1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: How do I determine the resultant vector R from R = A + B ?

  1. Apr 8, 2015 #1
    Hi everyone,
    I need steps and some guidance on how to solve questions like this. I know how to do right triangle questions using the Pythagorean method, but when it comes to these non right angle triangles I get so confused. Help!

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


    2. Relevant equations
    a) The magnitude of the component "R" along the x-axis?
    b) The magnitude of the component "R" along the y-axis?
    c) The magnitude of the resultant "R" ?
    d) The direction of the resultant "R" represented by the angle it makes with the x-axis?

    3. The attempt at a solution
    I don't know if i'm on the right track..

    a) 8cos23= 7.36 m
    b) 6sin23= 2.34 m
    c) R^2 = a^2 + b^2 - 2abcos23 = 12.24 m
    d) I have no clue where to begin ..
    Last edited: Apr 8, 2015
  2. jcsd
  3. Apr 8, 2015 #2
    Start by resolving A and B into components in the x and y directions. Even though the magnitudes of the vectors don't add algebraically, their components in each direction do. So when you are done adding their components, you will have the components of the resultant.

  4. Apr 8, 2015 #3

    Hmm I'm not quite sure how to do that... Do I have to solve for the x and y axis separately?

    for x axis do i have to do this:
    8cos23 =
    6cos23 =

    and y axis:
    8sin23 =
    6sin23 =

    And I'm also confusing myself because am I supposed to use 23 degrees or 67degrees? and why?
  5. Apr 8, 2015 #4
    Well, first of all, the vector A, length 6, is pointing exclusively in the y direction. It doesn't have a component in the x direction. So, what are its components in the x and y directions?

    Your components for vector B are more on target. But, there is something a little amiss with your y component. Which way is the y component of B pointing, up or down? What does that tell you about its sign?

  6. Apr 8, 2015 #5

    magnitude of R on the x axis:
    ax = 6cos90 = 0
    bx = 8cos23 = 7.36

    rx = ax + bx
    rx = 7.36

    along the y axis:
    ay = 6sin90 = 6
    by = -8sin23 = - 3.13

    ry = ay + by
    ry = 2.87

    B is pointing down, meaning its negative... but I still don't know what I'm supposed to do with that information .. Am i going in the right direction?
  7. Apr 8, 2015 #6
    Yes. You've already correctly answered parts a and b. For part c, the resultant R is obtained from the pythagorean theorem using rx and ry. Do you see how rx and ry are the components of the vector R in the figure?
  8. Apr 8, 2015 #7
    Yes now I do. Phew I went a long way from my original thread, I appreciate the help Chestermiller.

    So for part c,

    R = rx^2 + ry^2
    = (7.36)^2 + (2.87)^2

    As for part d,
    I would have to use the sine law? Im totally stuck here.

    R = 7.89
    angle = 23 degrees

    B = 8.0
    angle = ?

    angle R makes with the x-axis=
    cos23/7.89 = cos(theta)/8
    ( cos23 x 8 )/ 7.89
    = 21 degrees
  9. Apr 8, 2015 #8
    Drop a normal from the tip of the vector R to the x axis. This produces a right triangle. One side of this triangle is rx, a second side is ry, and the hypotenuse is R. From this right triangle, what is tanθ?

  10. Apr 8, 2015 #9
    ok i created a right angle triangle...
    tan(theta) = opp /adj
    hypotenuse is 7.89
    rx = 7.36
    ry = 2.87

    tan theta = 7.89 / 7.36
    = 46.99 degrees

    Does this look right? :nb)
  11. Apr 8, 2015 #10

    tan theta = 2.87 / 7.36
    = 21 degrees
  12. Apr 8, 2015 #11
    No. Opposite = 2.87, Adjacent = 7.36,

    tanθ = (2.87)/(7.36)

  13. Apr 8, 2015 #12
  14. Apr 8, 2015 #13

    I believe I can now do these questions on my own. Through all the errors and trials, I appreciate your help so much Chestermiller.
  15. Apr 8, 2015 #14
    I like your agressiveness, persistence, and determination. Great qualities.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted