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Torque Equilibrium HELP

  1. May 6, 2005 #1
    HELP Torque Equilibrium HELP!

    A scaffold is constructed by placing two supports under a 13 ft, 4 in (a) board weighing 90 lbs. The supports are each 1 ft, 10 in (b) from their respective ends. Two painters stand on the board; one weighs 175 lbs and stands 4 ft, 8 in (c) from the left end of the board. Where must the 225 lbs painter stand so that each support carries the same weight? (express your distance in feet from the right end of the board (d).)

    picture: http://upload.wikimedia.org/wikipedia/en/6/63/Mm_painters.gif
     
    Last edited: May 6, 2005
  2. jcsd
  3. May 7, 2005 #2

    Andrew Mason

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    To solve this, you have to put the centre of mass of the system in the exact middle between the supports. If you think of the board as the x axis with the centre of mass at 0, m1x1 + m2x2 = 0

    AM
     
  4. May 7, 2005 #3

    Pyrrhus

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    Well you need to apply the equilibrium conditions

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = 0 [/tex]

    [tex] \sum_{i=1}^{n} \vec{\tau}_{i} = 0 [/tex]

    You need to show a distance d where [itex] |\vec{N}_{1}| = |\vec{N}_{2}| [/itex]

    You can start by combining equilibrium condition number 1 with number 2 like

    [tex] 2 \vec{N} + \vec{W_{1}} + \vec{W_{2}} + \vec{W}_{board}= 0 [/tex]

    such our scalar equation (up-right positive)

    [tex] N = \frac{W_{1} + W_{2} + W_{board}}{2} [/tex]

    Now we take the 2nd Equilibrium condition with respect to point A which is rightmost part of the scaffold.

    [tex] \vec{r}_{1} \times \vec{W}_{1} + \vec{r}_{2} \times \vec{W}_{2} + \vec{r}_{3} \times \vec{N}_{1} + \vec{r}_{4} \times \vec{N}_{2} + \vec{r_{5}} \times \vec{W}_{board}= 0 [/tex]

    Thus our scalar equation (counter-clockwise positive)

    [tex] dW_{2} - bN_{2} + W_{1}(a-c) - N_{1}(a-b) + \frac{a}{2} W_{board}= 0 [/tex]

    Remember the condition [itex] |\vec{N}_{1}| = |\vec{N}_{2}| [/itex]

    [tex] dW_{2} - bN + W_{1}(a-c) - N(a-b) + \frac{a}{2} W_{board}= 0 [/tex]

    where

    [tex] N = \frac{W_{1} + W_{2} + W_{board}}{2}[/tex]

    [tex] dW_{2} - b(\frac{W_{1} + W_{2} + W_{board}}{2}) + W_{1}(a-c) - (\frac{W_{1} + W_{2} + W_{board}}{2})(a-b) + \frac{a}{2} W_{board} = 0 [/tex]

    Now solve for d.
     
    Last edited: May 7, 2005
  5. May 7, 2005 #4
    thanks Cyclovenom! that was sooooo incredibly helpful!
     
  6. May 7, 2005 #5

    Andrew Mason

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    Ok. But there is an easier way to do this.

    The torque about each support is determined by the total weight of the system x distance of the centre of mass of the (board + men) from that support + the upward force of the other support x the distance between supports. The sum of these torques = 0. Since the upward force of each support has to be the same (that is the condition), the torques from each support are equal and opposite. This means that the weight of the system x the distance of the centre of mass to the support is the same for both supports. So the centre of mass has to be in the middle.

    Since the centre of mass of the board is in the middle of the board, it drops out as well. You only have to deal with the centre of mass of the men. Length of the board = 160 inches. Middle is 80 inches from each end. L1 is the distance of the first man to the middle (80-56 = 24). L2 is the distance of the second man to the middle.

    [tex]M1L1 + M2L2 = 0[/tex]

    [tex]L2 = - M1L1/M2[/tex]

    So L2 = 175 * 24 / 225 = 18.67 inches. In other words, the heavy guy has to be 61.33 inches from the other end.

    AM
     
  7. May 7, 2005 #6

    Pyrrhus

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    Yes Andrew, definetly taking the Torque sum with respect to the center of mass of the board is easier, when i looked at the problem, i just went for the obvious approach. Good find!.
     
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