It's not entirely obvious to me that the body will rotate about A, it will depend on how the external forces are applied to it. So without more information I don't know if your intuition is correct.

Yes, it appears that there's missing (or hidden) information.

If the forces from the wiggly red arrows don't have tangential components, then they won't spin up the disk/cylinder around A.

If you aren't concerned about how fast it will get there, you don't need the masses and just need the component of the total force on the cylinder tangential to B (i.e. perpendicular to the A-B line, which will rotate about B). At equilibrium, that force will be zero and all the force will be parallel to the A-B axis (unless the force is strong enough to stretch or break things...).

At risk of messing it up because I'm doing geometry in my head before eating, it looks like F_perp=F_x*sin(theta)+F_y*cos(theta), where theta is the angle measured in the little interior angle that you show in the second pic. When F_perp=0, then all the force is along F_parallel. For completeness that should be F_parallel=F_x*cos(theta)+F_y*sin(theta).

You can check this by looking at the first picture, where theta=0. sin(0)=0 and cos(0)=1, so F_perp=F_x*0+F_y*1. If the linkage makes it all the way to 90 degrees, then F_perp=F_x*1+F_y*0, so I think that's right.

In reply to Re^2: [OT] Forces. by bitingduck
in thread [OT] Forces. by BrowserUk

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