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Tension in a rope suspending a hinged plate

  1. Nov 25, 2006 #1
    Determine the tension in the support rope and the forces of the hinge on he uniform rectangular mass below. The length on the rect. is 12 and the height is 6m. The angle that the rope makes with the rectangle is 30 degrees.

    I broke up the hinge forces into x and y components:

    Bx=Tcos 30
    By + Tsin 30= mg

    mg(6) <--(half of the length) = T sin 30 (12)

    T = mg(6)
    --------- = mg
    12(sin 30)

    Now that Ive calculated T, I use it to find out the forces of the hinge.

    Bx = mg(cos 30)
    By = (square root of) \[mg^2 - \[3/2] mg2]

    That's as far as I got. I don't know how to determine the actual forces since there are no number provided for the mass. And I don't think that the Tension calculated is correct.

    Also, does the space between the hinge and the object affect the tension in the rope? Are my calculations of the tension only determine it along the length of the object?

    Any help would be much appreciated!

    Attached Files:

  2. jcsd
  3. Nov 25, 2006 #2


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    See comments in red above.
  4. Nov 27, 2006 #3
    So this portion:

    T = mg(6)
    --------- = mg
    12(sin 30)

    mg - T cos 30
    T sin 30

    after adding the T cos30 (6)?

    Wouldn't that change the value of Bx and By though?
  5. Nov 27, 2006 #4
    There is also another part to this question that I need help with:

    If the cable snaps, the rectangular mass will rotate about the hinge. Assume the rectangle begins to fall starting from the rest the instant the rope snaps and the moment of inertia about the center of mass, perpedicular to the page is I(bar)z = 450 Kgm^2

    What is the moment of inertia of the rectangle about the hinge axis perpendicular to the page? and what will be the magnitude of the angular velocity of the rectangle as it swings to its lowest postion?

    I attempted to use the perpendicular axis theorum using the Bx and By forces of the pervious question, but since there are no actual numbers in those forces, I could not get a definate answer. Also, I could not set the I z to equal those of By and Bx since Bx and By are the forces of the hinge while Iz is the inertia of the center of mass of the rectangle.

    Any suggestions as to how I can approach this problem?

  6. Nov 27, 2006 #5


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    You're losing me with your algebra. It's
    mg(6) = Tsin30(12) + Tcos30(6), now divide both sides of eq. by 6,
    mg = 2Tsin30 + Tcos30
    mg = T(2sin30 +cos30)
    mg = T(1 + .866) = T(1.866)
    solve T = mg/1.866

    Then solve
    Bx=Tcos 30 = (mg/1.866)(.866) = .464mg
    and since
    By + Tsin 30= mg, then
    By = mg - Tsin30
    By = mg - (mg/1.866)(1/2)
    By = mg - .268mg
    By = .732mg
  7. Nov 27, 2006 #6


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    Are you sure the mass isn't given? You can calculate it by knowing the formula for the rotational moment of inertia of a rectangle about its c.m. Then use the parallel axis theorem to calculate I about the hinge, which is not a function of the hinge forces anyway. Are you familiar with the calcuations for I_cm and I_hinge? Once solved, you can calculate the angular speed at the bottom using conservation of energy prnciples.
  8. Nov 28, 2006 #7
    Okay, so I get:

    I(cm) = 1/12M (L^2 + W ^2)
    = 1/12M (12^2 + 6^2)
    = 15M

    The only other way I can think of to calcualte the speed is:
    Torque = mg (L/2)
    = m(9.8)(12/2)
    T= 58.8M
    T = I alpha
    58.8M/15 = 15M/15
    alpha = 3.92 rad/sec^2

    Then I used an angular kinematic eq.
    omega^2 = omega (initial) + 2(alpha)(theta initial - theta final)
    omega^2 = 2(3.92) (pi/2) <----(assuming/hoping that the angle the rect. rotates is 90 degrees?)
    omega = 3.51 rad/sec

    I tried using cons. of energy but that didn't get me anywhere since the mass and H are not given.
  9. Nov 28, 2006 #8


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    See comments above.
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