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Thread: 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 = $\displaystyle 2Y + Y = 120g $
    $\displaystyle 3Y = 120g $

    $\displaystyle y = 40g $

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


    taking moments about 'Q' $\displaystyle 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 = $\displaystyle 2Y + Y = 120g $
    $\displaystyle 3Y = 120g $

    $\displaystyle 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' $\displaystyle 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 $\displaystyle x$ is the distance QX? If so, then the $\displaystyle (0.8m - xm)$ term in your equation should simply be $\displaystyle xm$. Leave the weight of Arthur as $\displaystyle 60g$, and you get:

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

    $\displaystyle \Rightarrow 16 + 64 = 60x + 16$

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

    So QX is 1.07 m.

    Grandad
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