Hanging accordian balanced by opposing moments

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The discussion revolves around a hanging accordion system with five beams and six axles, focusing on balancing forces and moments. The system is symmetric, allowing for an equal distribution of forces at the ends of the beams. The tension in certain beams leads to directional forces, which must be countered for static equilibrium. When the lower axles are pinned, moments generated by the forces must also be balanced, leading to a need for a deeper understanding of how moments translate into forces. The solution emphasizes the law of conservation of momentum, stating that the sum of all torques must equal zero for equilibrium, illustrating the balance between opposing torques in the system.
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"hanging accordian" balanced by opposing moments

This is not a coursework problem, but it looks like one so I'm placing
it here. If this is in error, do correct me. I'm a newbie :-)

Assume an arrangement as below with a fixed surface, five ideal,
weightless beams B_1 through B_5 joined by six axles A_1 through A_6:

physics.png


All the vertexes are axles free to rotate in the plane of the diagram
as above. The system is symmetric, so Fn can be split equally between
the ends of B5 and act directly on axles A_5 and A_6. The downward
force results in tension in B3 and B4 and compression in B6:

physics2.png


with |F_B4| = |F_B3| = |Fn|*(PI/2-theta) and the compression force in
B6 opposing the x components of the tensions.

The tension in B3 and B4 is transformed into tension forces in B1 and
B2 plus directional forces F_x1 and F_x2:

physics4.png


...and F_x1 = 2*Fn*tan(theta), F_x2 = -F_x1. If F_x1 and F_x2 are
opposed the system is static as the tensions are opposed by the fixed
surface at top. I hope this is correct so far.

However, I'm interested in the case where the lower two axles A_5 and
A_6 are pinned (or exactly countersprung, eg, with torsion springs) so
that F_x1 and F_x2 are effectively opposed by moments in B_3 and B_4
about axles A_5 and A_6.

I understand how forces give rise to moments, but my understanding is
quite squishy when it comes to moments giving rise to forces.
Specifically, I know that a moment does not necessarily result in a
force perpendicular to the lever arm, but not really how to wield that
fact.

In any case, my naive solution for t1 is a torque that produces a
force normal to the B_3 lever arm with an x component matching F_x1:

physics5.png


t_1 = -|B_3|*|F_x1|/sin(theta)
with matching symmetric solution for A6.

...this torque is much higher than the torque actually exerted by
F_x1. How is this being balanced? Increased tension in B_1? I don't
know for certain what the actual superimposed direction and magnitude
of the force vector resulting from t_1 is.

I'd appreciate some guidance on the correct solution and even moreso
on the proper way to be approaching it.
 
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The correct solution for this problem is to use the law of conservation of momentum. This states that the sum of all of the torques acting on an object must equal zero for the object to remain in equilibrium. In this case, the torques from Fn, F_x1 and F_x2 must be balanced by the torques from t_1 and t_2. This means that:t_1 + t_2 = Fn * (PI/2 - theta) + F_x1 - F_x2Solving for t_1 gives:t_1 = Fn * (PI/2 - theta) + F_x1 - F_x2 - t_2This shows that the torque t_1 is being balanced by the torque t_2 acting in the opposite direction. This means that the force vectors resulting from both t_1 and t_2 will oppose each other, thus resulting in the equilibrium of the system.
 
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