Transcript math and vector review

Mathematical Fundamentals
• Need working knowledge of algebra
and basic trigonometry
• if you don’t have this then you must see
me immediately!
15
Algebra Review
• Exponents - Square Roots
5
1/2
2
exponent
5 * 5 = 25
25 = 25
=5
3
2 =2*2*2=8
16
Order of Operations
• Solve the following problem
(12 +
2
3
* 3)2-1 + 3 * 2 - (8/4) - 52 - 6/2 = ???
17
18
Order of Operations
• (1) parentheses, brackets, and braces
• (2) exponents, square roots
• (3) multiplication and division
• (4) addition and subtraction
19
Order of Operations Problem
SOLUTION
(12 +
2
3
* 3)2-1 + 3 * 2 - (8/4) - 52 - 6/2 = ???
1. parentheses
2
(12+ 3 *3)
1a.
2
3
*3=2
1b. 12 + 2 = 14
1c. 8/4 = 2
20
Order of Operations Problem
SOLUTION
(12 +
2
3
* 3)2-1 + 3 * 2 - (8/4) - 52 - 6/2 = ???
2. exponents
52 = 25
3. multiplication & division
2
(12 + 3 *3)2 = 14*2 = 28
NOTE: 14 was calculated in steps 1a and 1b.
6/2 = 3
3*2 = 6
21
Order of Operations Problem
SOLUTION
(12 +
2
3
* 3)2-1 + 3 * 2 - (8/4) - 52 - 6/2 = ???
Substitute into equation
28 -1 + 6 - 2 - 25 - 3 = 3
22
Trigonometry
• field of mathematics focusing on
relationships between sides of and the
angles within a right triangle
23
Trigonometry Review
a
a = “opposite” side
b = “adjacent” side
c = “hypotenuse”
q = angle
c
q
b
24
SOHCAHTOA
4 Basic Relationships
1. a2 + b2 = c2
c
(Pythagorean Theorem)
2. sin q = opp/hyp
= a/c
q
3. cos q = adj/hyp
b
= b/c
a = “vertical component”
4. tan q = opp/adj
b = “horizontal component”
= a/b
c = “resultant”
a
25
Two types of TRIG problems
Type A
Given
c&q
Type B
a&b
Solve For
a&b
c&q
26
TYPE A Problem
Given:
c = 10 m/s
q = 40 degrees
Find: a and b
a
o
40
b
a
sin 40 =
10 m/s
o
o
10 m/s * sin 40 =
a
10 m/s
10 m/s
o
a = 10 m/s * sin 40 = 6.43 m/s
o
cos 40 =
b
10 m/s
o
10 m/s * cos 40 =
b
10 m/s
10 m/s
o
b = 10 m/s * cos 40 = 7.66 m/s
27
Type B Problem
Given: a = 400 lb, b = 100 lb
Find: c and q
c
400 lb
q
100 lb
a2 + b2 = c2
(400 lb)2 + (100 lb)2 = c2
160000 lb2 + 10000 lb2 = c2
170000 lb2 = c2
c = 412.3 lb
a
tan q = b
400 lb
tan q = 100 lb
tan q = 4
tan-1 (tan q) = tan-1(4)
o
q = 76.0
28
Inverse Trig Functions
If sin is a trig function
then sin-1 is an
inverse trig function
:inverse trig functions simply “undo” trig functions
29
SOHCAHTOA
• SOH
 Sine = Opposite/Hypotenuse
• CAH
 Cosine = Adjacent/Hypotenuse
• TOA
 Tangent = Opposite/Adjacent
30
25
20
o
a
b
Calculate the vertical (a) and horizontal
sides of this right triangle.
31
25
20
a
o
b
o
sin 20 =
a
25
a = 25 (sin 20)
a = 8.55
o
cos 20 =
b
25
b = 25 (cos 20)
b = 23.49
32
c
15
Solve for the length
of the hypotenuse (c)
and the angle, q.
q
10
33
c = 152 + 102
c
15
q
10
c = 325
c = 18.03
15
tan q =
10
q = tan-1 (1.5)
o
q = 56.3
34
UNITS
• Use the SI system
– AKA Metric System
– 4 basic units
•
•
•
•
length
mass
time
temperature
-----
meter
kilogram
second
degree Kelvin
(Celsius)
35
Vector Resolution Example
Billy pulls on his new wagon with
50 lbs of force at an angle of 45 .
