Professor Martinez. COMMON CONVERSION FACTORS 1 ft = 0.3048 m 1 lb = 4.4482 N 1 slug = 14.5938 kg...
Transcript of Professor Martinez. COMMON CONVERSION FACTORS 1 ft = 0.3048 m 1 lb = 4.4482 N 1 slug = 14.5938 kg...
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Professor Martinez
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COMMON CONVERSION FACTORS
1 ft = 0.3048 m 1 lb = 4.4482 N 1 slug = 14.5938 kg
Example: Convert a torque value of 47 in • lb into SI units. Answer is 5.31026116 N • m
•Work problems in the units given unless otherwise instructed!
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Problems Convert 2 km/h to m/s. How many ft/s is
this?
Evaluate each of the following and express with SI units having an appropriate prefix:
(a) (50 mN)(6 GN)
(b) (400 mm)(0.6 MN)²
(c) 45 MN³/900 Gg
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THE INTERNATIONAL SYSTEM OF UNITS
(Section 1.4)
• No Plurals (e.g., m = 5 kg not kgs )
• Separate Units with a • (e.g., meter second = m • s )
•Most symbols are in lowercase.
•Some exceptions are N, Pa, M and G.
•Exponential powers apply to units, e.g., cm • cm = cm2
•Compound prefixes should not be used.
•Other rules are given in your textbook (pg. 10).
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NUMERICAL CALCULATIONS (Section 1.5)
Must have dimensional “homogeneity.” Dimensions have to be the same on both sides of the equal sign, (e.g. distance = speed time.)
• Be consistent when rounding off.
- greater than 5, round up (3528 3530)
- smaller than 5, round down (0.03521 0.0352)
- equal to 5, round up.
• Use an appropriate number of significant figures (3 for answer, at least 4 for
intermediate calculations). Why?
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PROBLEM SOLVING STRATEGY: IPE, A 3 Step Approach
1. Interpret: Read carefully and determine what is given and what is to be found/
delivered. Ask, if not clear. If necessary, make assumptions and
indicate them.
2. Plan: Think about major steps (or a road map) that you will take to solve a given
problem. Think of alternative/creative solutions and choose the best
one.
3. Execute: Carry out your steps. Use appropriate diagrams and equations. Estimate
your answers. Avoid simple calculation mistakes. Reflect on /
revise your work.
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ATTENTION Review
2. In three step IPE approach to problem solving, what does P stand for?
A) Position B) Plan C) Problem
D) Practical E) Possible
1. For a statics problem your calculations show the final answer as 12345.6 N. What
will you write as your final answer?
A) 12345.6 N B) 12.3456 kN C) 12 kN
D) 12.3 kN E) 123 kN
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FORCE VECTORS, VECTOR OPERATIONS & ADDITION COPLANAR FORCES
Today’s Objective:
Students will be able to :
a) Resolve a 2-D vector into components.
b) Add 2-D vectors using Cartesian vector notations.
In-Class activities:• Reading Review • Application of Adding Forces• Parallelogram Law • Resolution of a Vector Using Cartesian Vector Notation (CVN) • Addition Using CVN • Attention Review
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READING Review
1. Which one of the following is a scalar quantity?
A) Force B) Position C) Mass D) Velocity
2. For vector addition you have to use ______ law.
A) Newton’s Second
B) the arithmetic
C) Pascal’s
D) the parallelogram
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READING QUIZ
1. Which one of the following is a scalar quantity?
A) Force B) Position C) Mass D) Velocity
2. For vector addition you have to use ______ law.
A) Newton’s Second
B) the arithmetic
C) Pascal’s
D) the parallelogram
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APPLICATION OF VECTOR ADDITION
There are four concurrent cable forces acting on the bracket.
How do you determine the resultant force acting on the bracket ?
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SCALARS AND VECTORS (Section 2.1)
Scalars Vectors
Examples: mass, volume force, velocity
Characteristics: It has a magnitude It has a magnitude
(positive or negative) and direction
Addition rule: Simple arithmetic Parallelogram law
Special Notation: None Bold font, a line, an
arrow or a “carrot”
In the PowerPoint presentation vector quantity is represented Like this (in bold, italics, and yellow).
