Here Ax , Ay, and N are reactive forces on the rigid bodies. They will do no work during the virtual displacement considered.

P and Q are the active forces which will do virtual works. Internal forces will also do no work like reactive forces. Internal forces always are having action and reactions, hence the total work by action will cancel the total work done by reactions.

**SOLVED EXAMPLES**

**Example 9.1**. Find out the value of e for which the bar can be in equilibrium.

Given, OB = b, OA =a.

** Solution.**

For force θ

X_{θ} = b cos θ

dx_{ θ} = – b sin θ d θ

dx_{θ} = Q dx_{θ} = -Q b sin θ dθ

for force p

y_{p} = a sin θ

dy_{p} = a cos θ d θ

du _{p} = pdy p = ap cos θ d θ

du = du _{θ} + du_{p}

0 = -Q b sin θ d θ + ap cos θ d θ

ap / bQ = tan θ

θ = tan ^{-1} (ap / bQ)