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Math Help - Statics of rigid bodies

  1. #1
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    Statics of rigid bodies

    I'm stuck on the following question:

    ABCD is a uniform square lamina with side 4 and mass 8kg. It's hinged at A so that it is free to move in the vertical plane. It is mantained in equilibrium with B vertically below A by a horizontal force acting at C and a vertical force acting at D, each of magnitude F newtons. Find the value of F.

    I don't know how to resolve the forces when taking moments at A or if they even need to be resolved since a square has 90 degrees so the forces are penpendicular to the side AB. Can someone help please?
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  2. #2
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    Quote Originally Posted by free_to_fly View Post
    I'm stuck on the following question:

    ABCD is a uniform square lamina with side 4 and mass 8kg. It's hinged at A so that it is free to move in the vertical plane. It is mantained in equilibrium with B vertically below A by a horizontal force acting at C and a vertical force acting at D, each of magnitude F newtons. Find the value of F.

    I don't know how to resolve the forces when taking moments at A or if they even need to be resolved since a square has 90 degrees so the forces are penpendicular to the side AB. Can someone help please?
    The plate ABCD revolves around point A.

    I've attached a sketch of the situation as I understand it.
    Attached Thumbnails Attached Thumbnails Statics of rigid bodies-drehb_platte.gif  
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  3. #3
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    So taking moments at A:
    2*8*9.8cos45=2*4F

    Is that correct?
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  4. #4
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    You should have more moments than that.

    Ok here's how I would do it:

    \mbox{Assume CCW +}

    \sum M_A = 0

    -(8kg)(9.81m/s^2)(\cos(45))(2m) + (F)(4m) + (F)(4m) = 0

    2(F)(4m) = (8kg)(9.81m/s^2)(\cos(45))(2m)

    F = \frac{(8kg)(9.81m/s^2)(\cos(45))(2m)}{2\cdot4m} = \boxed{13.9 N}

    That would be the answer for BOTH forces assuming that both have to be the same.
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  5. #5
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    Quote Originally Posted by free_to_fly View Post
    So taking moments at A:
    2*8*9.8cos45=2*4F

    Is that correct?
    Yes, it is the right turning momentum.

    The left turning momentum is caused by the forces F_1 and F_2 and the effective lever arm l.

    The length of l = 4 \cdot \sqrt{2}
    and since |\overrightarrow{F_1}| = |\overrightarrow{F_2}| the resulting force of \overrightarrow{F_1} and \overrightarrow{F_2} is \vec  R:

    |\vec R| = |\overrightarrow{F_1}| \cdot \sqrt{2}

    Because \vec R is perpendicular to l you don't have to take any angle into account. The solid is in equilibrium if both momenti are equal:

    2 m \cdot 8 kg \cdot 9.81 \frac{m}{s^2} = 4 \cdot \sqrt{2}m \cdot |\vec R| = 4 \cdot \sqrt{2}m \cdot |\overrightarrow{F_1}| \cdot \sqrt{2} will yield:

    |\overrightarrow{F_1}| = 2 \cdot 9.81 \  N = 19.62 N
    Attached Thumbnails Attached Thumbnails Statics of rigid bodies-drehb_platte.gif  
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