qsin , cos(180 ) cos , tan(180 ) tan uadrant

U U U U U U

U U U U U U

U U U U U U

! ! !

! ! !

! ! !sin(90 ) cos , cos(90 ) sin

s

)

(1st quadrant)

(2nd quadr in(90 ) cos , cos an(90 ) sin t)

U U U U

U U U U

! !

! !

+

+

+

-

-

-

-

+

c os

sin




Double- & two-angle relations

2 2 2 2

2

sin(2 ) 2sin cos

cos(2 ) 1cos(2 ) cos sin 2cos 1 so: cos

2

2tantan(2 )

1 tan

U U U

UU U U U U

UU

U

!

¨ ¸! ! !© ¹

ª º

!

Two angle relations:

Double angle relations:

sin sin cos cos sincos cos cos sin sin

tan tantan

1 tan tan

E F E F E FE F E F E F

E FE F

E F

s ! ss !

ss !

m

m




Arcs and sectors

2 2

Arc ength,

Angle measured in radians, / arc length/radius

1Sector area2 2

s r

s r

r r

U

U

U T UT

!

!

¨ ¸!© ¹ª º

r s

U




Similar trianglesThe sides of two similar triangles are proportional and the angels are thesame. The respective heights of these triangles are also proportional to thesides.

a b c hd e f H

! ! !




Part 2

Vectors




Scalar: A quantity like mass or temperature, which only has amagnitude.Vector: A quantity like heat flux or force which has both amagnitude and a direction; denoted by a bold faced character (a),an underlined character (a), or a character with an arrow on it:

Vectors

ar

F

Fx

Fy

i

j

x

y

U

Resolution of a Vector: A vectorcan be resolved along differentdirections using the parallelogramrule. The figure shows how oneresolves vector F into componentsF x and F y which are along thegiven directions (i and j are theunit vectors; vectors of unitlength).

2 2

tan

x y

x y

y

x

F F

F F

F

F U

!

!

!

F i j

F




Vector additionAddition follows the parallelogram law described in the figure.

x x y y F E F E ! F E i j

F

Fx

Fy

x

y

EEy

Ex

F+E

E

FE+F




Part 3

Dot Product, Cross Product and

Triple Product




Dot product

The dot product of two vectors yields a scalar:

C ! A . B

Magnitude:

cosC AB U!

A

B




Right handed system of coordinates

thumb

indexmiddle




Cross productThe cross product of two vectors yields a vector:

v ! A B C

Magnitude:

sinC AB U!

Direction:Vector C has a direction

perpendicular to the plane containingA and B such that C is specified bythe right hand rule.




Cross productLaws of operations:The Commutative Law is Not Valid:

v { v A B B A

v v A B = -B A

Multiplication by a scalar:

a a a av ! v v ! v A B A B = A B A B

The Distributive Law:

v ! v v A B + D A B A D




Cross products of unit vectorsThe direction is determined using the right hand rule. As shown in thediagram, for this case the direction is k and the Magnitude is:

| i v j |=(1)(1)(sin90 °) = (1)( 1)( 1) =1so: i v j = ( 1) k = k

and: i v j = k i v k = - j i v i = 0 j v k = i j v i = -k j v j = 0 k v i = j k v j = -i k v k = 0

alphabeti c al order +




Cross product of two vectors

Cross product of two vectors in terms of their components:

A = Axi + Ay j +Az k

B = B xi + B y j + B z k

A v B = (Axi + Ay j + Az k ) v ( B xi + B y j + B z k )

= AxB x ( i v i) + AxB y ( i v j ) + AxB z ( i v k ) + Ay B x ( j v i) + Ay B y ( j v j ) + Ay B z ( j v k ) + Az B x ( k v i) + Az B y ( k v j ) + Az B z ( k v k )

= 0 + AxB y k ² AxB z j - Ay B x k + 0 + Ay B z i + Az B x j ² Az B y i + 0

Hen c e: A v B = (Ay B z ² Az B y ) i - (AxB z -Az B x) j + (AxB y - Ay B x ) k




Cross product of two vectorsThis equation may also be written in a compact determinant form:

x y z

x y z

A A A

B B B

v

i j k

A B =

For element i: ( i )(A y B z ± A z B y )

z y x

z y x

B B B

A A A

k j

BA !v

z y x

z y x

k j

BA !v

For element j: (- j )(A B z -A z B )

(notice the negative sign

here) z¡ ¢

z¡ ¢

B B B

A A A

k ji

!v

z£ ¤

z£ ¤

B B B

A A A

k ji

!v

For element k : ( k )(A¥ B y - A y B¥ )

¦ y§

¦

y§

B B B

A A A

k ji

BA !v

z y x

z y x

B B B

A A A

k ji

BA !v

Hence:

A v B = ( A y B z ± A z B y ) i - ( A x B z - A z B x ) j + ( A x B y - A y B x ) k





m

In summary, The cross product of vectors a and b is a vector perpendicularto both a and b and has a magnitude equal to area of the parallelogramgenerated from a and b .

The direction of the cross product is given by the right hand rule (fingers

from vector a to vector b and thumb is along vector c ). Order is important inthe cross product:





( , and are the unit vectors)

Area sin

hich is the unit vector along the line perpendicular to the p

x y z

x y z

x y z y z z y x z z x x y y x

x y z

a a a

b b b

a a a a b a b a b a b a b a b ab

b b b

U

v ! v

!

!

v ! ! !

a b b a

a i j k i j k

b i j k

i j k

a b = i j k m m

m lane o anda b

m




Triple product

.

x y z x y z

x y z x y z y z z y x x z z x y x y y x z

x y z x y y

a a a a a a

b b b b b b b c b c a b c b c a b c b c a

c c c c c c

® ! ±

! v ! ¯± ! °

a i j k

b i j k a b c =

c i j k

The volume of the parallelepiped constructed from the vectors a , b , and c is given bythe triple product of the three vectors:

volume sin cos

. cos

abc

a

U N

N

!

v ! va b c b c




Part 4

D

ifferentiation,I

ntegration andCentroids




Differentiation (common derivatives)d/d

x( c )= 0The der ivative of a constant is zer o.

Example: d(7) /dx = 0

d/dx( c × x )= c

The r ate of change of a linear function is its slope.

Example: d(3 × x ) /dx = 3

d/dx (xn) = n × x(n-1)

Example: d( x 4 )/dx = 4× x 3

d/dx (log x) = 1/x

The der ivative of the log of x is its inver se.

Example: d ( log ( x + 1) ) /dx = 1 / ( x + 1)

d/dx (eax) = a eax

Example: d (e3x ) /dx= 3 e3x

d/dx (sin cx) = c cos x

Example: d ( sin3x ) /dx = 3cos x

d/dx (cos x) = -sin x

Example: d ( cos t)= -sin t




Integral of a function

The integral of a functionf( x) over an interval fromx1 to x2 yield the areaunder the curve in this

interval.

Note: The integral represents the

( ) F x x(§ as 0 x( p




Some indefinite integrals to remember




Some indefinite integrals to remember

Note: Remember to add a constant of integration if you are not specifyinglimits. You evaluate the constant of integration by forcing the integral topass through a known point.




Definite integralNote: For definite integrals subtract the value of the integral atthe lower limit from its value at the upper limit. For example, if

you have the indefinite integral.Note: The following notation is common:

2

12 1( ) ( ) ( )

x x

x x F x F x F x!

! !

Integration by parts:

U dV U V Vd U ! ´ ´




Centroid of an area

The centroid of an area is the area weighted average location of thegiven area.

1 1 1,

OC OC OC

OC OC OC

A A A

x y

x y

d A x xd A y r d A A A A

!

!

! ! !´ ´ ´

r i j

r i j

r r




Centroid of an areaFor example, consider a shape that is a composite of n individual segments,each segment having an area Ai and coordinates of its centroid as xi and y i.The coordinates of the centroid of this composite shape is given by




Centroids of common shapes

Statics Math

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