Note that the angles start from zero and increase to 360°. By convention, angles are given relative to the positive x-axis. For this purpose, choose the positive x-axis as pointing toward 0°, and the negative x-axis as pointing towards 180°, the positive y-axis at 90° and the negative y-axis at 270°. (a) A: 200 g along +x axis B: 100 g 45° above -x axis II: (a) A: 150 g 60° along +y axis B: 200 g 45° above -x axis C: 100 g 30° below -x axis COMPONENT METHOD: (b) A: 100 g 30° above -x axis B: 150 g along -y axis (b) A: 150 g 30° below +x axis B: 200 g 60° above +x axis C: 150 g 60° above -x axis For both parts I and II, use the component method to find the resultants and their directions Using the component method (for two given vectors), we can write Ax+ Bx = Rx; Ay + By = Ry For three given forces in Part II, this can be rewritten as Ax + Bx + Cx = Rx; Ay + By + Cy = R₂ The magnitude of R in each case, is found using Pythagoras theorem: R = √ (R² + R₂²) and the direction (the angle q that R makes with the positive x-axis) can be found from q=tan ¹ [R, /R] Rewrite the angle as measured counterclockwise from the positive x-axis. Make sure to indicate what quadrant R is by noting the signs (+ or -) on R, and Ry.
Note that the angles start from zero and increase to 360°. By convention, angles are given relative to the positive x-axis. For this purpose, choose the positive x-axis as pointing toward 0°, and the negative x-axis as pointing towards 180°, the positive y-axis at 90° and the negative y-axis at 270°. (a) A: 200 g along +x axis B: 100 g 45° above -x axis II: (a) A: 150 g 60° along +y axis B: 200 g 45° above -x axis C: 100 g 30° below -x axis COMPONENT METHOD: (b) A: 100 g 30° above -x axis B: 150 g along -y axis (b) A: 150 g 30° below +x axis B: 200 g 60° above +x axis C: 150 g 60° above -x axis For both parts I and II, use the component method to find the resultants and their directions Using the component method (for two given vectors), we can write Ax+ Bx = Rx; Ay + By = Ry For three given forces in Part II, this can be rewritten as Ax + Bx + Cx = Rx; Ay + By + Cy = R₂ The magnitude of R in each case, is found using Pythagoras theorem: R = √ (R² + R₂²) and the direction (the angle q that R makes with the positive x-axis) can be found from q=tan ¹ [R, /R] Rewrite the angle as measured counterclockwise from the positive x-axis. Make sure to indicate what quadrant R is by noting the signs (+ or -) on R, and Ry.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![Note that the angles start from zero and increase to 360°. By convention, angles are
given relative to the positive x-axis. For this purpose, choose the positive x-axis as
pointing toward 0°, and the negative x-axis as pointing towards 180°, the positive y-axis
at 90° and the negative y-axis at 270°.
(a) A: 200 g along +x axis
B: 100 g 45° above -x axis
II: (a) A: 150 g 60° along +y axis
B: 200 g 45° above -x axis
C: 100 g 30° below -x axis
COMPONENT METHOD:
(b)
A: 100 g 30° above -x axis
B: 150 g along -y axis
(b) A: 150 g 30° below +x axis
B: 200 g 60° above +x axis
C: 150 g 60° above -x axis
For both parts I and II, use the component method to find the resultants and their
directions
Using the component method (for two given vectors), we can write
Ax + Bx = Rx; Ay + By = Ry
For three given forces in Part II, this can be rewritten as
Ax + Bx + Cx = Rx; Ay + By + Cy = R₂
The magnitude of R in each case, is found using Pythagoras theorem:
R = √ (R₂²+R₂²)
and the direction (the angle q that R makes with the positive x-axis) can be found from
q=tan ¹ [R, /R]
Rewrite the angle as measured counterclockwise from the positive x-axis. Make sure to
indicate what quadrant R is by noting the signs (+ or -) on Rx and Ry.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0b319816-3828-4c3a-a4b8-8b6da6d278f2%2F2e8a980b-6129-4d55-aea4-3de59867ead6%2Fia6rz6b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Note that the angles start from zero and increase to 360°. By convention, angles are
given relative to the positive x-axis. For this purpose, choose the positive x-axis as
pointing toward 0°, and the negative x-axis as pointing towards 180°, the positive y-axis
at 90° and the negative y-axis at 270°.
(a) A: 200 g along +x axis
B: 100 g 45° above -x axis
II: (a) A: 150 g 60° along +y axis
B: 200 g 45° above -x axis
C: 100 g 30° below -x axis
COMPONENT METHOD:
(b)
A: 100 g 30° above -x axis
B: 150 g along -y axis
(b) A: 150 g 30° below +x axis
B: 200 g 60° above +x axis
C: 150 g 60° above -x axis
For both parts I and II, use the component method to find the resultants and their
directions
Using the component method (for two given vectors), we can write
Ax + Bx = Rx; Ay + By = Ry
For three given forces in Part II, this can be rewritten as
Ax + Bx + Cx = Rx; Ay + By + Cy = R₂
The magnitude of R in each case, is found using Pythagoras theorem:
R = √ (R₂²+R₂²)
and the direction (the angle q that R makes with the positive x-axis) can be found from
q=tan ¹ [R, /R]
Rewrite the angle as measured counterclockwise from the positive x-axis. Make sure to
indicate what quadrant R is by noting the signs (+ or -) on Rx and Ry.
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