Thermodynamic generalized force and thermodynamic potential I have stumbled across these and have taken some interest. Are the meanings of generalized "force" and "potential" the analogous to the case of mechanics where the derivative of one with respect to a variable gives the other. In this case we are effectively considering statistical systems. What does potential and force mean here?
 A: Here I will focus on the notion of thermodynamic potential and how it relates to generalised thermodynamic forces.
From far away there is no obvious analogy between the rules and objects of mechanics and those of thermodynamics: while a model in mechanics relies on a set of forces combined with three laws of motion to determine the kinematics of a system, thermodynamics usually functions by applying two laws to a model system (often characterised by an equation of state relating the intensive variables of the system) so as to figure out which final state can or will be reached from a given initial state and external constraints. 
Nevertheless the two views can be connected to some extent by introducing the concept of thermodynamic potential.
I do not wish to be too general and will simply give a standard derivation in the case where a closed thermodynamic system undergoes some process at constant overall volume $V$ and temperature $T$. 
Let us characterise the progression of the process by the variable $x$; we want to know which values of $x$ can happen spontaneously starting from an initial value $x_i$.
The second law of thermodynamics tells us that $\Delta S_{univ}(x;x_i) \equiv S_{univ}(x)-S_{univ}(x_i) \geq 0 $.
As usual, we split the variation of entropy of the universe in two parts: one being the variation of entropy of the system $\Delta S_{sys}(x;x_i)$ and one being the variation of entropy of the outside of the system (which is just the "complementary" of the system) that I will denote here $\Delta S_{\overline{sys}}(x;x_i)$. This gives:
\begin{equation}
\Delta S_{sys}(x;x_i) + \Delta S_{\overline{sys}}(x;x_i) \geq 0
\end{equation}Now, if the temperature is constant, then the surrounding of the system can be modelled as a thermostat keeping the same temperature throughout. Its corresponding entropy variation is directly:
\begin{equation}
\Delta S_{\overline{sys}}(x;x_i) = -\frac{Q_{sys}(x;x_i)}{T}
\end{equation}Now, from the first law of thermodynamics we have that $\Delta U_{sys}(x;x_i) = W_{sys}(x;x_i)+ Q_{sys}(x;x_i)$, where $W_{sys}(x;x_i)$ refers to the work received by the system as a whole. Since the system has constant total volume, the work received is zero.
This lives us with $\Delta U_{sys}(x;x_i) = Q_{sys}(x;x_i)$ which, upon substitution in the inequality gives:
\begin{equation}
\Delta S_{sys}(x;x_i) -\frac{\Delta U_{sys}(x;x_i)}{T} \geq 0
\end{equation}This can be reshuffled a bit to give:
\begin{equation}
\Delta A_{sys}(x;x_i) \leq 0
\end{equation}where one would recognise right away the Helmholtz free energy $A_{sys}(x) = U_{sys}(x)-TS_{sys}(x)$.
What does this equation say? It says that starting from an initial value $x_i$, the system will be forced to go to values $x$ ensuring that the Helmholtz free energy is decreasing. This is not without similarity with potential energy in mechanics whose associated force drives a mechanical system towards low potential energy values. Thus, everything happens as if there was a thermodynamic force acting on the variable $x$ to drive it towards some values as dictated by the laws of thermodynamics.
The corresponding "force" appears in fact in out-of-equilibrium calculations looking at how does the variable $x$ evolve with time (these models are often of a Langevin type) and reads:
\begin{equation}
f_{V,T}(x) \equiv  -\left(\frac{\partial A_{sys}(x)}{\partial x} \right)_{V,T}
\end{equation}
Incidentally, it is also of great value for equilibrium calculations, for when the thermodynamic force is zero, then the thermodynamic system has reached at least a local stable thermodynamic equilibrium.
We can see its use in the following example:

As you see the overall system is quite complicated with a gas that can compress a spring with the overall volume and temperature fixed. In these conditions, we have derived previously that the right thermodynamic potential (i.e. the function of state of the system itself whose decrease guarantees automatically that the second law of thermodynamics is satisfied). 
The energy of system is simply $U_{sys}(x) = C_v T + \frac{1}{2}k x^2$ and, neglecting the entropy of the spring, the entropy of the system can be shown to be simply $S_{sys}(x)= \frac{P_i V_i}{T} \ln[1+ax/V_i]$ giving:
\begin{equation}
C_v T + \frac{1}{2}k x^2 - P_i V_i \ln \left(1 + \frac{a x}{V_i} \right)
\end{equation}
Plotting this function as a function of $x$ gives:

We see that starting from zero, the system is driven towards larger positive values of $x$ as it goes "downhill" along the free energy landscape. Eventually, it well settle down at the bottom of the free energy curve where no more net force is acting on it.
A: In mechanics, $E = \int F.dr$. That is, energy is given by a force integrated over a displacement. You can find analogous quantities in thermodynamics. You find a generalized "force" that with a generalized "displacement" can be used to find an energy. The analogy runs deeper in that if the "force" becomes unbalanced, a "displacement" will occur and the resulting transfer of energy can be found using $\int F.dr$.
If we look at the usual equation for internal energy, U:
$dU = TdS + PdV$
the forces and displacements are apparent. If pressure is unbalanced, a volume change will occur and the energy transfer is given by $\int P dV$. This energy is the mechanical work done by the system. We say that pressure is the "driving force" and volume is the "displacement"
It should be noted that the "force" is always an intensive quantity and the "discplacement" is always extensive and the pair of quantities are called conjugate variables.
