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    Part a answer is 3cm. Can anyone help with part b? Thanks a lot. I’ve worked out the distance from G to DC
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    (Original post by Sherryew)
    Part a answer is 3cm. Can anyone help with part b? Thanks a lot. I’ve worked out the distance from G to DC
    Can you please post what you've done so far? You should have a diagram showing where the centre of mass must be if AB is inclined at 25 degrees. Let us know if you don't understand this part.
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    (Original post by Notnek)
    Can you please post what you've done so far? You should have a diagram showing where the centre of mass must be if AB is inclined at 25 degrees. Let us know if you don't understand this part.
    What I get is the center of mass(G) from AB=3cm, and tan25=3/AN, therefore DN(Distance from G to DC)=6-3/tan25. And then I’m stuck
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    (Original post by Sherryew)
    What I get is the center of mass(G) from AB=3cm, and tan25=3/AN, therefore DN(Distance from G to DC)=6-3/tan25. And then I’m stuck
    What is N? I can't see it mentioned in the question. Do you have a diagram?
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    (Original post by Sherryew)
    What I get is the center of mass(G) from AB=3cm, and tan25=3/AN, therefore DN(Distance from G to DC)=6-3/tan25. And then I’m stuck
    Forgive me for barging in, Notnek , took me quite a while to edit this image and I just refreshed the page to see you've contributed awhile ago.

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    Sherryew If you look at the diagram I have edited, this is how it would look like after the lamina has been suspended from A. All you do is rotate the image and align the Centre of Mass with the suspended point.

    \frac{3}{\tan(25)} will give you the distance of {AK} from my diagram but you want the distance from the DC plane to point O.

    Think again, why is 6-\frac{3}{\tan(25)} wrong?
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    let F be the foot of the perpendicular from G to line AB.

    the distance GF = 3

    angle FAG = 25°

    distance DA = 6

    triangle AFG is right angled.

    you need distance DF.
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    (Original post by ManLike007)
    Forgive me for barging in, Notnek , took me quite a while to edit this image and I just refreshed the page to see you've contributed awhile ago.

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    Sherryew If you look at the diagram I have edited, this is how it would look like after the lamina has been suspended from A. All you do is rotate the image and align the Centre of Mass with the suspended point.

    \frac{3}{\tan(25)} will give you the distance of {AK} from my diagram but you want the distance from the DC plane to point O.

    Think again, why is 6-\frac{3}{\tan(25)} wrong?
    omg just realise that I made a silly mistake:// What I get is DK=3/tan25 -6. But I don’t know how to find distance of O from DC
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    (Original post by the bear)
    let F be the foot of the perpendicular from O to line AB.

    the distance OF = 3

    angle FAO = 25°

    distance DA = 6

    triangle AFO is right angled.

    you need distance DF.
    Nah. Angle GAD=25, not FAO, and FO=2
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    (Original post by Notnek)
    What is N? I can't see it mentioned in the question. Do you have a diagram?
    Could you pls have a look at the diagram below [email protected]? N is the point K in his diagram
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    (Original post by Sherryew)
    Nah. Angle GAD=25, not FAO, and FO=2
    i meant G, not O...

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    (Original post by Sherryew)
    Could you pls have a look at the diagram below [email protected]? N is the point K in his diagram
    I don't know what you've tried but at some point you need to take moments in the y direction (looking at the original diagram).

    So i.e. in part a) you took moments about AB. Go back to this original diagram and try taking moments about the horizontal that includes B. Remember that you now know this \bar{y} from your working in b).
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    These international papers have some tough questions. The 2013 (R) had an even harder COM question than this one (question 6)
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    There will be different approaches to this question so ManLike007 please share what you did (with a hint) if it's different to the hint I gave above.
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    (Original post by Notnek)
    I don't know what you've tried but at some point you need to take moments in the y direction (looking at the original diagram).

    So i.e. in part a) you took moments about AB. Go back to this original diagram and try taking moments about the horizontal that includes B. Remember that you now know this \bar{y} from your working in b).
    Ill try then
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    (Original post by Notnek)
    These international papers have some tough questions. The 2013 (R) had an even harder COM question than this one (question 6)
    But unfortunately I’ve done that one correctly but couldn’t solve this one😭😭
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    (Original post by Sherryew)
    But unfortunately I’ve done that one correctly but couldn’t solve this one😭😭
    Maybe this one is harder then The folded lamina was very unusual which is why I thought it was harder plus the working isn't easy either.
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    (Original post by Notnek)
    Maybe this one is harder then The folded lamina was very unusual which is why I thought it was harder plus the working isn't easy either.
    Sorry but I don’t know how to find the distance of O to the horizontal that includes B.
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    (Original post by Sherryew)
    Sorry but I don’t know how to find the distance of O to the horizontal that includes B.
    Actually it's probably simpler to take moments from DC : Call \bar{y_s} the y distance of the square's centre of mass from DC and call \bar{y_l} the y distance of the lamina's centre of mass from DC.

    48 \times 0 - 16\times \bar{y_s} = 32\bar{y_l}

    You've already done the working in b) to find y_l so you can find y_s which is the answer to the question.

    It's late so I hope I'm not overcomplicating this or making mistakes...
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    (Original post by Notnek)
    Maybe this one is harder then The folded lamina was very unusual which is why I thought it was harder plus the working isn't easy either.
    Also came up in this year's M2 paper back in June, oh boy was I in for a treat when I thought it wouldn't come up.
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    (Original post by ManLike007)
    Also came up in this year's M2 paper back in June, oh boy was I in for a treat when I thought it wouldn't come up.
    I was also surprised to see it because I thought folded laminas were only a special treat for international students They're not bad once you've done one but I remember struggling with it the first time I looked - trying to consider the folded section separately with double the mass is a bad idea.
 
 
 
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