# Why can we not set each applied force equal to zero?

With reference to page 17 of "Classical Mechanics" by Goldstein, Safko and Poole, the small paragraph after eq. 1.43, $$\sum_i \mathbf{F}^{(a)}_i \cdot \delta \mathbf{r}_i ~=~ 0.\tag{1.43}$$ I do not understand why we cannot set each applied force equal to zero if the $\delta \mathbf{r}_i$ are not independent but related by the constraint. Now, I figure that setting the coefficients (the forces) equal to zero is motivated by the consideration of a simpler equivalent scenario in which the work vanishes because the forces vanish one by one. However I cannot see how, in order to require the above, we need to change to generalised (and consequently independent) coordinates first.

In particular, Goldstein considers the case of system of particles indexed by $i$ such that the total force $\mathbf{F}_i$ on each particle vanishes. Then we have that the virtual work due to a change of coordinates $\delta \mathbf{r}_i$ is

$$\sum_i \mathbf{F}_i \cdot \delta \mathbf{r}_i ~=~ 0. \tag{1.40}$$

Then he separates the force on each particle into "applied" and "constraint" forces, $$\mathbf{F}_i ~=~ \mathbf{F}_i^{(a)} + \mathbf{f}_i\tag{1.41}$$ and restricts himself to systems for which the net virtual work of the forces of constraint is 0. Why can we not set each applied force equal to zero if the $\delta \mathbf{r}_i$ are not independent but related by the constraint?

• Not all of us have access to textbooks. Can you provide more context? Jul 26, 2017 at 16:28
• The paragraph derives Langrange equation of motion from D'alambert principle. We consider a virtual displacement of coordinates (so at an instant of time t). en.wikipedia.org/wiki/D%27Alembert%27s_principle. The displacements in question are the differentials in the formula as seen in the first equation in the page linked above. Jul 26, 2017 at 16:53
• Sorry I believe the paragraph was not very clear in conveying the message. In fact we are not setting the coefficients to be zero. It is a question of linear independence, in which case the coefficients are REQUIRED to be 0. Apologies again. Jul 26, 2017 at 16:57

In many physics problems of interest, the applied force $\mathbf{F}_i^{(a)}$ of the $i$'th point particle is not zero.
But your focus should not be on the applied forces $\mathbf{F}_i^{(a)}$. They are there and properly accounted for. Your focus should instead be on the other forces $\mathbf{f}_i$.
The non-trivial statement in eq. (1.43) (apart from Newton's 2nd law for statics $\mathbf{F}_i=0$) is that the other forces produce no virtual work, $$\sum_i \mathbf{f}_i \cdot \delta \mathbf{r}_i ~=~ 0,$$ which e.g. is not true if the other forces include sliding friction.
I think that your doubt can be overcome by reading the equation $(1.43)$ as a linear combination of vectors which are -since they are related by constraints- linear dependent. This fact means that they're not independent, i.e. you can't conclude that all the coefficients (in that case the applied forces) are equal to zero.