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Equilibrium: Minimum Cable Length

  1. Jun 15, 2011 #1
    1. The problem statement, all variables and given/known data
    The above figure shows a uniform iron beam of mass 253 kg and length L = 3 m. The cable holding the beam in place can take a tension of 1300 N before it breaks. Use the Young's Modulus for steel to be 200e9 N/m2. (You may ignore the small mass of the cable in this calculation.) a)What minimum length of the cable?
    Lengthmin= m

    HELP: How will the torque placed on the beam by the cable depend on the angle between the cable and the beam??
    b)Assume the cable is made of steel and has a diameter of 1". How much will it stretch before it breaks?
    Δmax = cm

    HELP: Young's modulus for steel is 200X109N/m2.


    2. Relevant equations



    3. The attempt at a solution

    for a)

    I set the forces in the x up like :
    mgcos(phi)-Fcos(theta)=0

    y:
    Fsin(phi)+Tsin(theta)-W=0

    torque
    -Wdsin(180)+TLsin(theta)=0

    where phi is the angle of the wall the normal force and theta is the angle of the cable and the beam. i dont know where x comes into play

    for part b i think it would be like this

    stress=1050/(1.27^2pi)

    stress/20000000=strain

    change in L will be strain*the L from part a
     

    Attached Files:

  2. jcsd
  3. Jun 15, 2011 #2

    Delphi51

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    Homework Helper

    I don't follow the first term. It should be the weight times the distance from the pivot to the center of mass, which is L/2. There is no angle involved since the weight acts perpendicular to the beam. I agree with your second term. Solve for the angle theta, from which you can calculate the length of the cable.
     
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