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Cylinder over a step -- Is the normal force radial?
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[QUOTE="Chestermiller, post: 5012492, member: 345636"] No. The x and y coordinates of the center of the cylinder are: ##x=-rcos(\pi/6 + \theta)## ##y=rsin(\pi/6 + \theta)## Yes, that's the net counterclockwise moment about A, but the force F and the gravitational constant g in this equation should not be indicated as vectors. They should be the scalar magnitude of F and the scalar magnitude of g. What is the clockwise moment about A? I asked for the moment [I]balance [/I]about A. All you have given is one side of the equation. What is the other side of the equation (in terms of time derivatives of θ)? Chet [/QUOTE]
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Cylinder over a step -- Is the normal force radial?
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