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Vector Components

  1. Feb 16, 2009 #1
    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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  3. Feb 16, 2009 #2

    tiny-tim

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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:
     
  4. Feb 16, 2009 #3
    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.
     
  5. Feb 16, 2009 #4

    tiny-tim

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    use sin(180º - θ) = sinθ, cos(180º - θ) = -cosθ :wink:
     
  6. Feb 16, 2009 #5
    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.
     
  7. Feb 16, 2009 #6

    tiny-tim

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

    can you please write it out again?​
     
  8. Feb 16, 2009 #7
    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.
     
  9. Feb 16, 2009 #8

    tiny-tim

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    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:
     
  10. Feb 16, 2009 #9
    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?
     
  11. Feb 16, 2009 #10

    tiny-tim

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    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:
     
  12. Feb 16, 2009 #11
    Okay thank you.
     
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