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Homework Help: Vector Components Problem

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Vector C has a magnitude of 5.0 units and makes an angle of -90.0º with the positive x-axis, vector D has a magnitude of 7.0 units and makes and angle of –120º with the positive x-axis. What is the magnitude of the vector sum of C + D? - I am assuming that this means the resultant.

    2. Relevant equations

    General for finding the components:

    [tex] A_x = Acos\theta [/tex]

    [tex] A_y = Asin\theta [/tex]


    [tex] A = \sqrt {{A_x}^2 + {A_y}^2} [/tex]

    3. The attempt at a solution

    I first start with [tex] \vec C [/tex]

    [tex] C_x = 5.0 cos (-90) = 0 [/tex]
    [tex] C_y = 5.0 sin (-90) = -5 [/tex]

    Move on to [tex] \vec D [/tex]

    [tex] D_x = 7.0 cos (-120) = -3.5 [/tex]
    [tex] D_y = 7.0 sin (-120) = -6.1 [/tex]

    I have all the components now, moving on to finding the A + B - which I am assuming means the resultant.

    [tex] R_x = 0 + (-3.5) = -3.5 [/tex]
    [tex] R_y = -5 + (-6.1) = -11.1 [/tex]

    [tex] R = \sqrt { (-3.5)^2 + (-11.1)^2} = 11.6 [/tex] - This is my answer to the question

    Now, I am not sure if my assumption that A+B = Resultant is true. Any ideas on where I am messing up. Thank you
    Last edited: Sep 19, 2009
  2. jcsd
  3. Sep 19, 2009 #2
    Are you sure that you copied this correctly? If you were given C and D I don't see why they would be asking you for A+B.
  4. Sep 19, 2009 #3
    Sorry m8, had a different problem in my head. I made the change.

    Any ideas on what I did wrong?
  5. Sep 19, 2009 #4
    Not sure. It looks good to me :confused:

    Why do you think you are wrong? Did you copy the numbers down correctly?

    I get the same answer using the numbers you gave.
  6. Sep 20, 2009 #5
    Ok well if that looks right, then great! Now there is a second part to this question which I did not post and it states the same variables and measurements except you have to find the direction of the vector sum C+D referenced to the positive x-axis.

    First I use this equation:

    [tex] tan \theta = \frac {A_y} {A_x} [/tex]


    so I solve for theta and get:

    [tex] \theta = tan^-1 \frac {A_y} {A_x} [/tex]


    [tex] \theta = tan^-1 \frac {-11.5} {-3.5} = 73.07... [/tex]

    Looks like the vector is in the third quadrant so I add 180 to [tex] \theta [/tex] and get 253.1. The problem is that none of the above answers are part of my answer choices. What did I do wrong?
  7. Sep 20, 2009 #6


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    Homework Helper

    Ry is 11.1 not 11.5
  8. Sep 20, 2009 #7
    yah another typo. That still doesn't make a difference, I still have it wrong.
  9. Sep 20, 2009 #8


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    Homework Helper

    What are the answer choices?
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