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VECTOR OPERATIONS (Section 2.2)
Scalar Multiplication
and Division
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VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE
Parallelogram Law:
Triangle method (always ‘tip to tail’):
How do you subtract a vector?
How can you add more than two concurrent vectors graphically ?
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Vector Subtraction
R’ = A – B = A + (-B) Example, pg. 19
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“Resolution” of a vector is breaking up a vector into components. It is kind of like using the parallelogram law in reverse.
RESOLUTION OF A VECTOR
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CARTESIAN VECTOR NOTATION (Section 2.4)
• Each component of the vector is shown as a magnitude and a direction.
• We ‘resolve’ vectors into components using the x and y axes system.
• The directions are based on the x and y axes. We use the “unit vectors” i and j to designate the x and y axes.
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For example,
F = Fx i + Fy j or F' = F'x i + F'y j
The x and y axes are always perpendicular to each other. Together,they can be directed at any inclination.
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ADDITION OF SEVERAL VECTORS
Step 3 is to find the magnitude and angle of the resultant vector.
• Step 1 is to resolve each force into its components
• Step 2 is to add all the x components together and add all the y components together. These two totals become the resultant vector.
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Example of this process,
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You can also represent a 2-D vector with a magnitude and angle.
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EXAMPLEGiven: Three concurrent
forces acting on a bracket.
Find: The magnitude and angle of the resultant force.
Plan:
a) Resolve the forces in their x-y components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
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EXAMPLE (continued)
F1 = { 15 sin 40° i + 15 cos 40° j } kN
= { 9.642 i + 11.49 j } kN
F2 = { -(12/13)26 i + (5/13)26 j } kN
= { -24 i + 10 j } kNF3 = { 36 cos 30° i – 36 sin 30° j } kN
= { 31.18 i – 18 j } kN
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EXAMPLE (continued)
Summing up all the i and j components respectively, we get,
FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN
= { 16.82 i + 3.49 j } kN
x
y
FRFR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN
= tan-1(3.49/16.82) = 11.7°
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GROUP PROBLEM SOLVING
Given: Three concurrent forces acting on a bracket
Find: The magnitude and angle of the resultant force.
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F1 = { (4/5) 850 i - (3/5) 850 j } N
= { 680 i - 510 j } N
F2 = { -625 sin(30°) i - 625 cos(30°) j } N
= { -312.5 i - 541.3 j } N
F3 = { -750 sin(45°) i + 750 cos(45°) j } N
{ -530.3 i + 530.3 j } N
GROUP PROBLEM SOLVING (continued)
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GROUP PROBLEM SOLVING (continued)
Summing up all the i and j components respectively, we get,
FR = { (680 – 312.5 – 530.3) i + (-510 – 541.3 + 530.3) j }N
= { - 162.8 i - 521 j } N
FR = ((162.8)2 + (521)2) ½ = 546 N
= tan–1(521/162.8) = 72.64° or
From Positive x axis = 180 + 72.64 = 253 °
y
x
FR
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ATTENTION Review1. Resolve F along x and y axes and write it in
vector form. F = { ___________ } N
A) 80 cos (30°) i - 80 sin (30°) j
B) 80 sin (30°) i + 80 cos (30°) j
C) 80 sin (30°) i - 80 cos (30°) j
D) 80 cos (30°) i + 80 sin (30°) j
2. Determine the magnitude of the resultant (F1 + F2) force in N when F1 = { 10 i + 20 j } N and F2 = { 20 i + 20 j } N .
A) 30 N B) 40 N C) 50 N
D) 60 N E) 70 N
30°
xy
F = 80 N
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Homework
Complete Problems:1-6 (pg. 15)1-8 (pg. 15)2-10 (pg. 28)*2-32 (pg. 38)*2-57 (pg. 42)*
Complete all work in pencil Grading on 1-5 scale Looking for thoroughness (don’t just write the
answer) Due next Thursday at beginning of class
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