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Adding 3 Vectors?

  1. Feb 12, 2013 #1
    Adding 3 Vectors????

    1. The problem statement, all variables and given/known data

    I am given Vector A which has a magnitude of 12, Vector B which has a magnitude of 5, and Vector C which has a magnitude of 8. I am given that the angle between Vector A and Vector B is 30 degrees and that the angle between Vector B and Vector C is 80 degrees. All angles are counterclockwise and Vector A is the HORIZONTAL..

    I have to find the magnitude of |A+B-C| and the angle, relative to Vector A


    2. Relevant equations
    Cosine and Sine Law


    3. The attempt at a solution

    What I did was added A and B first using an angle of 150 degrees in the cos law. This gave me an answer of 16.52... However from here I flip Vector C and draw the resultant from there... However I do not know what the angle opposite from the resultant of (A+B) + (-C)

    Please give me some guidance
     
  2. jcsd
  3. Feb 12, 2013 #2
    Re: Adding 3 Vectors????

    I tried it but I am not sure if I got it correct. Is the answer 16.8863 and θ=36.4 degrees?
     
  4. Feb 12, 2013 #3
    Re: Adding 3 Vectors????

    I dont have the answer.... What was the angle opposite to the resultant that you got?
     
  5. Feb 12, 2013 #4
    Re: Adding 3 Vectors????

    Someone has gotten an answer of 19.715..... Heres how he did it:

    Start by splitting the vectors into their x and y components.

    the angle of A is 0 degrees.
    the angle of B is 30 degrees.
    the angle of C is 80+30 = 110 degrees.

    so:
    A_x = 12
    A_y = 0

    B_x = 5cos(30) = 4.3301
    B_y = 5sin(30) = 2.5

    C_x = 8cos(110) = -2.7362
    C_y = 8sin(110) = 7.5175

    now we can add the vectors easily (call the resulting vector D):
    D_x = A_x + B_x - C_x
    D_y = A_y + B_y - C_y

    D_x = 12 + 4.3301 + 2.7362 = 19.0663
    D_y = 0 + 2.5 - 7.5175 = -5.0175

    now we can find the angle the vector makes by using tan^-1(D_y / D_x)
    tan^-1(-5.0175 / 19.0663) = -14.73 degrees from the horizontal.

    the magnitude is found by sqrt((D_x)^2 + (D_y)^2)
    sqrt(19.0663^2 + (-5.0175)^2) = sqrt(388.699)
    =
    19.715

    Anyone know how to do it through drawings instead of x and y components?
     
  6. Feb 12, 2013 #5
    Re: Adding 3 Vectors????

    Vectors are added head to tail. You can draw them out starting with A being horizontal to the right. B is added to the arrow end of A up 30° from horizontal, and C is added to the end of B at 80° from the line of vector B or an effective 110° from horizontal. You need to sum the X and Y components of each of these vectors and form a final triangle to solve for the ange and resultant magnitude. Equations would be as follows:

    Sum of X's: 12+[5cos(30)]-[8cos(110)] = Xsum
    Sum of Y's: 0+[5sin(30)]-[8cos(110)] = Ysum

    *Remember that a -(-) equals a positive and the -[8cos(110)] ends up being added to 5cos(30) due to it becoming negative, and similarly for the sines but in this case it end up being
    negative.

    Resultant vector triangle is formed from these summed X and Y values, and you solve it with pythagorean theorum. Start with Xsum horizontal to the right, and YSum being added vertically down from the end of the XSum vector arrow. Resultant will be Tan-1(-YSum/XSum) = -X° down from horizontal, and magnitude is =Sqrt(Xsum^2+Ysum^2)
     
    Last edited: Feb 12, 2013
  7. Feb 12, 2013 #6
    Re: Adding 3 Vectors????

    I think I got the angle wrong. I got the angle between the resultant and A from the dot product. Sorry! :cry:
     
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