1. Not finding help here? Sign up for a free 30min 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!

Finding the resultant of two forces in 3d

  1. Sep 20, 2014 #1
    Hi, For the following image, we have to calculate the resultant of those two forces. I have broken them down to their components. Did I do this correctly because when I actually solve for the force I am not getting the correct answer of 940N in the back of my book.

    Also, I tried calculating the direction of force P in the x plane , Q in the z plane and when I used the cosine formula I got
    cos^2 theta = -0.4 ( a negative number) does that imply that our vector does not exist in the x plane and therefore is 0 for P? and 0 for Q in the z plane?

    s5924p.jpg
    2hx14zn.jpg
     
  2. jcsd
  3. Sep 21, 2014 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    The photograph does not clearly define the vectors, far as I can see.
     
  4. Sep 21, 2014 #3
    They are not defined by x,y,z co-ordinates but by their coordinate angles?
     
  5. Sep 21, 2014 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Yeah, but the angles are not well defined as I see it. It's not clear whether the angles are with the x, y or z axes. Maybe with some effort I could figure it out, but a verbal description would have been a big help.
     
  6. Sep 21, 2014 #5

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Your calculations show the Qx = Q cos 30. This is incorrect. The projection of Q onto the x-z plane makes an angle of 30 degrees to the x-axis.
    The magnitude of the projection of Q onto the x-z plane is not equal to the magnitude of Q.

    Your calculation of Qy is correct.

    You have a similar situation with the force P. The projection of P onto the x-z plane makes an angle of 25 degrees to the z-axis.
    Again, the magnitude of the projection of P onto the x-z plane is not equal to the magnitude of P.
     
  7. Sep 21, 2014 #6

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Doesn't look too confusing but I think...

    Qx = 450*Cos55*Cos30
    Qz = 450*Cos55*Cos60

    Haven't checked them all.

    Cross posted with the reply by Steamking.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted