2 Rods, 1 Spring

Simple trigonometry gives the length of the spring as $3-2\cos{\theta}$ , and so the extension is $2-2\cos{\theta}$ , and the tension in the spring is $100(1-\cos{\theta})$ .

Now consider (say) the left hand rod. Since it is in equilibrium the total turning moment about the hinge must be zero. The only relevant forces are the weight of the rod, acting vertically through the centre of gravity, and the tension force in the spring which acts horizontally. Simple trigonometry gives the length of the lever arm for each of these forces, and we find

$\frac{g}{2}\cos{\theta}=100(1-\cos{\theta})\sin{\theta}$

Substituting in $g=10$ and rearranging gives

$\theta=\cos^{-1}\left(1-\frac{1}{20\tan\theta}\right)$

I solved this quickly by starting at $\theta=50$ and iterating this formula. it quickly converges to

$\boxed{26.107}$

note.

I tried three other rearrangements of the formula first, which did not converge. However this numerical approach is so easy that even with these false starts it is (at least for me!) much faster than finding an exact solution using trig and algebra!

With a pocket calculator just press "50=" then enter $\cos^{-1}\left(1-\frac{1}{20\tan{ANS}}\right)$ and keep tapping "=" until the numbers converge!

@Steven Chase sir, I did something and got the same answer. After solving, i got a equation like $\cot \theta = 20(1-\cos \theta)$ . Am I correct? Did it also came for you?

The hinge can supply reaction forces to satisfy the force equilibrium requirements. However, the hinge cannot supply any moments. Thus, the gravity and spring moments must cancel each other:

Spring Length

$l = 3 - 2 cos\theta$

Spring Stretch:

$s = l - 1 = 2 - 2 cos\theta$

Spring Force:

$F_s = k s = 100(1 - cos\theta)$

Spring Moment:

$M_s = F_s * sin\theta = 100sin\theta (1 - cos\theta)$

Gravity Force:

$F_g = 1 \times g = 10$

Gravity Moment:

$M_g = F_g \frac{1}{2} cos\theta = 5 cos\theta$

Equating the two:

$100sin\theta (1 - cos\theta) = 5 cos\theta \\ cot\theta = 20(1 - cos\theta)$

Solving this equation (by whatever means) for the angle yields an answer of approximately 26.107 degrees

Equilibrium is achieved when the horizontal force on the bar (F = k•d) is equal in strength to the vertical force on the bar (F = m•g = 1•10 = 10 N). Therefore, since d = F/k = 10/50 = 0.2 m (and because of the symmetry of the scenario) the ends of the spring stretch 0.1 m horizontally toward each of the hinges. This results in the 1 m bars projecting a length of 0.9 m onto the line defined by the two hinges. Therefore, "theta" = arccos(0.9/1) = 25.842°.