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Finding the force needed for equilibrium.

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data

    5. Find the magnitudes of the forces F2 and F3 that you would need to balance a 2.45 N force at 0.0 degrees If the angle of F2 is 35.0 degrees and F3 is 50 degrees.

    2. Relevant equations

    F2 cos 35 degrees - F3 cos 50 degrees =0

    F2 sin 35 degrees -F2 sin 45 degrees -mg= 0



    3. The attempt at a solution

    -.25 =m
    g= 9.80 m/s2

    F2 cos 35 degrees - F3 cos 50 degrees =0
    F2 sin 35 degrees -F2 sin 45 degrees = 2.5

    Is this set up right? Thanks.
    Is there a better way to calculate the forces to find equilibrium?
     
    Last edited: Sep 14, 2011
  2. jcsd
  3. Sep 14, 2011 #2

    PeterO

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    I would need to know what your angles are measured with respect to??
     
  4. Sep 14, 2011 #3
    angle between f2 and f3 are 85 degrees. f2 is 35 degrees, and f3 is 50 degrees. 3 forces are acting on each other. f1 is just a straight line between them.
     
    Last edited: Sep 14, 2011
  5. Sep 14, 2011 #4

    PeterO

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    OK the angle between f2 and f3 you have stated.

    f2 is 35 degrees. 35 degrees to what?
    f3 is 50 degrees. 50 degrees to what?

    Try giving True bearings. 0 degrees is North, 90 degrees is East, 180 degrees is South, 270 degrees is West.
     
  6. Sep 14, 2011 #5
    35 degrees north of west
    50 degrees south of west
     
  7. Sep 14, 2011 #6

    PeterO

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    OK then

    I am assuming then that f1 it due East.

    The vertical components of f2 and f3 must be equal in magnitude - they are already opposite in direction
    The horizontal components of f2 and f3 must add to 2.45 N to balance f1.
     
  8. Sep 14, 2011 #7
    Would mg/sq. root 2 work?
     
  9. Sep 14, 2011 #8

    PeterO

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    I don't thing so. You need sine and cosine functions to work out the components
     
  10. Sep 14, 2011 #9
    ok thanks!
     
  11. Sep 14, 2011 #10
    I received 3.87 N for both f2 and f3 magnitude.
     
    Last edited: Sep 14, 2011
  12. Sep 15, 2011 #11

    PeterO

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    I can't see that they would be equal.

    Can you confirm that 2.45 N East is the only other force involved.

    It this whole thing on a horizontal plane or is there a weight hanging somewhere/some how?
     
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