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I'm trying to do the following problem:

A lever $ABC$ (see figure) has weights $W_1$ and $W_2$ at distances $a_1$ and $a_2$ from the fixed support $B$. Using the principle of virtual work, prove that a necessary and sufficient condition for equilibrium is $$W_1a_1 = W_2a_2$$ My attempt:

Now, the principle of virtual work states that $$\sum_{i=1}^{N}{\bf F}^{(a)}_i\cdot \delta{\bf r}_i = 0$$ I know the forces are equal to the given weights. However, I have a slight problem with the $\delta r_i$. I know it is a variation of the position vector $r$ of the $i$th particle. So, taking the origin as B, we have $\delta r_1=-\delta a_1$ and $\delta r_2=\delta a_2$, and my equation becomes: $${W_1}\cdot(-\delta a_1) + {W_2}\cdot \delta a_2= 0$$ Is my work up to this point correct? And is it correct to simply conclude from here that $W_1a_1=W_2a_2$? I would be more comfortable if I could get rid of the $\delta$s, but given that I'm not comfortable with working with them, I don't quite know how to do this.

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The key is in the dot product in your equation $$\sum_{i=1}^{N}{\bf F}^{(a)}_i\cdot \delta{\bf r}_i = 0.$$ That means that only displacements in the direction of the forces - in this case, vertical - count towards the equilibrium condition. The lever arms come in since the vertical displacements must be related geometrically for the bar to remain straight.

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