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# Angle, equilibrium and centre of mass watch

1. Question 3 part iii from 7 June 2005 MEI Structured Mathematics Mechanics 2
Here is a svg drawing I made to represent part ii on the paper, the object is perpendicular to Z axis.
The centre of mass is (x,y,z)=(31/14,31/14,-4/7)≈(2.21,2.21,-0.57)

In part iii, it changes and is now hanging off in equilibrium.
"The object is now freely suspended from A and hangs in equilibrium with AC at alpha degrees to the vertical"
I take the vertical to be the Z axis?
AC=√(32)=4*√(2)
I'm confused when it says "equilibrium", do I have to find the moments about A?
2. Well I do have the rest of Easter but I'd prefer an answer soon.
3. If it is fixed at A and allowed to hang freely, then equilibrium can only be achieved if the centre of mass is directly below A. Otherwise there would be an unbalanced moment.
4. (Original post by Primus2x)
I'm confused when it says "equilibrium", do I have to find the moments about A?
If you let the centre of mass be at point M, then when the object hangs freely AM will be vertical. Therefore, the angle AC makes with the vertical will be the angle that AC makes with AM.

I can't really make sense of the diagram, so the rest is up to you.
5. AC is always equal to AM or only equal when the system is in equilibrium?

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