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Virtual Work and Static Equilibrium
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[QUOTE="TSny, post: 6642214, member: 229090"] In applying the principle of virtual work, you don't need to think about the tension forces. All you need is the idea that for any infinitesimal virtual displacement of the system, the total work done by the external gravitational forces is zero. Or, as Feynman expresses it, "the sum of the heights times the weights doesn't change". You don't need Newton's laws. I don't know how you arrived at your equation (1). That is, I don't know if you got it from the principle of virtual work. I think you took the virtual displacement of the system to be a downward displacement of the 1 kg block by a distance dh[SUB]1[/SUB] with consequent horizontal and vertical displacements dx[SUB]A[/SUB] and dh[SUB]A[/SUB]of block A and a vertical displacement dh[SUB]B[/SUB] of block B. Now there is a peculiarity here. If all I know is that the 1 kg block moves downward by dh[SUB]1[/SUB], there is no way to determine the other displacements dx[SUB]A[/SUB], dh[SUB]A[/SUB], and dh[SUB]B[/SUB] without making additional assumptions. To see this, note that I can make the 1 kg block move downward by displacing the knot only horizontally to the left by dx[SUB]A[/SUB] so that dh[SUB]A[/SUB] = 0. Or, I can make the 1 kg block move downward by displacing the knot only vertically by dh[SUB]A[/SUB] so that dx[SUB]A[/SUB] = 0. So, dx[SUB]A[/SUB] and dh[SUB]A[/SUB] are not determined by just knowing that the 1 kg block moves downward. But this is actually to our advantage. This means that you can take the virtual displacement of the system to be the result of just a horizontal displacement of the knot by dx[SUB]knot[/SUB] with no vertical displacement of the knot. Now you should be able to work out the displacements of the three blocks in terms of dx[SUB]knot[/SUB] and set up the principle that the total gravitational work is zero. This will give you one equation to work with. Next, you can take the virtual displacement of the system to correspond to a vertical displacement of the knot by dh[SUB]knot[/SUB] with no horizontal displacement of the knot. The principle of virtual work will give you a second equation to work with. The horizontal and vertical displacements of the knot are two independent virtual displacements. You can imagine other possibilities. For example, you could have a virtual displacement of the system where the knot moves in such a direction that block B doesn't move at all. Or, you could have a virtual displacement such that block A doesn't move. [/QUOTE]
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Virtual Work and Static Equilibrium
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