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Simple rotational mechanics problem

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data

    Cliff and Will are carrying a uniform 2.0 m board of mass 71 kg. Will is supporting the board at the end while Cliff is 0.6 m from the other end as shown. Cliff has attached his lunch to the end of the board, and the tension in the string supporting the lunch is 200 N.

    24g7no0.jpg

    A) Find the force that Cliff exerts on the board.

    B) Find the force that Will exerts on the board.

    2. Relevant equations



    3. The attempt at a solution

    It seems easy enough, I will set Will's Normal force to be the pivot point and solve for Cliff's.

    [tex](1.4m)N_{Cliff} = (200N)(2m) + (710N)(1m)[/tex]

    [tex]N_{Cliff}\approx 793N[/tex]


    To find Will's, I will subtract the Cliff's (793N) from the total (1110N) to get 317N.

    I then wanted to check my answer by solving for Will's by using Cliff as the pivot point, but came up with different numbers.

    [tex]N_{Will}(1.4m)=(200N)(.6m) + (710N)(.4m)[/tex]


    [tex]N_{Will}\approx 289N[/tex]

    Where did I go wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 27, 2010 #2

    kuruman

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    How did you get that the total down force is 1110 N? All I see directed down is 200 N + 710 N.

    Also, the right side of the second calculation should have torques with a relative minus sign because the weight of the board and the weight of the lunch are on either side of the pivot (Cliff).
     
    Last edited: Apr 27, 2010
  4. Apr 27, 2010 #3
    So my calculation for Cliff's force (part A) looks correct? I just need to subtract 910N by 793N to get Will's force?

    So the second part should have, say,

    (-200)(.6) + (710)(.4) = N(1.4)

    N = 117
     
  5. Apr 28, 2010 #4

    kuruman

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    Yes and yes. The two calculations agree now, don't they?

    For future reference, it is easier to see what is going on if you put all the torques on the same side and say that their sum equals to zero. You can adopt the convention that "counterclockwise" torques are positive and "clockwise" torques are negative. For example, for the second part, you would write

    -(N*1.4 m) + (710 N)*(0.4 m) - (200 N)*(0.6 m) = 0

    which is the same equation that you have.
     
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