Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

enter image description here

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.