Calculate Crumple Zone Rigidty?

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The discussion centers on calculating the rigidity of a vehicle's crumple zone using kinetic energy and force equations. The user calculates the initial kinetic energy from a collision and applies the equation FcosxΔd to determine the force, questioning the angle used in the calculation. They conclude that the angle is 180 degrees, leading to a calculated force of 691 kN, which they equate to crumple zone rigidity. There is uncertainty about whether this theoretical calculation aligns with practical measurements, suggesting that real-world rigidity might be assessed using a hydraulic ram. The conversation highlights both theoretical and practical approaches to understanding crumple zone rigidity.
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Homework Statement
Given the change in kinetic energy and distance, find the constant force needed in the crumple zone to stop the vehicle.
Relevant Equations
Work Energy Theorem: W = ΔEk
W = FcosxΔd
The vehicle comes to a stop after the collision so my kinetic energy is equal to (2000kg)(22m/s)2 / 2.
(Ekfinal -Ekinital)
I used the equation FcosxΔd = ΔEk. Knowing that the kinetic energy is -484000J and the length of the crumple zone is 0.70m, I can substitute those values into the equation.
Fcosx(0.70) = -484000.
My question is what would be the angle in the equation of W = FcosxΔd. During a collision the force from a wall acts in the opposite direction from the force of the crumple zone? So the angle is cos180 degrees = -1. Is this right?
So the constant force would then be F = -484000/-0.70 = 691Kn?
That means this number is also the crumple zone rigidity.
 
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All looks good.
 
Is that how crumple zone rigidity is calculated??
 
AJ22 said:
Is that how crumple zone rigidity is calculated??
In practice? I don’t know. It might be measured simply by squeezing it with a hydraulic ram.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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