Chapter 6 - Vectors
Scalar Quantities - express only magnitude ie. time, distance, speed Vector Quantities - express magnitude and direction. ie. velocity 80 km/h, 58 displacement 10 km (E) acceleration 4.0 m/s 2, 27 force 100 N, 110
Vector Quantities represented by an arrow. length of arrow indicates magnitude (drawn to scale), arrowhead indicates direction.
Graphical Analysis of Vectors 1. Add vectors by placing the tail of one vector at the head of the other vector. 2. A third vector is drawn connecting the tail of the first vector with the tip of the last vector. This vector, the resultant, represents the sum of the vectors. 3. Order of addition does not matter.
Ex. 1 A = 200 m(e), B = 400 m(e) 200 m 400 m 600 m
Ex. 2. A = 400 m(e), B = 200 m(w) 400 m 200 m 200 m
Ex. 3. A = 200 m(e), B = 400 m(w) 200 m 200 m 400 m
Vector Addition in 2-Dimensions Resultant a) magnitude = measure the length from tail of the 1st vector to the tip of the last vector. b) direction = measure the angle with a protractor from the horizontal vector, measured counter-clockwise.
Vector Addition in 2-Dimensions Ex.1 A = 95 km,e B = 55 km, N 110 km 55 km 30 o 95 km
Several Vectors Ex.1: A hiker travels 7 km N, then 18 km E, and 3 km S. What is the hiker s final displacement? A = 7 km,n B = 18 km, E C = 3 km, S 7 km 18 km 3 km 77 o 18 km
Several Vectors Ex.2: An ant travels 3cm, 20 o, then travels 7cm, 60 o, and finally 5cm, 110 o. What is the final displacement? A = 3km, 20 o B = 7km, 60 o C = 5km, 110 o 13 km 5 km 110 o R = 13km, 70 o 7 km 70 o 20 o 3 km
Vector Addition Order of addition does not matter the resultant still has the same magnitude and direction!
Analytical Method of Vector Addition Length of the Vector: Pythagorus Theorum a 2 + b 2 = c 2 4.0 2 + 6.0 2 = c 2 c = (4.0 2 + 6.0 2 ) 4.0 m/s (a) (c) = 7.2 m/s 6.0 m/s (b)
Analytical Method of Vector Addition Angle of the Vector: Trigonometric Functions sin θ = opposite hypotenuse cos θ = adjacent hypotenuse tan θ = opposite adjacent Opposite REMEMBER: SOH/CAH/TOA Adjacent Hypotenuse θ
Ex#1. An airplane flying toward 0 o at 90 km/h is being blown toward 90 o at 50 km/h. What is the resultant velocity of the plane? a 2 + b 2 = c 2 90 2 + 50 2 = c 2 90 km/h 50 km/h c = (90 2 + 50 2 ) tan θ = 50 90 = 103 m/s θ = 29 o R = 103 m/s, 29 o
Ex#2. A swimmer jumps into a river and swims straight for the other side at 3.0 km/h(n). There is a current in the river of 4.0 km/h (W). What is the swimmer s velocity relative to the shore? a 2 + b 2 = c 2 4.0 km/h 3.0 2 + 4.0 2 = c 2 θ 3.0 km/h c = (3.0 2 + 4.0 2 ) = 5.0 m/s tan θ = 4.0 3.0 R = 5.0 m/s, N53 o W θ = 53 o
Independence of Vectors perpendicular vector quantities are independent of each other.
Ex#1. A motorboat heads east at 8.0 m/s across a river that flows north at 5.0 m/s. a) Calculate the resultant velocity. 8.0 2 + 5.0 2 = c 2 c = (8.0 2 + 5.0 2 ) γ θ 8.0 m/s 5.0 m/s tan θ = 5.0 8.0 = 9.4 m/s θ = 32 o γ = 90 o 32 o = 58 o R = 9.4 m/s at N58 o E
Ex#1. A motorboat heads east at 8.0 m/s across a river that flows north at 5.0 m/s. b) If it takes the boat 10 s to cross the river, what is the length of the river? γ θ 8.0 m/s 5.0 m/s The boat travels at 8.0 m/s east for 10 s. d = vt = (8.0 m/s)(10s) = 80 m
Ex#1. A motorboat heads east at 8.0 m/s across a river that flows north at 5.0 m/s. c) How far down the river did he travel? γ θ 8.0 m/s 5.0 m/s The river flows at 5.0 m/s north for 10 s. d = vt = (5.0 m/s)(10s) = 50 m
Ex#2. boat travels 3.5 m/s and heads straight across a river 240 m wide. a) If the river flows at 1.5m/s, what is the resultant speed of the boat relative to the shore? 3.5 2 + 1.5 2 = c 2 c = (3.5 2 + 1.5 2 ) γ θ 3.5 m/s 1.5 m/s tan θ = 1.5 3.5 = 3.8 m/s θ = 23 o γ = 90 o 23 o = 67 o R = 3.8 m/s at 67 o
Ex#2. boat travels 3.5 m/s and heads straight across a river 240 m wide. b) How long does it take the boat to cross the river? γ θ 3.5 m/s 1.5 m/s The boat crosses at 3.5m/s a distance of 240m. v = 3.5m/s d = 240 m t =? t = d v = 240m 3.5m/s = 68.57s = 69s
Components of Vectors Two vectors acting in different directions may be replaced by a single vector the resultant. Therefore: a single vector is the resultant of 2 vectors. C2 B1 A B2 A is a result of B and C C1 Vector Resolution finding the magnitude of a component in a given direction.
Ex#1. A person pulling a sled exerts a force of 58N on a rope held at an angle of 30 o with the horizontal. What is the vertical component? What is the horizontal component? F R 30 o F H F V F H = F R cos 30 o = (58N)(0.866) = 50N F V = F R sin 30 o = (58N)(0.5) = 29N
Ex#2. A wind with a velocity of 40.0 km/h blows towards 30.0 o. a) What is the component of the wind s velocity toward 90.0 o? v 90 = v R sin 30 o 30 o v R v 90 = (40.0km/h)(0.5) = 20.0 km/h v 0 b) What is the component of the wind s velocity towards 0 o? v 0 = v R cos 30 o = (40.0km/h)(0.866) = 34.6 km/h
Ex#3. Beth attempts to pull a stake out of the ground by pulling a rope that is attached to the stake. The rope makes and angle of 60.0 with the horizontal. Beth exerts a force of 125N on the rope. What is the magnitude of the vertical component of the force acting on the stake. F R = 125 N F R F V F V = F R sin 60 o 60 o F H F V = (125 N)(sin 60 o ) = 108 N
Vectors At Any Angle 1. Resolve each vector into perpendicular components. F 2 = 8.0 N 120 o 10 o F 1 = 12.0 N F 1H = F 1 cos 10 o = (12.0N)(0.9848) = 11.8 N F 1V = F 1 sin 10 o = (12.0N)(0.1736) = 2.1 N F 2H = F 1 cos 120 o = (8.0N)(-0.5) = -4.0 N F 2V = F 1 sin 120 o = (8.0N)(0.8660) = 6.9 N
Vectors At Any Angle 2. Add horizontal components together. F 2 = 8.0 N 120 o 10 o F 1 = 12.0 N F H = F 1H + F 2H = 11.8 N + -4.0 N = 7.8 N
Vectors At Any Angle 3. Add vertical components together. F 2 = 8.0 N 120 o 10 o F 1 = 12.0 N F V = F 1V + F 2V = 2.1 N + 6.9 N = 9.0 N
Vectors At Any Angle 4. Add vertical and horizontal components to obtain the final resultant. F R 2 = F V 2 + F H 2 = 9.0 2 + 7.8 2 F 2 = 8.0 N 120 o 10 o F 1 = 12.0 N F R = 11.9 N tanθ = 9.0/7.8 θ =49 o 7.8 N 9.0 N
Example: Two ropes are pulling a log. Rope A pulls at an angle of 128 o at 10 N. Rope B pulls at angle of 15 o at 15 N. What is the net force on the log? 10 N 128 o 15 N 15 o F x = F Ax + F Bx = F A cos128 o + F B cos15 o = (10N)cos128 o + (15N)cos15 o = -6.2N + 14.5N = 8.3 N F y = F Ay + F By = F A sin128 o + F B sin15 o = (10N)sin128 o + (15N)sin15 o = 7.9N + 3.9N = 11.8 N F A = 10 N @ 128 o F B = 15 N @ 15 o F = (8.3 2 +11.8 2 ) = 14.43N = 14N Θ = tan-1(11.8/8.3) = 55 o F R = 14N at 55 o