Solving vectors using component method

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Homework Help Overview

The problem involves adding vectors using the component method, specifically focusing on three vectors defined by their magnitudes and directions. The subject area is vector addition in physics, particularly in two dimensions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the x and y components of each vector but expresses confusion regarding the contribution of the third vector, which is directed west. Some participants question the interpretation of vector directions and angles, while others clarify the conventions for measuring angles in vector components.

Discussion Status

The discussion includes attempts to clarify the component calculations for the vectors. Some guidance has been provided regarding the angles associated with each vector, and there is acknowledgment of the original poster's confusion about the third vector's direction. The conversation reflects a mix of interpretations and clarifications without reaching a definitive consensus.

Contextual Notes

Participants note the importance of consistent angle measurement and the implications of vector direction on component calculations. The original poster's confusion highlights the need for clarity on vector definitions and conventions in the context of the problem.

yety124
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Homework Statement


Add the following vectors using component method:
Vector A=175km; 30 degrees north of east
Vector B=153km; 20 degrees west of north
Vector C=195km; West
a)Find x and y component of each vector
b)Solve for the magnitude of resultant vector
c) Solve for direction(angle)

2. The attempt at a solution
So i these are what i got
Vector A: x=175cos 30...x=151.56 y=175sin 30...y=87.5
Vector B: x=153cos 20...x=-143.77 y=153sin 20...y=52.33
(are these correct?)
so after getting this i added all the x's and y's but then i realized that the question had said that vector C goes 195km west, so i now got confused what to do next.

please help thank you for your time.
 
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There is nothing special about "in a westerly direction". A westerly vector contributes a component D.cos 180o in the x-direction, and D.sin 0o in the y-direction.
 
well that was stupid of me, thanks for the reply finally solved it :D
 
If \theta is measured counterclockwise from the positive x-axis then a vector of length r and angle \theta has components rcos(\theta) and r sin(\theta). Strictly speaking, you are free to choose the "positive x-axis" any way you want as long as you are consistent but the usual convention is that the positive x-axis points East.

For the first vector you are given that r= 175km and the directon is 30 degrees north of east. "north of east" is counterclockwise from east so the angle is \theta= 30 degrees.

For the second vector you are given that r= 153km and the direction is 20 degrees west of north. West is clockwise of north but north itself is 90 degrees clockwise of east. The angle is \theta= 90+20= 110 degrees.

For the third vector you are given that r= 195km and the direction is west. West is exactly opposite east so the angle is \theta= 180 degrees.
 

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