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I am currently really stuck on this problem and I am not sure how to actually solve it, I think I can reason it out logically but I do not know how to show it. Here it is and under I will show what I know: enter image description here

So this is where I am:

I can assume that the neutral state occurs at θ=0 because the spring is completely unstretched and the force of gravity would cancel out with the normal force of the blocks, leaving the total work=0 and then from there we can say that $\frac{dv}{dθ}=0$ so $\frac{d^2v}{dθ^2}=0$ meaning this point is neither a stable or unstable point making it the neutral point (process of elimination)

From then I can only guess that the two other equilibrium points occur when A and B are completely horizontal (only potential from gravity? θ=90?) and when A and B are in the state in the picture (Potential gravity=Potential spring?)

For Equilibrium $\frac{dv}{dθ}=0$ (Which needs to be satisfied for all three cases)

For stable equilibrium $\frac{d^2v}{dθ^2}>0$

For unstable equilibrium $\frac{d^2v}{dθ^2}<0$

For neutral state, all I can find in my lecture notes is "higher order derivative must be examined" so I assume that if $\frac{d^2v}{dθ^2}=0$ I will call it neutral for this problem because it does not satisfy the stable or unstable equilibrium states.

Now I am not sure how to proceed but this is my intuition on this problem.

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  • $\begingroup$ Neutral state = Equilibrium? $\endgroup$ – HDE 226868 Nov 16 '14 at 22:45
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    $\begingroup$ I think neutral state means a state of equilibrium where the state of the object will always be the same after a disturbance, for eg ball on smooth horizontal floor. $\endgroup$ – Skawang Dec 12 '15 at 16:46
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When $\theta = 0$ the spring extension is zero and take this to be the zero of gravitational potential energy.

With the rod at an angle $\theta$ from the geometry of the system find the extension $x$ of the spring and the vertical height $h$ through which the centre of mass of the rod has fallen.

The potential energy of the system $V = \frac 12 k x^2 - mgh$

Differentiate $V$ with respect to $\theta$.

$\frac {dV}{d\theta} =0$ is the condition for equilibrium and solve the resulting equation for three value of $\theta$, one of which is easy to find but the other two are a little harder.

Differentiate again to get $\frac {d^2V}{d\theta^2}$ and put in your three values of $\theta$ to decide which type of equilibrium it is for each of the values of $\theta$.

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The 'easiest' way to solve this is by using energy, as suggested by the OP and Farcher. Start by summing the gravitation and spring potential energies as follows:

$$E = -mg\frac{l}{2} sin\theta + \frac 12 kx^2$$

x is given by the following relationship (the spring will always be in tension):

$$ x=lsin\theta+lcos\theta-l=l(sin\theta+cos\theta-1) $$

This gives the following equation for the total energy:

$$ E = -mg\frac{l}{2} sin\theta + \frac 12 kl^2(sin\theta+cos\theta-1)^2 $$

Differentiating this gives:

$$ \frac{dE}{d\theta} = -mg\frac{l}{2} cos\theta + kl^2(cos^2\theta-sin^2\theta-cos\theta+sin\theta) $$

(I arrived at this same equation by balancing the forces, so this seems to be correct.) Where this equates to zero will give the equilibrium points. Using the following relations:

$sin^2\theta=1-cos^2\theta $

$sin\theta=\sqrt{1-cos^2\theta} $

the above can be re-written as:

$$ 0 = -mg\frac{l}{2} cos\theta + kl^2(2cos^2\theta+\sqrt{1-cos^2\theta}-cos\theta-1) $$

This can be solved to find $cos\theta$, but it is a bit tricky. If you multiply everything out, $cos\theta=0$ drops out as a solution, which is expected ($\theta=\frac\pi2$). However, this then leaves a cubic equation to be solved for $cos\theta$, which is a bit unpleasant (it can be solved by hand, but it doesn't seem like the answer is going to drop out nicely).

It wasn't my idea of a fun night in, so I cheated and plotted out the energy equation vs $\theta$ using a graphing tool:

System Energy vs $\theta$

This looks plausible, since it passes through zero energy at $\theta=0$ and has the expected equilibrium point at $\theta=\frac \pi2$. The other two equilibrium points are at $\theta=0.169$ radians (9.68 degrees), which is stable, and $\theta=0.591$ radians (33.9 degrees), which is unstable.

So, at 9.68 degrees, there is a stable equilibrium point, but if you push the beam past 33.9 degrees (unstable), then it will drop to the next stable equilibrium at $\theta=\frac\pi2$.

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My guess:
$\theta = 0$ might not be equilibrium cause if you stop holding the A box, it will fall.
at $\theta = 90$ might be neutral equilibium (I assume the spring can pass through the wheel to the side of box A), since there is no tension in the spring, and the normal force on the B box balances the forces of gravity(assuming the B box is restricted to the horizontal plane only and will not fall).

Now if we assume the spring can pass through the wheel to the side of box A, (another way to look at it is that the whole string and spring assembly is just all spring), Then the tension (tension due to the stretched spring) at side A (upward) will be equal to the tension in side B (toward the left). Now with that in mind, $T = F = k(x-x_0)$ where x is the total length of the string-spring assembly, and $x_0$ is the unstretched distance l. The total length can be found by adding length of side A and length of side B (which is just the sides of a right triangle with hypotenuse l).

That might be a helpful clue.
So if you agree with my idea, it might be that when A falls, the spring is not compressed but instead stretched.

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  • $\begingroup$ This is not just a guess but an educated guess, so take it into consideration. You can actually find 3 equilibrium states here. one when $\theta = 90$ and the other two when $0<\theta <90$. Hint: the stable equilibrium will be at lesser $\theta$ than the unstable equilibrium. Once at the unstable equilibrium, the configuration can snap toward the neutral or stable equilibrium. $\endgroup$ – philip_0008 May 28 '16 at 14:33
  • $\begingroup$ Qualitatively, this is because in the unstable equilibrium, the string-spring is more stretched hence greater tension, and disturbing the system will shift it towards lesser tension. $\endgroup$ – philip_0008 May 28 '16 at 15:08

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