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Homework Help: Minimum length of cable help

  1. Jan 2, 2004 #1
    Picture is attached.

    The above figure shows a uniform iron beam of mass 254 kg and length L = 3 m. The cable holding the beam in place can take a tension of 1300 N before it breaks. (You may ignore the small mass of the cable in this calculation.)

    a)What minimum length of the cable?
    b) Assume the cable is made of steel and has a diameter of 1". How much will it stretch before it breaks?

    Okay. I am stuck on part a so far. I know that this is a static equilibrium, therefore, the summation of all the forces in x and y direction equal zero.
    Also, the summation of torques must also equal zero.

    But I'm not sure how to write out the equations. Can someone give me a lending hand?

    Attached Files:

  2. jcsd
  3. Jan 3, 2004 #2


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    Let me try to label some forces without drawing a picture:

    The weight of the bar, W, acts at the center of mass of the bar (1.5 m from the left end (set your origin at the left end)).

    The tension in the cable, T, has components Tx (clearly pointing to the left) and Ty (pointing up).

    The mounting of the bar (pin joint) can support a vertical force, Ry (pointing up) and a horizontal force, Rx (pointing to the right).

    Some equations for static equilibrium:

    [tex] \Sigma F_y = R_y + T_y - W = 0 [/tex]

    [tex] \Sigma F_x = R_x - T_x = 0 [/tex]

    [tex] \Sigma M_o = T_y\cdot L - \frac{WL}{2} = 0 [/tex]

    You can use these to find the y-component of the tension (actually, you only need the moment equation), but not much else.

    Say you've solved for Ty and you want to find the minimum length of cable. You know that for the minimum length of cable, the cable tension will be at its maximum (not a great plan from an engineering standpoint, but nonetheless...). So:

    [tex] T = \sqrt{T_x^2+T_y^2} = 1300 [/tex]

    use that to solve for Tx.

    The length of the cable is given by:

    [tex] L_c = \frac{L}{\cos \theta} [/tex]

    where Lc is the length of the cable, L is the length of the bar (3 m), and θ is the angle between the cable and the bar, so that:

    [tex] \tan\theta = \frac{T_y}{T_x} [/tex]

    That should allow you to find Lc

    For part b, you should be able to use Hooke's law, σ = εE, to find the length it stretches (you have the tension, the cross sectional area, the unstretched length, and you can look up the Young's modulus).
  4. Jan 3, 2004 #3
    I'm not sure as I haven't actually worked it through, but because both θ and Lc are unknown, you might also have to use
    Ty = Ry

    (This must be true since otherwise the bar would rotate about its center of mass.)
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