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Homework Help: Equilibrium Problem

  1. Mar 22, 2006 #1
    A uniform beam weighs 2.00 x 10^2 N and holds a 4.50 x 10^2 N weight. Find the magnitudes of the forces exerted on the beam by the two supports at its ends.

    There is a diagram shown... the beam is horizontal, and a vector is drawn from the centre of the beam, pointing down. The weight is located 1/4 of the beam's length from the right hand supporting beam. The supporting beams are represented as F1 and F2. The beam appears to be resting on the supports, with no hinges or any other thing.

    Any help? :cry:
  2. jcsd
  3. Mar 22, 2006 #2
    I'll call the weight of the beam FWB and the weight of the actual weight being supported FWW. L is the length of the beam.
    This is what I've done:

    [-FWB(L/2)] + [-FWW(L/4)] + F1 + F2 = 0

    I have two unknowns, F1 and F2. I don't know what to do with them!
  4. Mar 22, 2006 #3
    Number 1, the center of the object where the force is drawn is the object's weight.
    Number 2, 1/4 from RS (right side) is the weight.
    Number 3, 3/4 from LS (left side) is the weight.

    So, Newton's Law states that the sum of the forces must always be zero for an object to be in equilibrim (my understanding of it).

    So, the weight of the object and the weight of the beam are the only vectors pointing downwards. The only vectors pointing upwards are the force done by the supports.

    Seeing how we use vectors, by convention, take the upward direction as positive. Downward as negative.

    Draw a force-body diagram.

    Sum of all forces pointing upward = sum of all forces pointing downward.

    For some reason, you will have to use torque, if you don't know it already X_X

    T is force times distance. So you are given everything pretty much :-)

    See if that helps... If you don't know torque, let me know.
  5. Mar 23, 2006 #4


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  6. Mar 23, 2006 #5
    Thanks so much for your help! I got some help from a student who took the course last semester, and it all makes sense to me now!

    Thanks again :smile:
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