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Help with Force problem

  1. Oct 27, 2005 #1
    Four masses are positioned at the corners of a rectangle, as indicated in Figure 12-23 (not to scale).

    Figure 12-23 (attachment)

    (a) Find the magnitude and direction of the net force acting on the 2.0 kg mass if x = 0.20 m and y = 0.15 m.

    (b) How do your answers to part (a) change (if at all) if all sides of the rectangle are doubled in length?

    I don't even know where to start on this problem, any help would be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Oct 28, 2005 #2
    Hi wr1015.
    Imagine a rectangle ABCD in counterclockwise direction, from botton to top. At vertice A there is mass m1, at B, m2, at C, m4 and at D m3.
    I'm putting the origin of coordinate axis at m1. The vector force F on m1 is:
    X -> x axis versor
    Y -> y axis versor
    x -> rectangle basis
    y -> rectangle height

    F=m1m2/x^2 X + m1m3/y^2 Y + m1m4/(x^2+y^2)(x X+y Y)/(x^2+y^2)^1/2

    this is the net force done on m1 by all others including their direction. Grouping:

    F=( m1m2/x^2 + m1m4x/(x^2+y^2)^3/2 ) X +
    + ( m1m3/y^2 + m1m4y/(x^2+y^2)^3/2 ) Y

    This is the vector force. When you substitute your values, try to see if you can write the module of the vector times one linear combination of versors, so the direction apears explicity. Depending on the values, the direction can coincide with the diagonal of the rectangle.

    In part (b), just take x->2x and y->2y

    I hope this help you.
    Cheers
     
    Last edited: Oct 28, 2005
  4. Dec 11, 2009 #3
    I was wondering in what physics book was this problem located?


     
  5. Dec 14, 2009 #4

    st0

    User Avatar

    Im still confused about how to solve this problem. Isnt the force gravitational force, so you multiply by G on top? Also, how is that you find the distance between 1 and 4?
     
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