Chapter 3 - Kinematics in Two Dimensions (Vectors) PDF

# Chap 3 ## Kinematics in Two Dimensions (Vectors) ### 3-1: Vectors and Scalars * A vector is a quantity which must have both magnitude and direction (velocity, force, etc.) * A scalar is a quantity which must have magnitude only (mass, temperature, energy, time) ### 3-2: Addition of Vectors - G...

# Chap 3 ## Kinematics in Two Dimensions (Vectors) ### 3-1: Vectors and Scalars * A vector is a quantity which must have both magnitude and direction (velocity, force, etc.) * A scalar is a quantity which must have magnitude only (mass, temperature, energy, time) ### 3-2: Addition of Vectors - Graphical Methods * If we have two vectors A and B in the same axis, then 3m -> 4m -> AB -> 7m The result of the addition of vectors is called Resultant R * But if < 3m -> 4m -> then 1m -> A B A+B = R * If one vector A in the x-axis and the other vector B in the y-axis then A -> +B -> B then R = A + B * To find the Resttant R we measure the length of R and using the same scale for A and B, we get the magnitude of |R|. Then, we measure the angle θ which is the angle between R and the x-axis. We have determined the resultant of the addition of A and B, the magnitude and the direction. The magnitude can also be determined by using the Pythagoras theorem |R|=√A²+B² * If A = 10Km to the east and B = 5 Km to the north, then R =√10² + 5² = 11.2 Km * The rules for getting the resultant by a graphical method are 1. Draw a diagram, draw A to scale 2. Next, draw the second vector to the same scale and placing its tail at the tip of the first vector directed at the same original direction 3. The vector (Arrow) drawn from the tail of the first vector A to the tip of the second vector (B) represents the sum of resultant of the two vectors R=A+B * If we have more than two vectors. We do the same as for two vectors. e.g. if we have three vectors, A+B+C then A -> B -> C -> R ### 3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar. * Negative Vector: if A is a vector then -A is defined as: it has the same magnitude of A but is in the opposite direction to it. * Then subtraction of vectors is equal to the sum of first plus the negative of the second. * Then A - B is given by A + -B = A + -B = A - B * ### 3-4 Adding Vectors by Components * Any vector can be resolved into two components along X and Y axes * If we have a vector A then its components are if A makes an angle θ, then the X-component is given by Ax= A* cos θ, and the y-component is given by Ay * = A *sin θ where |A| is the magnitude of the vector |A| and θ must be measured from the positive x-axis counter clockwise. * Example: Suppose A represents a displacement of 500m in a direction 30 north of east, then y North Ax=A*cos θ = 500* cos(30) = 500 (0.866) = 433m (east) Ay Ay=A*sin θ = 500* sin (30) = 500(0.5) = 250m (north) x East * There are two ways to specify a vector in a given coordinate system 1. We can give its components Ax and Ay 2. We can give its magnitude A and the angle θ it makes with the positive x-axis ### 3-5 * The above discussion is to shift from the second descriptions to the first. But, the following discussion is to shift from the first to the second description, i.e., from components to find the magnitude and the direction of the Vector. So, if we have the components Ax and Ay for a vector A, then we can get the magnitude and direction of the vector using the following relations A=|A| =√Ax² + Ay² (magnitude) Ay tan θ = Ax (direction) θ=tan-¹(Ay/Ax) (tan read tan inverse) ### 3-6 Adding Vectors ### 3-7 * We can use the components to add vectors, which is called the analytical method of adding vectors. * Suppose we have two vectors A and B, then the sum (resultant) R= A+B is given by Rx and Ry *Rx = Ax + Bx *Ry = Ay + By * or, |R| = R = √Rx² + Ry² * and the direction of R is θ = tan -¹(Ry/Rx) where θ is measured from the x-axis ### 3-8 * **Example 3-2: Mail Carrier's Displacement:** She leaves the post office and drives 22 Km north. She then drives 47 Km in a direction of 60° south of east. What is her displacement from the post office? y North Solution - 60° * Dix = D₁*cos θ = 22(0) = 0 Km D₂ * Diy = D₁*sin θ = 22(1) = 22Km * D₂x = D₂cos(-60) = 47(0.5) = 23.5 Km * D₂y = D₂sin(-60) = 47(-0.866) = -40.7Km, we use the angle (-60°) since it is measured clockwise * Then the Resultant Components are: - DRx = D₁x + D₂x = 0+ 23.5 = 23.5 Km - DRy = D₁y + D₂y = 22 + (-40.7) = -18.7Km * This specifies the resultant displacement completely or magnitude or direction: - DR = √DRx² + DRy² = √(23.5)² + (-18.7)² = 30 Km - θ = tan-¹(DRy/DRx) = tan-¹(-18.7/23.5) = tan-¹(-0.796) = -38.5° * The minus sign means θ is measured clockwise, θ is below the x-axis. So, the resultant displacement is 30 Km directed at 38.5° Southeast ### 3-9 * Example 3-3: Three Short Trips An airplane trip involves three legs. The first leg is due east for 620 Km; the second leg, southeast (45°) for 440 Km and the third leg is at 53° south of west for 550 Km. As shown, what is the plane's total displacement? Solution: 1. Drwa a diagram 2. Choose axes (figure) 3 & 4 Calculate the components: * D₁: D₁x = D₁*cos θ = 620 Km * D₁y = D₁*sin θ = 0 Km * D₂: D₂x = D₂*cos(45°) = 440(0.707) = 311Km * D₂y = D₂*sin(45°) = 440(-0.707) =-311Km * D₃: D₃x = D₃*cos(53°) = 550(0.602) = -331Km * D₃y = D₃sin(53°) = 550(0.799) = -439Km * Note: if the components point in the -x or -y direction, we give a minus sign to the component * 5. Add the components: - We add the x-components together and the y-components together: * DRx = D₁x + D₂x + D₃x = 620 + 311 + (-331) = 600 Km * DRy = D₁y + D₂y + D₃y = 0 + (-311) +(-439) = -750 Km * This is one way to give the answer ### 3-10 * 6. Magnitude and Direction: We can give the answer as; * DR = √DRx² + DRy² = √(600)² + (-750)² = 960 Km * θ = tan-¹(DRy/DRx) = tan-¹(-750/600) = tan-¹(-1.25) = -51° Southeast