# sol03 - 6 50.9967 N Question 1 chap 3 sect 1 part 1 of 1 10...

This preview shows pages 1–4. Sign up to view the full content.

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 anti-parallel ( 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 counter-clockwise 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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. Question 4, chap 3, sect 1. part 1 of 1 10 points A person walks 25 m west and then 45 m at an angle of 60 north of east. What is the magnitude of the total dis- placement? 1. 31.0 m 2. 67.8 m 3. 45.7 m 4. 26.5 m 5. 39.1 m correct 6. 61.4 m 7. 40.9 m 8. 43.6 m 9. 56.3 m 10. 60.2 m
Explanation: In a triangle with three sides of a, b, and c , if the angle between a and b is θ , then c 2 = a 2 + b 2 2 ab cos θ .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern