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Beam is suspended on two ropes

  1. Jan 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Homogeneous beam shaped in letter L is suspended on two ropes of the same length l.
    The beam is of negligible thickness and its longer arm is three times longer than the shorter arm. Its overall length is l as well as the length of one rope. Look at each situation in the figures. How can I determine the forces acting on the rope at points A and B in both situations?
    There is picture on first example: http://fyzikalniolympiada.cz/archiv/55/fo55a1_z.pdf

    2. The attempt at a solution
    A)
    So, I have some equalities...
    1) the resultant force is zero
    T_1+T_2-F_G = 0
    T_1+T_2 = F_G = mg
    2) the resultant moment is zero
    M_1 + M_2 = 0
    ... where is the axis of rotation? in C?
    and if the axis of rotation in C, where is the arm of forces? (the d from equation M=Fd)
    ... it is a perpendicular line from C to lengthened shape?
    What is different in the second situation? apply the same equation?

    Thank you
     
  2. jcsd
  3. Jan 7, 2014 #2
    Good, so far.

    There is no rotation. Any point can be used to compute moments. But because your equation has only the moments of the tension in the ropes, that point must be C (otherwise, you would have the moment of weight).

    It is the distance between the line of force and the line parallel to the force passing through C.

    Same equation, but note that the line of force is different.
     
  4. Jan 7, 2014 #3
    Ok, thanks.
    AC = 2/3 l; BC = 1/3 l
    (AB)^2 = (AC)^2 - (BC)^2 = (2/3 l)^2 - (1/3 l)^2 = 3/9 l^2
    AB = √(3/9 l^2) = √3 /3 * l = 1/√3 * l

    I'll add a point on pictures
    X ... point making right triangle XBC
    I can make two equations by Pytgagoras equation.
    1) (AB - XC)^2 = AC^2 - AX^2
    (1/√3 * l - XC)^2 = (2/3 * l)^2 - AX^2
    2) XC^2 = BC^2 - AX^2
    XC^2 = (1/3 * l)^2 - AX^2

    I have two unknows in the equations... So, I can solve it.
    I have to know XC anc AB-XC fordetermining moments.
    Than M= T_1*(AB-XC) - T_2*XC
    And then I will have some system of equations for determinate T_1 and T_2.
    It is true?

    However, what with B? Same solution, but no AB-XC an XC, but different progress? So, XC an YC.
     
  5. Jan 7, 2014 #4
    I misdirected you in my previous post. I said that if you choose C as the point about which the moments of force are computed, then the moment of weight will be zero. That is not correct.

    I think it will be better to find the location of the center of mass in both cases then compute moments about this center. Are you familiar with the formula for the moment of force that uses the sine function?
     
  6. Jan 7, 2014 #5
    Why it isn't correct?
    Hmm, no, I know only basic formulas :-(
     
  7. Jan 7, 2014 #6
    There are three forces: two in the ropes and the weight of the beam.

    The weight of the beam acts as if it was applied to the center of mass of the beam. To compute its moment properly, you have to know its center of mass.
     
  8. Jan 7, 2014 #7

    haruspex

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    This is a repost of https://www.physicsforums.com/showthread.php?t=730319.
    Numeriprimi, why didn't you respond to me on that thread? There, I suggested you take moments about one of the points of attachment. For the moment of the beam, you could first find its centre of mass as Voko suggests, but I think you might as well just treat the two arms as separate loads. E.g. consider the section AC. What is its mass? Where is its mass centre? What moment does it have about A?
     
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