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Math Help - mechanics help (moments)

  1. #1
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    mechanics help (moments)

    I am stuck on question 4b, if anyone can have a look at my attachment and help ?

    I keep getting a negative answer. Here's my working;

    the reactions at point Q and R =  2Y + Y = 120g
      3Y = 120g

     y = 40g

    so the reaction at point q = 80g and reaction at point r = 40g, is this bit correct?


    taking moments about 'Q'  40g \times 0.4m + 40g \times 1.6m = 588(0.8m-xm) + 20g \times 0.8m

    is this moments equation correct? cause when I solve for 'xm' it gives me -0.26 ?
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  2. #2
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    Moments

    Hello Tweety
    Quote Originally Posted by Tweety View Post
    the reactions at point Q and R =  2Y + Y = 120g
      3Y = 120g

     y = 40g

    so the reaction at point q = 80g and reaction at point r = 40g, is this bit correct?
    Yes
    Quote Originally Posted by Tweety View Post
    taking moments about 'Q'  40g \times 0.4m + 40g \times 1.6m = 588(0.8m-xm) + 20g \times 0.8m

    is this moments equation correct? cause when I solve for 'xm' it gives me -0.26 ?
    No, it's not. I assume that x is the distance QX? If so, then the (0.8m - xm) term in your equation should simply be xm. Leave the weight of Arthur as 60g, and you get:

     40g \times 0.4m + 40g \times 1.6m = 60g\times xm + 20g \times 0.8m

    \Rightarrow 16 + 64 = 60x + 16

    \Rightarrow x = \frac{64}{60}=1.07

    So QX is 1.07 m.

    Grandad
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