physics i unit 4 vectors & motion in two dimensions astr.gsu.edu/hbase/vect.html#vec1 web sites
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Physics I Unit 4VECTORS& Motion in TWO Dimensions
http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec1
Web Sites
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• Objectives:Calculate and / or measure Velocities Vectors using graphical means
Calculate and / or measure Velocities Vectors using algebraic means
Do Now Page 117 #1
Homework
Unit 4 Lesson 1
IN CLASS:Page 121 Example 1
Pg 121 – 125 Practice Problems: 1-9 ODD:
Pg 141 – 142 Problems:
80 – 89 all 80 – 88 EVEN
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Vectors and Scalars
A vector has magnitude as well as direction.
Some vector quantities: displacement, velocity, force, momentum
A scalar has only a magnitude.
Some scalar quantities: mass, time, temperature
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Addition of Vectors – Graphical Methods
For vectors in one dimension, simple addition and subtraction are all that is needed.
You do need to be careful about the signs, as the figure indicates.
8 km + 6 km = 14 km East
8 km - 6 km = 2 km East
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Addition of Vectors – Graphical Methods
Even if the vectors are not at right angles, they
can be added graphically by using the “Head to Tail” method:
1. Draw V1 & V2 to scale.2. Place tail of V2 at tip of V13. Draw arrow from tail of V1 to tip
of V2This arrow is the resultant V (measure length and the angle it makes with the x-axis)
Same for any number of vectors involved.
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Addition of Vectors – Graphical Methods
The parallelogram method may also be used;
1. Draw V1 & V2 to scale from common origin.2. Construct parallelogram using V1 & V2 as 2 of the 4 sides.Resultant V = diagonal of parallelogram from common origin (measure length and the angle it makes with the x-axis)
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Addition of Vectors – Graphical MethodsIf the motion is in two dimensions, the situation
is somewhat more complicated. Here, the actual travel paths are at right angles to one another; we can find the displacement by using the Pythagorean Theorem.
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Objectives:Word Problems with VECTORS
Do Now• Page 117 #2• Page 121 #4
Homework• Pg 141 – 142 Problems:
80 – 89 all
Unit 4 Lesson 2
IN CLASS:
Pg 125 Section Review Problems: 11 -15 ALL
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Adding Vectors by Components
Any vector can be expressed as the sum of two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicular to each other.
Unit 4 Lesson 2
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Adding Vectors by Components
If the components are NOT perpendicular,
they can be found using trigonometric functions.Vx = V Cosθ Vy = V
Sin θ1) __________ _________2) ___________ _________3) ___________ _________+Total Vx Total Vy
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Objectives:-Determine Equilibrium Forces-Determine the motion of an object on an inclined plane with and without friction
Do Now Page 128 #17 Next SLIDE(s)
Homework Page 133-135
Practice Problems: 33- 41 odd
Unit 4 Lesson 3 Nov 2
IN CLASS:Pg 133 Example 5Pg 134 Example 6
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DO NOW 3 Vector Problem Graphically Analytically
Unit 4 Lesson 3
Find the resultant vector (Magnitude and Direction of the THREE FORCE vectors acting on the Object at the origin
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Solving the components
• X Components Vx = V Cosθ• Ax = 44.0 * COS (28.0) = 38.85• Bx = 26.5 *COS (124) = - 14.82• Cx = 31.0 * COS (270.0) = 0.00• X total = 24.031
• Y Components Vy = V SINθ• Ay = 44.0 * SIN(28.0) = 20.66• By = 26.5 *SIN(124) = 21.97• Cy = 31.0 * SIN(270) = - 31.0• Y total = 11.63
V = √{[24.031]2 + [11.63]2}V = √{[24.031]2 + [11.63]2}V = √{[712.748]} = 26.6973 m/s
= 26.7 m/s
3 Vector Problem
Θ = Tan-1 {11.63/24.031}Θ = Tan-1 {0.483958} Θ = 25.8225 degrees Θ = 25.8 degrees
Unit 4 Lesson 3
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Objectives:Two Dimensions in Motion
Do Now If F┴ = Fapp Cosθ
If F ⁄⁄ = Fapp SINθ And If Fg = 2.55 Kg * 9.8 m/s2 = 25
N Θ = 15.0 ° What is the Velocity after 10
seconds of the object rolling down a 15.0 degree ramp?
Homework page 135 #’s 38 -45 ALL
Unit 4 Lesson 5 NOV 5
IN CLASS: A 62 kg skier is going down a 37 degree slope. The coefficient of friction is 0.15. How fast is the skier going after 5.0 sec?
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Grade quiz Now A 62 kg skier is going
down a 37 degree slope. The coefficient of friction is 0.15.
How fast is the skier going after 5.0 sec?
Vf = ______________
solution FN = F┴ = Fg Cosθ
= mg(Cosθ) = 62*9.8*(Cos37) = 485.2509N
If F ⁄⁄ = Fg SINθ = mg(Sinθ) = 62*9.8*(Sin37) = 365.6628N
Fnet = Fapp - Ffric
Fnet = F ⁄⁄ - μmg(Cosθ) Fnet = 365.6628N – μ 485.2509N
Fnet = 365.6628N – (0.15) 485.2509N Fnet = 298.8752 N = ma a = Fnet /m = 298.8752 /62 = 4.7238
m/s2
Vf = Vi + at = (4.7238 m/s2 )* (5.0) Vf = 23.6190 = 2.4 X 101 m/s
Unit 4 Lesson 5A NOV 6
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Objectives:Multiple Vectors
Do Now Vector 1 = 42.5N @ 56.0 Degrees Vector 2 = 75.1 N @ 245.0 deg Vector 3 = 10.5 N @ 315.0 deg Mass = 15.5 Kg Find F net= ___________ Find Vel (5 sec)= _________ Find Distance (5 sec) = _________
Homework Page 159 # 23, 25
Unit 4 Lesson 6 Nov 6
IN CLASS: Solve**
Vx = Vi COS Vx Total =
Vy = Vi SIN Vy Total =
V resultant = √{[Vx Total]2 + [Vy Total]2}
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Objectives:Adding Vectors ALGEBRAICALLYForces in Equilibrium
Do Now 1 A vector has a Magnitude
of 10.0 Newton’s, 30.0 from the horizontal.
Find the “X” component of the vector :
Find the “Y” component of the vector :
Homework Page 142 #’s 88, 90, 92, 94, 95,
96 97 HONORS
Unit 4 Lesson 7 Nov 8
DO NOW 2:What is the actual magnitude and the direction of a boat if it is heading due NORTH (0 degrees) at 3.0 m/s on a river that is flowing due EAST (90 degrees) at 2.0 m/s and a wind that is blowing @ 5.0 m/s to the North East (30 degrees)?
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DO NOW: What is the Frictional Force required to keep a 10.0 Kg block from sliding down a 25 degree incline?
Unit 4 Lesson 7 Nov 9IN CLASS 1:
Romac hangs his 50.0 Kg Repair sign from two wires that make a 90.0 degree angle between themselves and a 45.00 degree angle from the horizontal. What is the tension on each wire? Equilibrium
50.0*9.801 = 490.0 N490 / 2 = 245.0 Sin (45) = 174.94785 / HypHyp = 174.94785 / Sin (45) =
Romacs
MGBRepair
45.00
90.0
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Objectives:Exam Review
Do Now
Homework Putting it all
together
Unit 4 Lesson #
IN CLASS: