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    Name:  help.png
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Size:  20.8 KBIn the Mark scheme they have took moments about C but I don't know how they get 3gcos30? how is that the force x perpendicular distance from c? I c
    e
    cant see it..
    edit: would it not be 3g(cos30)(2)?
    thanks in advance!
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    The 3g cos(30) is the force 3g * 1 cos(30), because the perpendicular distance is the same as the arrow marked "R" (even though they're unrelated in the diagram, it's a useful marker for me to use on the diagram.)
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    (Original post by Smaug123)
    The 3g cos(30) is the force 3g * 1 cos(30), because the perpendicular distance is the same as the arrow marked "R" (even though they're unrelated in the diagram, it's a useful marker for me to use on the diagram.)
    but is R not 2m away from c?
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    (Original post by upthegunners)
    but is R not 2m away from c?
    Yes, it is - I was only using "R" as a conveniently drawn arrow on the diagram, not as representative of the force R. g is cos30 away from c, because the "perpendicular distance" from g's action to c is cos30; imagine drawing a dotted line along what is currently marked 3g, all the way off the top of the page. At one point, the dotted line draws level with C; at that point, the distance between the dotted line and C is cos30, and that's the perpendicular distance. (because the distance between BC and the dotted line is constant, and at the point B that distance is cos30.)
    2m might be called the "parallel distance".

    EDITED TO ADD: See ghostwalker's diagram, which is exactly what I was trying to describe; a picture speaks a thousand words!

    EDITED AGAIN TO ADD: Said diagram seems to have disappeared - maybe I'm going mad…
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    See, in green.

    Reposted.
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    (Original post by ghostwalker)
    See, in green.

    Reposted.
    thanks for the diagram but I dont dont get why it is? can we not just do 3g and not have to times it by anything since (as you have showing in your diagram) 3g is already perpendicular to c?
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    (Original post by upthegunners)
    thanks for the diagram but I dont dont get why it is? can we not just do 3g and not have to times it by anything since (as you have showing in your diagram) 3g is already perpendicular to c?
    As Smaug123 explained. The moment of a force is the product of the force times the perpendicular distance of it's line of action from the fulcrum.

    In this case the perpendicular distance is represented by the line in green, which is 1\times\cos 30
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    (Original post by ghostwalker)
    See, in green.

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    (Original post by Smaug123)
    EDITED AGAIN TO ADD: Said diagram seems to have disappeared - maybe I'm going mad…
    I deleted and reposted it. Keep calm.
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    (Original post by ghostwalker)
    As Smaug123 explained. The moment of a force is the product of the force times the perpendicular distance of it's line of action from the fulcrum.

    In this case the perpendicular distance is represented by the line in green, which is 1\times\cos 30
    so if 3g was at point A in the diagram it would be 2 x cos30?
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    (Original post by upthegunners)
    Name:  ssdsdsdsd.png
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    Yes, the cos 30 comes from where the arrow is.

    Or you could use sin 60, as more obvious.
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    (Original post by upthegunners)
    so if 3g was at point A in the diagram it would be 2 x cos30?
    That's correct.
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    (Original post by ghostwalker)
    That's correct.
    you're a brilliant help

    thanks very much
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    (Original post by ghostwalker)
    I deleted and reposted it. Keep calm.
    Phew thank goodness for that!
 
 
 
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