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2

A state is thermodynamically stable when its Gibbs free energy is at a minimum. $$G=U-TS+PV$$ Holding all variables but $P$ and $V$ fixed, it means that: $$dG=dP V+PdV=0 \implies {dP\over dV}=-{P\over V}$$ Since neither of $P$ nor $V$ could be negative ${dP\over dV}$ must be negative.

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From a purely thermodynamical point of view, why does that entropy have to be a maximum at equilibrium? Say there is equilibrium, i.e. no net heat flow, why can the entropy not be sitting at a non-maximal value? The statement that entropy has maximum possible value in equilibrium means this: given imposed constraints (volume, total energy, molar ...

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Look at a free body diagram. With red and the contact normal forces, with pink the friction forces, and with gray the gravity forces. If in the end any of the friction forces come up to being negative, then flip the orientation.

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The definitions are equal: Sum of external forces zero, sum of external torques is zero. This comes from classical mechanics. For a perfect ideal fluid, the external force density is the pressure gradient: $\mathbf f = -\nabla p$, and therefore, uniform pressure in a fluid means no external force on it, and then it is in mechanical equilibrium. So, its more ...

5

Okay, the system is in equilibrium so all the forces must be balanced. First consider the weights We have $W$ downwards from the centre of each of the 3 balls (assuming their centres of mass are at their geometric centres). We also have $N_A = N_B = \frac{3}{2}W$ upwards from the points of contact between the plane and balls $A$ and $B$ respectively. ...

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