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Still having problems with vectors!

  1. Sep 25, 2005 #1
    Vector V1 is 8.94 units long and points along the -x axis. Vector V2 is 4.13 units long and points at +35.0° to the +x axis.
    (a) What are the x and y components of each vector?
    V1x =
    V1y =
    V2x =
    V2y =
    (b) Determine the sum V1 + V2.
    ______° (counterclockwise from the +x axis is positive)

    so i know that V1x=-8.94 and V1y = 0. i did the other components, but i got them wrong. i solved for V2y by 4.13cos(35) = 3.38. then solved for V2x by 4.13sin(35) = 2.37.

    So for part B i did Vx = 8.94+2.37=11.31 and Vy=0+3.38=3.38. Then found the magnitude by V=the square root of (11.31^2) + (3.38^2) = 11.80. then for the direction i did tan (angle) = 3.38/11.31=.299. the inverse of that is 16.63 degrees.

    Can anyone tell me what i did wrong
  2. jcsd
  3. Sep 25, 2005 #2


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    Staff: Mentor

    You lost the minus sign on -8.94. Minus signs count when you're adding components.
  4. Sep 25, 2005 #3
    ohh ok thanks! but what about the V2x and V2y components. what did i do wrong? because i got those components wrong too so i cant do part B until i finish part A...help please!
  5. Sep 26, 2005 #4


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    Homework Helper

    Draw the triangle, with V2 as the hypotenuse.
    Can you see now why V2y is not 4.13cos(35) but 4.13sin(35) ?
  6. Sep 26, 2005 #5
    still having trouble. so heres what i got..but i got them wrong!

    V1x= -8.94
    V1y = 0
    V2x= 4.13cos(35) = 3.38
    V2y= 4.13sin(35) = 2.37

    Vx= -8.94 +3.38 = -5.56
    Vy= 0 + 2.37 = 2.37

    V = [the square root of (-5.56)^2 +(2.37)^2] = 6.04

    tan (angle)= 2.37/-5.56 = -.426

    inverse tan (-.426) = -23.09

    so what did i do wrong. I got the V1x & V1y correct, but everything else wrong. PLEASE HELP!!!
  7. Sep 27, 2005 #6


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    Staff: Mentor

    Your magnitude for V looks OK to me. What is it supposed to be?

    As for the direction (angle) of V, draw a diagram that shows the x and y axes, and has Vx and Vy pointing in the proper directions, based on their signs. Look for a right triangle that has Vx and Vy as its two sides, and see where your angle of 23.09 degrees fits in (don't worry about the - sign here). From the diagram you should now be able to read off the angle that you were asked for, which is the angle measured counterclockwise from the +x axis to V.
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