How to Find the X Component of a Vector | Decomposing Vectors Homework

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Homework Help Overview

The discussion revolves around finding the x component of a vector with a magnitude of 6 m, directed 35 degrees north of west. Participants are exploring the appropriate angle to use in the cosine function for this calculation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are considering two methods for calculating the x component: one involves directly applying a negative sign, while the other uses an adjusted angle of 145 degrees. There is confusion regarding which method is preferable and the reasoning behind angle selection.

Discussion Status

Some participants have expressed that both methods are valid as long as the reasoning is sound. There is a suggestion to avoid memorizing rules without understanding, indicating a focus on conceptual clarity. However, no consensus has been reached on a preferred method.

Contextual Notes

Participants are discussing the angles associated with cardinal directions and their relevance to the problem, indicating a potential gap in understanding the relationship between vector direction and angle measurement.

Mr Davis 97
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Homework Statement


The magnitude of the vector is 6 m, and points 35 degrees north of west. Find the x component of the vector.

Homework Equations



x = |R|cos(theta)

The Attempt at a Solution



I am confused about what degrees to input into the cosine function. I know that the vector is pointing 35 degrees north of west, which means that the x component will be negative, but there are two ways I could do it am I am not sure which one I should do. First, I could tack on a negative and calculate -(6 m)cos35 = -4.9 m. Another way is not tack on a negative and calculate (6 m)cos145 = -4.9 m. Which method is better, and which one should I use on a regular basis?
 
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Both ways are fine, as long as you don't try to memorize a rule and blindly apply it. Personally, I like the first approach, and just tack on the negative sign when I know there should be one.
 
Mr Davis 97 said:

Homework Statement


The magnitude of the vector is 6 m, and points 35 degrees north of west. Find the x component of the vector.

Homework Equations



x = |R|cos(theta)

The Attempt at a Solution



I am confused about what degrees to input into the cosine function. I know that the vector is pointing 35 degrees north of west, which means that the x component will be negative, but there are two ways I could do it am I am not sure which one I should do. First, I could tack on a negative and calculate -(6 m)cos35 = -4.9 m. Another way is not tack on a negative and calculate (6 m)cos145 = -4.9 m. Which method is better, and which one should I use on a regular basis?
Well, you could learn the angles for the cardinal points of the compass, but why do that when you can just make something up?

East = 0°
North = 90°
West = 180°
South = 270°
 
SteamKing said:
Well, you could learn the angles for the cardinal points of the compass, but why do that when you can just make something up?

East = 0°
North = 90°
West = 180°
South = 270°

Huh?
 
Mr Davis 97 said:
Huh?
You've never seen the following diagram (or a similar one)?:
ucad.gif
It should have been used when you studied trigonometry.

After all,
cos (0°) = 1.0
sin (0°) = 0.0

cos (90°) = 0.0
sin (90°) = 1.0

cos (180°) = -1.0
sin (180°) = 0.0

cos (270°) = 0.0
sin (270°)= -1.0

etc., etc.
 

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