Determining x & y Components of Vectors in xy Plane

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Homework Help Overview

The original poster is attempting to determine the x and y components of two vectors in the xy plane, specifically a velocity vector and a force vector, each defined by their magnitudes and angles relative to specific axes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of trigonometric functions to find vector components, with some expressing confusion about applying these functions when vectors are directed along negative axes. Questions arise about the correct application of formulas for different quadrants.

Discussion Status

Participants are exploring various interpretations of how to apply trigonometric identities to vectors in different quadrants. Some guidance has been offered regarding the use of sine and cosine functions, but there remains uncertainty about specific cases and the application of formulas.

Contextual Notes

There is mention of confusion regarding the application of formulas for vectors in the negative x and y directions, as well as the need for clarification on how to handle angles in different quadrants.

Susanem7389
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I am having some trouble with this question.

Determine the x and y components of the following two vectors in the xy plane. (A) A 25- m/s velocity vector that makes an angle of 40 degree counterclockwise from the -x direction. (B) A 40lb force vector that makes an angle of 120 degree counterclockwise from the -y direction.
 
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Hi Susanem7389! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
I drew the graph and and tried the Ax=AcosDegree and Ay=AsinDegree, however I never get the correct answer and I'm not sure how to go about getting the answer when it is in the -x direction since the equation two lines before is for the +x direction.
 
Susanem7389 said:
… I'm not sure how to go about getting the answer when it is in the -x direction since the equation two lines before is for the +x direction.

use sin(180º - θ) = sinθ, cos(180º - θ) = -cosθ :wink:
 
Okay. Thank you. It is correct. So if you have a vector that is in the -x direction, then you could you the formula that you gave? Also, what about if you have a vector the is in the -y direction and +y direction and need to use the Ax=A sin Degree and Ay= A cos Degree in the +x direction, how would you go about solving those? I'm new to vectors and trying to figure out how to do them.
 
Susanem7389 said:
So if you have a vector that is in the -x direction, then you could you the formula that you gave? Also, what about if you have a vector the is in the -y direction and +y direction and need to use the Ax=A sin Degree and Ay= A cos Degree in the +x direction, how would you go about solving those?

erm … sorry … i don't understand any of that :confused:

can you please write it out again?​
 
Sure. Sorry about the confusion. I am just learning vectors and I'm trying to understand how to find the components. The equation in the book for finding components is Ax=Acosθ and Ay=Asinθ. However that is only the case if the vector is in the +x direction. How would you go about finding the component if the vector is in the -y and +y direction? Also, if you have a vector that is in the -x direction, do I always use sin(180º - θ) = sinθ, cos(180º - θ) = -cosθ to find the answer.
 
still not sure what you mean by the +y and -y directions …

are you talking about the third and fourth quadrants? :confused:

anyway, the formulas Ax=Acosθ and Ay=Asinθ work for any vector at any angle …

you simply have to understand how to find sin and cos for angles between 90º and 360º. :smile:
 
Yes, I meant about the third and fourth quadrants. Thank you for all your help. Do you have any more information or know of any websites that could help me understand how to find sin and cos for angles between 90 and 360 degree?
 
  • #10
Susanem7389 said:
Do you have any more information or know of any websites that could help me understand how to find sin and cos for angles between 90 and 360 degree?

sorry, i don't know of anything that would help …

but it's pretty obvious what the angle is in any particular case, just by drawing it :smile:
 
  • #11
Okay thank you.
 

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