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    Hello there again, I've got another weird question with a weird layout. If anyone can have a look and point me in the right direction, I would be very grateful!

    I've attached the question & my response, which I'm stuck on.

    Cheers, in advance!
    Name:  uploadfromtaptalk1364233502070.jpg
Views: 136
Size:  39.6 KB

    Name:  uploadfromtaptalk1364233514312.jpg
Views: 101
Size:  67.1 KB
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    (Original post by KiddingMe)
    Hello there again, I've got another weird question with a weird layout. If anyone can have a look and point me in the right direction, I would be very grateful!

    I've attached the question & my response, which I'm stuck on.

    Cheers, in advance!
    Name:  uploadfromtaptalk1364233502070.jpg
Views: 136
Size:  39.6 KB

    Name:  uploadfromtaptalk1364233514312.jpg
Views: 101
Size:  67.1 KB
    Hint:

    Those are equilateral triangles.

    Therefore, the distance from each vertex to the opposite side is

    a\sqrt{3}
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    (Original post by KiddingMe)
    Hello there again, I've got another weird question with a weird layout. If anyone can have a look and point me in the right direction, I would be very grateful!

    I've attached the question & my response, which I'm stuck on.

    Cheers, in advance!
    Name:  uploadfromtaptalk1364233502070.jpg
Views: 136
Size:  39.6 KB

    Name:  uploadfromtaptalk1364233514312.jpg
Views: 101
Size:  67.1 KB
    Just looking at the moments about D, the perpendicular distance of the 60N force is \sqrt{3}a, so the moment due to that force is 60\sqrt{3}a

    I don't know how you worked yours out.
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    Sorry but I can't see the latex output, only the code, as I'm on tablet/mobile app of studentroom!

    But yep, I know they're equilateral, but I'm having a re-go at it. I'll keep you updated guys! : )
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    Okie, this is my second attempt! Still stuck though
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    • Thread Starter
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    Yea, I got the moment of force P from D to be -2Pa. Did you get the same?

    I got perp. distance to be 2a when working out moment of P about D not a!

    Confused haha


    Edit: I'm taking clockwise to be negative and anticlockwise to be positive Btw! Did I make a mistake somewhere with the signs? Cry!
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    Both (50N and PN) are going clockwise direction, so I gave both a -ve sign in front. Still unsure
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    Hmm, I got (80a)root3. Could you give us a clue where I went wrong? Cheers boss!
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    Yes, my bad, but

    (i) for M(D), the perp dist to P is 2a.

    (ii) you haven't got the right perp. dist to the 50N force

    drop a line from D (length a root 3 as I said) and work out the perp. dist of the 80N force from B using pythagoras.
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    Cool! Just to confirm, the 50N calculations I got for both M(D) and M(B) are both wrong?

    Will continue to work out moment of 50N. Cheers for helpful info!

    Edit: I've got (80a)root3 for one of the moments of about B. Is this right? I still got perp. Distance to be root3 as all perps from any corner of the triangle are equal because it is equilateral.

    Edit 2: isn't the perp distance just the horizontal distance: a (if we're talking about the 50N force)?
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    Its cool guys, I've managed to work it out! Cheers for all the help u 2!!

    But, haha I'm stuck on the 2nd part, now that we've got a value of P. Here is my response: Btw, part b image is in the first post.
    Name:  uploadfromtaptalk1364244957057.jpg
Views: 60
Size:  28.4 KB

    Edit: i have worked it out guys after I made some mistakes (look at above Pic and see if you can realise them!)

    Cheers all & good night!
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    (Original post by KiddingMe)
    Its cool guys, I've managed to work it out! Cheers for all the help u 2!!

    But, haha I'm stuck on the 2nd part, now that we've got a value of P. Here is my response: Btw, part b image is in the first post.
    Name:  uploadfromtaptalk1364244957057.jpg
Views: 60
Size:  28.4 KB

    Edit: i have worked it out guys after I made some mistakes (look at above Pic and see if you can realise them!)

    Cheers all & good night!
    Damn, was about to tell you
 
 
 
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