410 likes | 459 Views
Explore how vectors and scalars guide you to pirate treasure using scalar and vector components for direction and magnitude. Learn two methods of vector addition for accurate navigation.
E N D
NW Vectors • Imagine that you have a map that leads you to a pirates treasure. • This map has instructions such as 10 paces to the northwest of the skull. • The 10 paces is a distance, 10 is the magnitude of that distance, and northwest is a direction. • Since this instruction has both magnitude and direction, it is a vector.
Vectors vs. Scalars Quantities that require a magnitude (how big or small) and direction for specification are called vectors. Those quantities that have magnitude but no direction are called scalars.
Examples of scalars are: distance (length) speed time mass temperature energy All vectors have a scalar component: The displacement vector includes both distance and direction. The velocity vector includes both speed and direction. Examples of vectors are: displacement velocity acceleration force momentum electric and magnetic fields Vectors and Scalars
Vector Addition The math associated with scalars is familiar to everyone (2+2=4). The math associated with vectors is more involved. You will need to know how to do both the graphical addition of vectors and the mathematical addition of vectors at right angles. Vectors are represented graphically as arrows. The length of the arrow is proportional to the magnitude of the vector. The arrow points in the direction of the vector. The arrow end is called the head and the other end is called the tail. In text, a vector is indicated by using a bold font. Some books use an arrow over the letter to indicate that it is a vector. The magnitude of a vector can be indicated by a plain or italic font. head Vector V has a magnitude of V. tail
Two Methods of Vector Addition • The head-to-tail method of vector addition is more useful when dealing with displacements. • The parallelogram method of vector addition is more useful when dealing with forces on an object. • When adding two vectors, you can move one of the vectors to either the head or the tail of the other vector so long as you do not change either the magnitude or direction of a vector you are moving. • Since vectors can be moved around for addition purposes, either method will give the right answer to a problem.
5 km 3 km 4 km Vector Addition Let’s use a treasure map again as an example of the addition of vectors. Let’s imagine the map tells you to go 4 kilometers east then 3 kilometers north to find the treasure. Your final position is 5 kilometers northeast of your original position. It would have saved time if that had been the one distance and one direction traveled in the first place (the smartest pirate gets the treasure).
5 km 3 km 4 km Resultant Vector The 5 km northeast arrow is the vector sum of the 4 km east vector and the 3 km north vector. The vector sum vector is called the resultant vector. How long the arrow of the resultant vector is and in what direction the arrow should be oriented can be found by using either method of vector addition, but in this case it is more visually accurate using the head-to-tail method.
5 km 5 km 3 km 4 km 4 km 3 km Head-to-Tail Method • The head-to-tail method is simple: • Place the two vectors you are adding together head to tail. • Draw the resultant vector from the open tail of one vector to the open head of the other vector. • It does not matter which head and tail you choose to match up when adding, you will get the same answer either way.
Parallelogram Method Recall that a parallelogram is afour sided figure where the opposite sides are parallel with each other. Squares and rectangles are also parallelograms. When using the parallelogram method of vector addition, you place the two tails of the vectors together. How long the arrow of the resultant vector is and in what direction the arrow should be oriented can be found by completing a parallelogram using the two vectors as sides, then drawing the resultant vector from the vector tails to the opposite corner of the parallelogram.
Parallelogram Method • Place the two vectors you are adding together tail to tail. • Complete the parallelogram. • Draw the resultant vector from the vector tails to the opposite corner of the parallelogram.
a2 + b2 = c2 (14 units)2 + (9 units)2 = c2 196 units2 + 81 units2 = c2 277 units2 = c2 16.64 units = c Graphing the Resultant Vector • It is easier and more accurate to complete the parallelogram and draw the resultant vector when using graph paper. • Example: Complete the parallelogram and draw the resultant vector of the two vectors below – i.e. add the vectors. • The red arrow is the sum of the two blue arrows. But what is it’s magnitude (value)? • Since it is draw on graph paper, we can get clever. Let’s just look at the red arrow. • We can use the red arrow as the hypotenuse of a right triangle. • By using the Pythagorean Theorem, we find the value of our resultant vector is 16.64 units.
Example: Force Vectors What is the resultant force? a2 + b2 = c2 (4 N)2 + (3 N)2 = c2 16 N2 + 9 N2 = c2 25 N2 = c2 5 N = c 5 Newtons NE 3 Newtons north 4 Newtons east Since the vectors are at right angles, the Pythagorean theorem can be used to find the magnitude of the resultant vector. If the vectors are not at right angles, the Pythagorean theorem CANNOT be used.
