Final Resting Place of a broken wood stick loaded with potential energy

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When a 102-foot, 2500-pound semi-rigid wood stick is loaded with potential energy and breaks two feet above a vice, the distribution of its landing will depend on several factors, including the stick's deflection and the elongation of the rope used to pull it. The stick bows more at the top, creating a significant deflection that influences the forces at play. Given the rope's 6% elongation and the nature of the stick's break, it is theorized that a portion of the stick will land on the side opposite the pull, despite the primary motion being directed forward. The application of Newton's equations and Hooke's Law can help determine the trajectory and force distribution upon breaking. Ultimately, it is likely that some of the stick will land on both sides of the vice, but the majority will fall towards the side being pulled.
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If a 102 foot 2500 lbs. semi rigid straight wood stick (aka bow) is placed in a vice (a really big one with no give) and loaded with energy by pulling at the top, 50 feet above the vice at a 60 degree angle to the ground with a rope ("0" lbs.), where would the stick land if the stick instanteously broke in half two feet above the vice?

I am intersted to know if the pressure applied in the pulling direction will produce motion in the opposite and how much (e.g. how many feet of the stick will land on each side of the vice or will it all land on one side and by how far?)
 
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When you say "semi-rigid" can you quantify that at all? It seems the answer would depend on both the degree to which the beam was deflected and the "stretching" of the chain or cord or whatever you're using. If the beam is perfectly rigid and the chain doesn't stretch at all then all forces are relieved the instant the beam fractures and so it just topples over.
 
Your answer, in my words says all of the stick will be on one side. I follow that.

Answer to your questions

1. Rope has 6% elongation
2. The stick bows more at the top then bottom (like a tree) - Let's say it bows 10 feet from vertical center, measured at the 102 foot mark.

Now, will any of the stick land on the back side of the vice unlike the situation you described. If yes, what %?
 
I can't help you with the details but the deflection of the beam will tell you what force was applied to it. In turn, that force will give you the effective "spring constant" of the cord. Since you know how much the cord was stretched you can determine the force being applied to the "free" beam (Hooke's Law) while the cord remains stretched. You should be able to apply Newton's equations of motion to find the trajectory. I think it's likely the beam will land on the cord side since it's being pulled in that direction with no torque!
 
Since the process creates a spring and momentum is always conserved, a portion of the stick must fall behind the vice opposite the side it was being pulled on. Agree?

I don't need to know or do the equations, it has been a while, just theoretically will the stick land on both sides of the vice?
 
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