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Equilibrium. Vectors and resultant forces.

  1. Jul 3, 2010 #1
    1. The problem statement, all variables and given/known data
    An object is at equilibrium if all the forces on it add up to zero. There's a beam which weights 124N 40degrees above the horizontal that is supported in equilibrium by a 100-N pull 30degrees above the beam and a force F at the floor. The third force on the beam is the 124-N wight that acts vertically downward.
    The question is use vector components to find the magnitude and direction of F.

    They give the weight of the beam vector A124N which is at 40 degrees above the horizontal
    and the Vector B100-N force pulling on it 30 degrees above the beam.
    There's another vector pulling on itVector F.

    2. Relevant equations
    vector F= VectorA+Vector B

    3. The attempt at a solution
    I found the vector components of vectors A and B which was for A 124cos40 and 124 sin40 and for B 100Cos70 and 100sin70 wich were Ax=94.9 and Ay79.7 and Bx=34 By=93.9 then I subtracted Ax-Bx and Ay-By to find the resulting forces acting on the floor and I get a resultant vector with a magnitude of 62.5 N but in the back of the book they get 49N. Can somebody tell me where am I wrong or what else do I need to do. Any help will be useful
  2. jcsd
  3. Jul 4, 2010 #2
    The question seems sort of confusing..

    In any case you made a component error. You are using 70degrees to find Bx, so it lies parallel to the ground. The problem says that force A is directly downward so it's x component (parallel to the ground) should be zero.
    You were adding components that lie on a different set of axis.

    That being said.. I still think you must have made a mistake in writing the question. Because I don't think it is possible to have the beam in equilibrium regardless of F.

    The torque exerted down by gravity is smaller than the torque exerted up by the 100N pull, so the beam would be rotating upward. If F is just the the normal force on the ground there is no way the ground is going to stop that rotation.

    if you find the answer please post it here, I'm curious now.
  4. Jul 4, 2010 #3
    Thanks for the help, the wording of the question also got me confused, but here's a link to the exact wording of the question and a diagram which also got me confused wether the force was directly downward or not. http://faculty.jsd.claremont.edu/jhigdon/phys33/phys33_07_HW_1.pdf [Broken]
    I already solved it but I don't know if I did it the right way, although I did get the right answer.
    Here's how I did it.
    1. Since the force is going downward there's a vector A with a magnitude of 124 going down. Using vector components I found the components of B using 70 degrees as the angle. which gave me Bx=34.20 and By=93.96
    2. From this I just subtracted By from Ay which gave me Cy=30.04 and from Bx=34.20 I found that Cy had to equal 34.20 in the opposite direction which gives me a vector with a magnitude of 45.5 which rounds off to 46 and got an angle of 41 degrees but since it is in the second quadrant it becomes 138.7 degrees which rounds off to 139 degrees which are the answers in the back of the book.
    Last edited by a moderator: May 4, 2017
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