How much of this resultant force is
actually working to pull the wagon
horizontally?
Radio Flyer
50 lbs
o
45
F = Fx + Fy
F = magnitude of F = 50 lbs
o
cos 45 =
F
45
Fy
o
Fx
o
sin 45 =
Fx
F
Fy
F
F = Fx + Fy
o
Fx = F cos 45
o
F
45
Fy = F sin 45
Fy
o
Fx
o
Fx = 50 lbs (cos 45 ) = 50 lbs * 0.707 = 35.4 lbs
o
Fy = 50 lbs (sin 45 ) = 50 lbs * 0.707 = 35.4 lbs
Sometimes the magnitude of a force is
written more simply as
50 lbs
o
Fx = 35.4 lbs
Fy = 35.4 lbs
45
Radio Flyer
Only the force acting in the x-direction acts to move
the wagon forward
Vector Decomposition
aka Vector Resolution
Any vector can be expressed as a pair of two
component vectors
these vectors
1) must be perpendicular to each other
2) are usually horizontal and vertical
40
Vector Decomposition
10sin(40)
Given the polar notation
of a vector, decompose it
into vertical and horizontal
components (Cartesian
coordinates).
q = 40o
x
10cos(40)
y
•sx = |S|cosq = 10(.766) =
7.66 m
• sy = |S|sinq = 10(.643) =
6.43 m
41
Vector Composition
(aka Vector Addition)
• to add 2 vectors must consider both
magnitude and direction
• the sum of 2 or more vectors is known as a
resultant vector
• if the vectors have the same direction then
you may add the magnitudes directly
+
=
• vectors are in opposite direction
– resultant vector points in direction of longer
vector
– size of resultant vector is the difference
between the component vectors
+
=
• vectors are pointed in different, nonparallel, direction
• graphical solution - TIP-TO-TAIL
method
+
• TIP-TO-TAIL method
+
=
Resultant vector is
the diagonal of the
resulting parallelogram
resultant
vector
place the tail of the 2nd vector at the tip of the 1st vector
connect the tail of the 1st vector to the 2nd vector
• TIP-TO-TAIL method is the preferred
method when adding more than 2
vectors
– include more vectors by attaching their tail
to the open tip in the diagram
+
+
+
+
+
+
+
+
Two Forces
Acting on the Hip
Vector Example
body
weight
muscle
W
Graphically compute the
resultant force acting on
the femoral head.
R = Fm+W
W
Fm
R
W
resultant force
acting on the
femoral head
Vector Addition
• Vectors can be added by
placing the tail of each
vector at the tip of the
previous one.
• The sum of all of these
vectors is called the
resultant vector. It connects
the tail of the first vector to
the head of the last vector.
50
Vector Addition
• Finding the horizontal
and vertical components
of each vector makes it
easy to find the resultant.
51
Vector Addition
• Simply add all of the
vertical lines for the
vertical component and
add all of the horizontal
lines for the horizontal
component. Be sure to
pay attention to the sign
of each of the lines.
52
Vector Addition
• Use the following formulas to
convert the coordinates into
polar notation:
Sy
q
Sx
•
s  s s
2
x
2
y
Opp
• q = arctan
Adj
53
S2 = 3m, 165
o
y
S1 = 6m, 40
x
o
54
S2 = 3m, 165
o
y
S1 = 6m, 40
x
o
55
Sx1 = |S1|cosq1 = 6(.799) = 4.60 m
S2 = 3m, 165o
Sy1 = |S1|sinq1 = 6(.643) = 3.86 m
Sx2 = |S2|cosq2 = 3(-.966) = -2.90 m
y
Sy2 = |S2|sinq2 = 3(.259) = .78 m
S1 = 6m, 40o
x
Sx = 4.60 - 2.90 = 1.70 m
Sy = 3.86 + .78 = 4.64 m
56
Polar Notation
• |S| =
x y
2
2
= 4.94 m
Opp
• q = arctan
= 69.9o
Adj
57