- #1
Soren4
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Homework Statement
Consider a block placed on a surface, in two different configuration, a and b. Explain the condition for which the mass is in equilibrium and describe qualitatively the rotation it follows when it falls.
Homework Equations
Center of mass theorem [itex] \sum F = M a_{cm} [/itex]
The Attempt at a Solution
Of course the condition for the equilibrium is that the vertical line passing through the CM is inside the basis of the mass (situation a). In both the situation the weight [itex]P[/itex] equals the normal reaction force in magnitude [itex]R[/itex], but when the condition of equilibrium is not satisfied, weight [itex]P[/itex] and the normal reaction force [itex]R[/itex] becomes a force couple and they exert a torque, which makes the block rotate. The problem is: about which point does it rotate? On my book it says that the rotation is about point [itex]B [/itex] but I don't see how can this be possible, since, as said before, [itex]P=R[/itex], so [itex]\sum F=0=M a_{cm} [/itex]. The center of mass does not accelerate, while it should do if the block would rotate about [itex]B [/itex] (centripetal acceleration would be needed in that case). How can that be?