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Static Equilibrium

  1. Dec 19, 2009 #1
    1. The problem statement, all variables and given/known data

    A 60 kg man hangs one meter from the end of an 8 m, 100 kg beam. A steel cable is attached to the end of the beam near the man, and the opposite end is supported by a frictionless pin (or a hinge). The system is originally at rest. The cable makes a 35 degree angle between the horizontal and the hypotenuse.
    Find the tension in the cable.



    2. Relevant equations

    T=Fd

    Net Force=0
    Net Torque=0
    3. The attempt at a solution

    I draw my free body diagram and get the weight force of the steel rod being exerted at 4 m. I get the hanging man’s normal force at 7 m. I get a tension force at 35 degrees on one end, and then I get a vertical and horizontal normal force from the pin on the other side.

    Horizontal forces
    Pin(x)-Cable(x)=0

    Vertical forces
    Pin(y) + Cable(y) -Beam- Man=0

    I am having trouble incorporating torque into this problem and I know I need to. Where should I place my reference frame to find torque?
     
  2. jcsd
  3. Dec 19, 2009 #2
    Take moments about the pin/hinge.
    If the pin is to the left of the man and the cable to the right
    clockwise: the man and the weight of the beam
    counterclockwise: moment of the tension about the pin
     
  4. Jan 14, 2010 #3
    ok I just retried this problem...

    I added the torque's of the rod and the man, and figured the vertical component of the tension in the cable should be equal to the torque.


    (4)(100)(9.81)+7(65)(9.81)=8387.55=Torque

    then to get the tension in the actual cable I did...

    8387.55/sin(35)=14623.24717=tenion

    This numbers seem pretty high to me. Is this wrong? If sowhere did I go wrong.

    Thank you.
     
  5. Jan 15, 2010 #4
    The first part is correct. (Is the man 60 or 65 kg?)
    The moment(torque) of the tension about the pin is given by the tension T in the cabel times its perpendicular distance from the pin.
    If you draw the triangle (rod(8m), cable, 35deg) you will see that this distance is given by 8 sin (35)
    If you equate the the two torques now you should get the answer.
     
  6. Jun 16, 2010 #5
    (4)(100)(9.81)+7(60)(9.81)=8044.2=Torque on the pin
    Torque exerted by cable=8F=8044.2
    F=1005.525

    I figured that the Force involved was the vertical component so I divided it by sin(35) to get the tension in the cable.

    Tension(cable)=1753.079?

    I am not sure what you were saying to do with the triangle. Is this what you were saying?
     
  7. Jun 16, 2010 #6
    The torque from the steel cable is the tension in it (T) multiplied by the perpendicular distance (OP) of the line of force from the pin. Here's how it looks.
    planckmoments2.png
    The perpendicular distance is r in the diagram and is 8 sin 35

    You have the correct clockwise torque
    4 x Wp + 7 x Wm [weight of beam and man]
    This is balanced by the torque of the tension, T, of the cable, and this is
    T x OP = T x 8 sin 35
     
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