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For moment about D , how could the FGF sin26.6 generate moment about D ?

FGF sin26.6 is located above point D , right ? How could FGF sin26.6 generate moment about D ? fsdfsf

ddada

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  • $\begingroup$ Is your name kelvinmacks? $\endgroup$ – AndyT Mar 24 '17 at 11:45
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F.GF sin (26.6) is the vertical component of the force in the top member, but you can't simply shift the point of application of that force. You can shift it where you like along the dotted line (ie, the line of action of the force), so you can shift it to O, at which time the moment of that force about D is indeed F.GFxsin(26.6)x6m.

It's a singularly clumsy way of doing it though.

For moment of F.GF about D I would do F.GF x 3m x cos(26.6), ie full F.GF force times the perpendicular distance between D and the line of action of the force.

Of course, 3 cos(atan(0.5)) is exactly the same number as 6 sin(atan(0.5)), so it makes no difference to the answer.

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