mechanics help (moments)

**Tweety** · Jan 21st 2009, 06:31 AM

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 ?

**Grandad** · Jan 21st 2009, 07:07 AM

Hello Tweety

Originally Posted by Tweety

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

Originally Posted by Tweety

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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