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Old 02-02-2009: 2nd February 2009 16:09 #1 
logic123 logic123 is offline Male
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Default M2
 
Stuck on this question

http://www.thepaperbank.co.uk/papers...20Jan%2002.pdf
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Old 02-02-2009: 2nd February 2009 16:12 #2 
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which question?
 
Old 02-02-2009: 2nd February 2009 16:13 #3 
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Which one?
Old 02-02-2009: 2nd February 2009 16:14 #4 
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4 a pls
Old 02-02-2009: 2nd February 2009 16:17 #5 
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you have to find the mass of the triangle and its centre, then the square and its centre, and then subtract them from each other using SIGMA Mxbar = SIGMA mixi
 
Old 02-02-2009: 2nd February 2009 16:21 #6 
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done that still dont get the answeer i did
(48x8/3)+(4x2)=44xX stilldont gert the answeer.
Old 02-02-2009: 2nd February 2009 16:26 #7 
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any offerds.
Old 02-02-2009: 2nd February 2009 16:27 #8 
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you need to take them away cos you are cutting out the square
 
Old 02-02-2009: 2nd February 2009 16:29 #9 
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i did take them away though 48-4=44
Old 02-02-2009: 2nd February 2009 16:39 #10 
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(8a/3 *1/2*12a*8a)-(4a*4a^2)/(12a*8a-4a^2)
sum of AiXi/sumai
 
Old 02-02-2009: 2nd February 2009 16:40 #11 
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Default Re: M2
 
Originally Posted by logic123
done that still dont get the answeer i did
(48x8/3)+(4x2)=44xX stilldont gert the answeer.
Shouldn't that be "Total mass = Sum of separate masses"? You are adding the wrong quantities
Spoiler:
Old 02-02-2009: 2nd February 2009 16:43 #12 
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green clover im not sure what you mean
Old 02-02-2009: 2nd February 2009 16:46 #13 
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Originally Posted by logic123
done that still dont get the answeer i did
(48x8/3)+(4x2)=44xX stilldont gert the answeer.

no it should be (48x8/3)-(4x2)=44xX
 
Old 02-02-2009: 2nd February 2009 16:49 #14 
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(8a/3 *1/2*12a*8a)-(2a*4a^2)/(12a*8a-2a^2) sorry made mistake
 
Old 02-02-2009: 2nd February 2009 16:51 #15 
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Default Re: M2
 
- For total mass before the square was removed: 48a^2 \times \frac{8a}{3}
- For separate masses:
The square: 8a^3
The template: 44a^2 \overline x
Use these to form an equation and remember to put them on the right side
If you still can't get it look at the spoiler in my post above.
 
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