Rod ABC is attached to blocks A and B that can move freely in the guides shown. The constant of the spring attached at A is k = 3 kN/m, and the spring is un-stretched when the rod is vertical. For the loading shown, determine the value of Θ corresponding to equilibrium.

Rod ABC is attached to blocks A and B that can move freely in the guides shown. The constant of the spring attached at A is k =3 kN/m, and the spring is un-stretched when the rod is vertical. For the loading shown, determine the value of Θ corresponding to equilibrium.

Given,
\( K = 3 \, \text{kN/m} \),
\( X_c = 0.4 \sin \theta \), \( Y_A = 0.2 \cos \theta \),
\( \delta X_c = 0.4 \cos \theta \delta \theta \), \( \delta Y_A = -0.2 \sin \theta \delta \theta \)

For the spring, initially at the vertical position, it is unstretched, so the change in mean position is \( \Delta x = 0.2 - 0.2 \cos \theta \). Then,

\( F_s = K \Delta x = 3000 \times (0.2 - 0.2 \cos \theta) = 600(1 - \cos \theta) \)

From the principle of virtual work,

\( \delta U = 0 \)

Or, \( 150 \times \delta X_c + F_s \times \delta Y_A = 0 \)

(Since the spring is compressed in the virtual displacement, force is exerted downward relative to the initial mean position, so force and displacement are both directed downward.)

Or, \( 150 \times 0.4 \cos \theta \delta \theta + 600(1 - \cos \theta) \times (-0.2 \sin \theta \delta \theta) = 0 \)

Or, \( 60 \cos \theta - 120 \sin \theta (1 - \cos \theta) = 0 \)

Or, \( \frac{1}{2} = \tan \theta (1 - \cos \theta) \)

Solving on calculator, we get:

∴ \( \theta = 52.22^\circ \)