Physics I Unit 4VECTORS& Motion in TWO Dimensions

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec1

Web Sites

• 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

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

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

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.

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)

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.

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

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec1

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

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

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

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

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

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?

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

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}

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)?

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

Objectives:Exam Review

Do Now

Homework Putting it all

together

Unit 4 Lesson #

IN CLASS: