Physics I Unit 4 VECTORS & Motion in TWO Dimensions 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 V1 3. Draw arrow from tail of V1 to tip of V2 This 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 Methods If 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: ALL 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 θ 1) __________ _________ 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 Practice Problems: odd Unit 4 Lesson 3 Nov 2 IN CLASS: Pg 133 Example 5 Pg 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) = Bx = 26.5 *COS (124) = Cx = 31.0 * COS (270.0) = 0.00 X total = Y Components Vy = V SINθ Ay = 44.0 * SIN(28.0) = By = 26.5 *SIN(124) = Cy = 31.0 * SIN(270) = Y total = V = √{[24.031] 2 + [11.63] 2 } V = √{[ ]} = m/s = 26.7 m/s 3 Vector Problem Θ = Tan -1 {11.63/24.031} Θ = Tan -1 { } Θ = degrees Θ = 25.8 degrees Unit 4 Lesson 3
Objectives: Two Dimensions in Motion Do Now If F ┴ = F app Cosθ If F ⁄⁄ = F app SINθ And If F g = 2.55 Kg * 9.8 m/s 2 = 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 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 How fast is the skier going after 5.0 sec?
Grade quiz Grade quiz Now A 62 kg skier is going down a 37 degree slope. The coefficient of friction is How fast is the skier going after 5.0 sec? Vf = ______________ solution F N = F ┴ = F g Cosθ = mg(Cosθ) = 62*9.8*(Cos37) = N If F ⁄⁄ = F g SINθ = mg(Sinθ) = 62*9.8*(Sin37) = N F net = F app - F fric F net = F ⁄⁄ - μmg(Cosθ) F net = N – μ N F net = N – ( N F net = N – ( 0.15) N F net = N = ma a = F net /m = /62 = m/s 2 Vf = Vi + at = ( m/s 2 )* (5.0) Vf = = 2.4 X 10 1 m/s Unit 4 Lesson 5A NOV 6
Objectives: Multiple Vectors Do Now Vector 1 = 56.0 Degrees Vector 2 = deg Vector 3 = 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** V x = Vi COS V x Total = V y = Vi SIN V y Total = V resultant = √{[V x Total] 2 + [V y Total] 2 }
Objectives: Adding Vectors ALGEBRAICALLY Forces 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 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 9 IN CLASS 1: Romac hangs his 50.0 Kg Repair sign from two wires that make a 90.0 degree angle between themselves and a degree angle from the horizontal. What is the tension on each wire? Equilibrium 50.0*9.801 = N 490 / 2 = Sin (45) = / Hyp Hyp = / Sin (45) =
Objectives: Exam Review Do Now Homework Putting it all together Unit 4 Lesson # IN CLASS: