Moment on a a semi quarter plate

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The discussion focuses on calculating moments in a 3D system involving forces and their components. Participants highlight the importance of choosing an appropriate axis for moment calculations, suggesting that selecting points with unknown forces can simplify the equations. There is confusion regarding the extra terms in the equations, but it's noted that they cancel out and do not affect the outcome. The conversation emphasizes that moments should be treated as vectors and calculated separately for each axis rather than added as scalars. Ultimately, the correct approach involves resolving moments about specific axes to achieve accurate results.
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Homework Statement


I have 750+CD +200=200-Fay -Fby = 0
then can someone suggset at which point or axis i should calculate the moment ? seems like no point / axis is suitable to do that ...

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I don't lnow why you have an extra 200 on each side, but since they cancel it doesn't matter.
You mean Faz and Fbz, no?
Which way does the force act in CD?

For choice of axis, it usually helps to pick a point with one or several unknown forces acting through it, so they do not feature in the equation. You will need equations for more than one axis, but they can pass through the same point.

You should be able to knock off Fax and Fay straight away.
 
haruspex said:
ch side, but sin
haruspex said:
I don't lnow why you have an extra 200 on each side, but since they cancel it doesn't matter.
You mean Faz and Fbz, no?
Which way does the force act in CD?

For choice of axis, it usually helps to pick a point with one or several unknown forces acting through it, so they do not feature in the equation. You will need equations for more than one axis, but they can pass through the same point.

You should be able to knock off Fax and Fay straight away.
for moment about A , i have 350(1) +CD(3) +200(3)+200(3√2)-FBZ(3√2) = 0
1798.5=-3CD+FBZ(3√2)
for moment about B , i have -200(1.55)-3CD-350(√13)+FAZ(3√2) =0
-1572=3CD-FAZ(3√2)
for moment about C, i have -350(2)+FAZ(3)+200(3)+200(3)-3FBZ= 0
500=3FBZ-3FAZ

I have tried to plug in the ans to see whether my working is correct or not . but , unfortunealy , my ans is wrong ...
the given ans is FBZ= 373N , FAZ=333N , CD=43.5N
 
In a 3D world, moments are not about a point but about an axis. Follow Haru's advice and look at moments about AC and about AB
 
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BvU said:
In a 3D world, moments are not about a point but about an axis.
To clarify, you can take moments about a point, but the moment you get will have an axis. That is, it will be a vector.
You can also choose to take moments about specific axes through the point. In that case each moment will be the component of the total moment parallel to the chosen axis. It's analogous to finding the net force as a vector, or finding the components of the net force along chosen axes.
werson tan said:
for moment about A , i have 350(1) +CD(3) +200(3)+200(3√2)-FBZ(3√2) = 0
As BvU indicates, this does not work. Those individual moments do not all have the same axis, so you cannot add them as scalars. Either find them as vectors, or take moments separately about the x, y, z axes through the point.
 
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