Results 1 to 2 of 2

Math Help - four displacement vectors problem.

  1. #1
    Senior Member
    Joined
    Jan 2007
    Posts
    477

    four displacement vectors problem.

    heres another physics displacement problem i'm having trouble with. I also included a drawing diagram to make it easier to see what is going on.

    The magnitudes of the four displacement vectors shown in the drawing are A = 11.0 m, B = 11.0 m, C = 12.0 m, and D = 32.0 m.
    what is the magnitude and directional angle for the resultant that occurs when these vectors are added together.

    magnitude is:
    and the directional angle is (in degrees)
    (from the +x axis)

    http://img339.imageshack.us/img339/2716/p142xd0.gif

    thanks for any help you can offer here.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,909
    Thanks
    329
    Awards
    1
    Quote Originally Posted by rcmango View Post
    heres another physics displacement problem i'm having trouble with. I also included a drawing diagram to make it easier to see what is going on.

    The magnitudes of the four displacement vectors shown in the drawing are A = 11.0 m, B = 11.0 m, C = 12.0 m, and D = 32.0 m.
    what is the magnitude and directional angle for the resultant that occurs when these vectors are added together.

    magnitude is:
    and the directional angle is (in degrees)
    (from the +x axis)

    http://img339.imageshack.us/img339/2716/p142xd0.gif

    thanks for any help you can offer here.
    You have the given coordinate axes, so break the vectors down into their components:
    A_x = -11.0 \cdot cos(20^o) and A_y = 11.0 \cdot sin(20^o)

    B_x = 11.0 \cdot cos(90^o) and B_y = 11.0 \cdot sin(90^o)

    etc.

    Then add all the x components and y components together.
    R_x = A_x + B_x + C_x + D_x = 0.40276
    and
    R_y = A_y + B_y + C_y + D_y = -16.6341

    The magnitude of the resultant vector is
    R = \sqrt{R_x^2 + R_y^2} = 16.639

    The reference angle is
    \theta = tan^{-1} \left (  \left | \frac{R_y}{R_x} \right | \right ) = 88.613^o
    and is in the fourth quadrant (as R_x is positive and R_y is negative.)

    So your vector is 16.6 m at 88.6 degrees below the +x axis.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Displacement Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 22nd 2010, 05:38 AM
  2. displacement problem
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: January 24th 2010, 06:32 PM
  3. Adding vectors to find displacement
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: September 30th 2009, 01:11 PM
  4. angle between displacement vectors
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 17th 2009, 11:15 AM
  5. vectors/displacement problem
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: September 2nd 2008, 02:28 PM

Search Tags


/mathhelpforum @mathhelpforum