# Hamilton's Principle and the Principle of Virtual Work

1. Dec 16, 2013

### SheikYerbouti

I understand that accepting Hamilton's principle will yield identical results as accepting Newton's laws. However, simply accepting that the integral of the difference between kinetic and potential energies is an extrema seems not intuitively obvious. The textbook that I used for my classical mechanics class (Fowles) states that Lagrange developed his mechanics through the use of the principle of virtual work. Elsewhere, I have read that the acceptance of the principle of virtual work is equivalent to the acceptance of Newton's first law. Since the whole concept of virtual displacements is somewhat counter-intuitive (at least initially), I am trying to understand its derivation from the Newton's first law. From the definition of virtual work we have:
$$\delta W = \sum_{i=1}^m \vec{F_i} \centerdot \delta \vec{r_i}$$
Where $\vec{r_i}$ is a function of n generalized coordinates and time. Since $\delta t = 0$, it follows from the properties of virtual displacement that $$\delta \vec{r_i} = \sum_{j=1}^n \frac{\partial \vec{r_i}}{\partial q_j} \delta q_j$$
After substituting this into the expression for virtual work and doing some rearrangement we find that
$$\delta W = \sum_{j=1}^n (\sum_{i=1}^m \vec{F_i} \centerdot \frac{\partial \vec{r_i}}{\partial q_j}) \delta q_j$$
We define the term in parentheses to be the generalized force $Q_j$. The principle states that the virtual work is zero for a static system only when the generalized forces are all zero, and this readily clear here. However, I do not see how this follows from Newton's first law, which was only initially applied the forces $\vec{F_i}$. Since the virtual displacements in the first sum are those caused only by the corresponding force and are completely arbitrary, I don't see how this relationship between the forces must carry on to the generalized forces, which are essentially a sum of dot products of arbitrary vectors. I feel like this is where my mistake is; the virtual differential operator is not present in the partial derivative of the position with respect to a generalized coordinate. Does this mean that $\frac{\partial \vec{r_1}}{\partial q_j} = ... = \frac{\partial \vec{r_m}}{\partial q_j}?$ Sorry for the lengthy post, I would greatly appreciate some clarification on this topic and/ or how to derive the principle of virtual work and Hamilton's variational principle from Newton's laws. (Simply showing that they yield equivalent results does not give me the deep, theoretical understanding that I would like to have.)

2. Dec 16, 2013

### SheikYerbouti

That realization I had about the partial derivative was the key to solving the problem. Although I haven't figured out how to derive Hamilton's principle from the principle of virtual work, I figured that I'd post the solution in case anyone else happens to encounter my difficulties. The virtual displacements caused by each force are completely arbitrary; we cannot assume any relationship between them. However, the point upon which the forces act is not arbitrary. For most cases, the locations that the forces act upon are identical, but when they are not, the partial derivatives of the interaction locations with respect to a particular generalized coordinate are equal for each force (the locations differ by a constant translation term, at least for rigid bodies). We can thus drop the index on the position vector in the derivative and slide it out of the sum. Thus the generalized force can be re-written:
$$Q_j = (\sum_{i=1}^m \vec{F_i}) \centerdot \frac{\partial \vec{r}}{\partial q_j}$$
For a system in static equilibrium, the sum of the vector forces was initially assumed to be zero. It follows directly that each generalized force must also be zero. This implies that the virtual work on a system in static equilibrium is zero.