This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Question 1, chap 3, sect 1. part 1 of 1 10 points A vector of magnitude 3 CANNOT be added to a vector of magnitude 4 so that the magnitude of the resultant is 1. 3. 2. 7. 3. 1. 4. 5. 5. 0. correct Explanation: The smallest magnitude of the resultant occurs when the vectors are antiparallel ( R = 1); the largest occurs when they are parallel ( R = 7). Therefore all listed values are possible except R = 0. Question 2, chap 3, sect 1. part 1 of 2 10 points Given two vectors vector F 1 , and vector F 2 . Where the magnitude of these vectors are F 1 = 77 N , and F 2 = 72 N . And where θ 1 = 21 ◦ , and θ 2 = 80 ◦ . The angles are measure from the positive x axis with the counterclockwise angular direc tion as positive. Draw the vectors to scale on a graph to determine the answer. What is the magnitude of the resultant vec tor bardbl vector F bardbl , where vector F = vector F 1 + vector F 2 ? 1. 109 . 458 N 2. 78 . 4916 N 3. 129 . 706 N correct 4. 87 . 5005 N 5. 121 . 553 N 6. 50 . 9967 N 7. 124 . 715 N 8. 136 . 289 N Explanation: Drawing a diagram to scale of the vectors, we have θ F 1 F 2 F Scale: 10 N where : F 1 = 77 N , θ 1 = 21 ◦ , F 1 x = 71 . 8857 N , F 1 y = 27 . 5943 N , F 2 = 72 N , θ 2 = 80 ◦ , F 2 x = 12 . 5027 N , F 2 y = 70 . 9062 N , and F = 129 . 706 N , θ = 49 . 4123 ◦ , F x = 84 . 3884 N , F y = 98 . 5005 N . The x components of the forces vector F 1 and vector F 2 are F 1 x = F 1 cos(21 ◦ ) = 71 . 8857 N F 2 x = F 2 cos(80 ◦ ) = 12 . 5027 N . And the y components are F 1 y = F 1 sin(21 ◦ ) = 27 . 5943 N F 2 y = F 2 sin(80 ◦ ) = 70 . 9062 N . The x and y components of the resultant vec tor vector F are F x = F 1 x + F 2 x = (71 . 8857 N) + (12 . 5027 N) = 84 . 3884 N F y = F 1 y + F 2 y = (27 . 5943 N) + (70 . 9062 N) = 98 . 5005 N . Hence the magnitude of the resultant vector bardbl vector F bardbl is bardbl vector F bardbl = radicalBig F 2 x + F 2 y = radicalBig (84 . 3884 N) 2 + (98 . 5005 N) 2 = 129 . 706 N . Question 3, chap 3, sect 1. part 2 of 2 10 points Note: Give the angle in degrees, use coun terclockwise as the positive angular direction, between the limits of − 180 ◦ and +180 ◦ from the positive x axis. What is the direction of this resultant vec tor vector F ? 1. 49 . 4123 ◦ correct 2. − 76 . 3995 ◦ 3. 104 . 994 ◦ 4. 131 . 792 ◦ 5. 89 . 2517 ◦ 6. − 134 . 449 ◦ 7. − 177 . 375 ◦ 8. − 65 . 1028 ◦ Explanation: The angle is given by θ = arctan parenleftbigg F y F x parenrightbigg = tan − 1 parenleftbigg 98 . 5005 N 84 . 3884 N parenrightbigg = 49 . 4123 ◦ . The “arctan” function is defined between lim its − 90 ◦ and +90 ◦ . Thus you must check to see in which quadrant your resultant vector lies. You should use the “atan2(y,x)” in your calculator, if available, since it does not have this ambiguity....
View
Full
Document
This note was uploaded on 12/04/2008 for the course PHY 302K taught by Professor Kaplunovsky during the Fall '08 term at University of Texas.
 Fall '08
 Kaplunovsky

Click to edit the document details