Example 1: Velocity Vectors What is the actual velocity of an eastbound plane with an airspeed of 100 km/h that is in a 60 km/h NW crosswind? Wind: 60 km/h northwest Plane: 100 km/h east Since the vectors are NOT at right angles, the Pythagorean theorem CANNOT be used to find the magnitude of the resultant vector. Unless this problem is on a graph where you can measure the resultant with a ruler, you cannot answer this problem with the information given.
Example 2: Velocity Vectors What is the actual velocity? Wind: 60 km/h northwest Plane: 100 km/h east If appropriate units are given on the graph, you can use a ruler to find the magnitude of the resultant vector.
7 cm 3 cm 6 cm 2 cm 4 cm 1 cm 5 cm 5 cm 1 cm 3 cm 6 cm 4 cm 7 cm 2 cm Graphing the Resultant Vector • Practice: Visualize the process of graphing the resultant vector of the two blue vectors below (add the two blue vectors). • Does it look like you visualized? If not, where did you go wrong? • What is the magnitude of the resultant vector? • Notice that this graph has a scale on the x and y axis. • The resultant vector stretches from 0.5 cm to 7.5 cm on the x-axis. • The magnitude of the resultant vector is 7 cm.
Components of Vectors y x All vectors can be broken down into horizontal and vertical (x and y) components. Consider the vector R below. Imagine that the tail of the vector is located at the origin of a Cartesian coordinate system. Now draw lines from the head of the vector to the x and y axis. Fill in the horizontal and vertical components as shown: R The red arrows are the horizontal and vertical components of the vector R. The horizontal and vertical components of any vector will always be at right angles to each other. vertical component horizontal component
o p p o s i t e hypotenuse sin θ = opposite hypotenuse adjacent hypotenuse opposite adjacent cos θ = θ adjacent tan θ = Quick Trig • When resolving vector components, some knowledge of trigonometry is needed. • Consider this right triangle. • If you are given an angle, θ, and the length of one side, you can use trig to find the remaining sides. • Just remember soh cah toa.
o h sin θ = 4 h sin 53.13° = 4 sin 53.13° h = Example 1: Quick Trig • Solve for the hypothenuse. • Since we are given the angle and the side opposite the angle, we want to use sin θ = opposite/hypotenuse. hypotenuse 4 h sin 53.13° = 4 θ = 53.13° adjacent h = 5
a h cos θ = 3 h cos 53.13° = 3 cos 53.13° h = Example 2: Quick Trig • Solve for the hypothenuse. • Since we are given the angle and the side adjacent the angle, we want to use cos θ = adjacent/hypotenuse. o p p o s i t e hypotenuse h cos 53.13° = 3 θ = 53.13° 3 h = 5
Example 3: Quick Trig • Solve for the angle, θ. • Since we are given the opposite and adjacent sides, and asked for the angle, we want to use tan θ = opposite/adjacent. • But in order to solve for the angle, we need to use the inverse of tangent. 4 hypotenuse o a tan-1 θ = 4 3 tan-1 θ = θ = ? 3 θ = 53.13°
Consider the vector A. Place the tail of the vector at the origin of a Cartesian coordinate system. The horizontal, or x-component, of A is found by Ax = A cos θ. The vertical, or y-component, of A is found by Ay = A sin θ. By the Pythagoream Theorem: Ax2 + Ay2 = A2 Every vector can be resolved using these formulas, such that A is the magnitude of vector A, and θ is the angle the vector makes with the x-axis. Each component must have the proper sign (+,-) according to the quadrant the vector terminates in (more on this later). Resolving Vector Components +y A Ay θ Ax +x
A bus travels 13 km on a straight road that is 22.62° north of east. What are the east and north components of its displacement? Sketch the problem. List known values: A = 13 km θ = 22.62° Solve for Ax and Ay. Ax = A cos θ = (13.0 km)(cos 22.62°) = 12 km (east) Ay = A sin θ = (13.0 km)(sin 22.62°) = 5 km (north) Check your answer: (12 km)2 + (5 km)2 = (13.0 km)2 144 + 25 = 169 +y (north) A Ay θ +x (east) Ax Example: Resolving Vector Components ?
Algebraic Method of Vector Addition Y • Two or more vectors (A, B, C, …) may be added by first resolving each vector to its x- and y-components. • The x-components are added to form the x-component of the resultant: Rx = Ax + Bx + Cx + … • The y-components are added to form the y-component of the resultant: Ry = Ay + By + Cy + … C Cy B By A Ay x Cx Bx Ax Y C B Ry A x Rx
Algebraic Method of Vector Addition Y • Because Rx and Ry are at a right angle, the magnitude of the resultant vector can be calculated using the Pythagorean theorem: R2 = Rx2 + Ry2 • To find the angle or direction of the resultant, recall that the angle will be the angle between the x-axis and the resultant and that tan θ = o/a. • Since the side opposite the angle is equal to Ry and the side adjacent is equal to Rx, we have: tan θ = Ry / Rx • Using the inverse of tangent we have: θ = tan-1 (Ry / Rx) C B Ry A x Rx Y R Ry θ x Rx
A GPS receiver told you that your home was 15.0 km at a direction of 40° north of west, but the only path led directly north. If you took that path and walked 10 km, how far, and in what direction would you then have to walk to reach your home? Sketch the problem. List the known values and the unknown: A= 10.0 km, due north R= 15.0 km, 40° north of west B= ? Find the components of R and A (remember that the angle is from the positive x-axis to the resultant. 180° – 40° = 140°). Rx = R cos θ = (15.0 km)(cos 140°) = –11.5 km Ry = R sin θ = (15.0 km)(sin 140°) = 9.64 km Ax = 0.0 km Ay = 10.0 km Example: Algebraic Method north B A R 140° 40° west x Y
Use the components of R and A to find the components of B. R = A + B, so B = R –A Bx = Rx– Ax = –11.5 km – 0.0 km = –11.5 km By = Ry– Ay = 9.64 km – 10.0 km = –0.36 km north Y B A R 140° 40° west x Example: Algebraic Method Y NE (+,+) NW (–,+) Use the signs to determine the direction. Imagine that the tail of B is at the origin of a Cartesian coordinate system. Recall the signs of each quadrant of this system. See that B points southwest. x SW (–,–) SE (+,–) north west B
Now use the components of B to find the magnitude of B. Bx2 + By2 = B2; B = √(–11.5 km)2 + (–0.36 km)2 = 11.5 km Use the inverse tangent to find the angle. θ = tan-1 (By / Bx) = tan-1 (–0.36 km / –11.5 km) = 1.79° Finally, our answer is: B = 11.5 km 1.79° south of west Example: Algebraic Method north west Y 1.79° B A R 140° 40° west x
Check Question • A weather station releases a balloon that rises at a constant 15 m/s relative to the air, but there is a wind blowing at 6.5 m/s toward to west. What are the magnitude and direction of the velocity of the balloon?
Extra Credit – 10 points • A weather station releases a balloon that rises at a constant 15 m/s relative to the air, but there is a wind blowing at 6.5 m/s toward to west. What are the magnitude and direction of the velocity of the balloon? 152 x 6.52 = R2 R = 16.35 m/s, 113° Or 16 m/s 67° north of west Ɵ = tan-1 (opp/adj) = tan-1 (6.5/15) = 23.4 degrees plus 90 degrees = 113 degrees
Vector Addition Find the Resultant vector. 652 + 502 = R2 4225 + 2500 = R2 82 6725 = R2 50 R = 82 65
Vector Addition Find the Resultant vector. 602 + 252 = R2 3600 + 625 = R2 4225 = R2 R = 65 65 25 60
Vector Addition Find the Resultant vector. 652 + 352 = R2 4225 + 1225 = R2 5450 = R2 R = 74 35 74 65
Vector Addition Find the Resultant vector. 452 + 252 = R2 2025 + 625 = R2 2650 = R2 R = 52 52 25 45
Vectors 552 + 102 = R2 Find the Magnitudes of the Vectors. 10 55 3025 + 100 = R2 15 3125 = R2 57 55 R = 56 56 552 + 152 = R2 76 40 R = 57 652 + 402 = R2 R = 76 65
Vectors Find the Magnitudes of the Vectors. 76 56 65
Vector Addition The planes new vector is 269 mph WNW 269
Vector Addition Find the Resultant vector.
Vector Addition The planes new vector is 335 mph NNW 335
Vector Addition Find the Resultant vector.
Vector Addition Find the Resultant vector.