What is the Average Force During a Collision?

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SUMMARY

The discussion focuses on calculating the average force during a collision involving an eighteen-wheeler weighing 95,000 lbs, impacting a concrete wall at 80 mph with a deformation of 0.1 m. The average acceleration can be derived using the equations of motion, specifically the formula Vf = Vo + 2a(Xf - Xo). The average force can be estimated using the kinetic energy approach, where force multiplied by distance equals energy, leading to the conclusion that average force is calculated over distance rather than time. The discussion emphasizes the complexity of accurately determining average force due to the nature of collision dynamics.

PREREQUISITES
  • Understanding of Newton's Second Law (Sum of F = ma)
  • Familiarity with kinematic equations (Vf = Vo + 2a(Xf - Xo))
  • Knowledge of energy conservation principles in collisions
  • Ability to convert units (e.g., lbs to kg, mph to m/s)
NEXT STEPS
  • Research the concept of kinetic energy and its relation to force and distance
  • Learn about the dynamics of collisions and the force-time relationship
  • Explore advanced topics in collision mechanics, including impulse and momentum
  • Study real-world applications of crash testing and safety engineering
USEFUL FOR

Students in physics or engineering, safety engineers, and professionals involved in vehicle design and crash testing will benefit from this discussion.

FrostyMan
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Homework Statement


As part of a safety test, an eighteen wheeler is being shot down a track by a rocket into a solid concrete wall. The truck is fully loaded, for a total weight of 95,000lbs. If the truck hits the wall at 80mph and causes a deformation of .1m in the wall before coming to a halt:

a. Draw two free body diagrams, one of the truck and the other of the wall, during the crash.
b. What is the average acceleration of the truck during the collision?
c. what is the average force on the truck during the collision?

Homework Equations


SumofF=ma
Vf=Vo+2a(Xf-Xo)
Xf=Xo+volt+(1/2a)t^2

The Attempt at a Solution



I drew two free body diagrams: (_ is for formatting purposes)

______^________________________________^
______| Fn(Ground->Truck)________________|Fn(Ground-->Wall)
____Truck-->Fp(Rocket->Truck)___________Wall-->Fp(Truck-->Wall)
______|Fg(Earth->Truck)__________________|Fg(Earth-->Wall
______v________________________________v

I'm having trouble finding the average acceleration and the average force, however. I converted the lbs to kg (95,000lbs to 43091.24kg) and the mph to m/s (80mph to 35.7632m/s). I tried treating it as a 1D problem to find acceleration and wrote a table of variables but did not have enough knowns to finish the equations.

Any suggestions would be greatly appreciated.
 
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I often see this question in different guises. It is an error on the part of the poser of the question.
You are given a stopping distance and asked about a force. Since force x distance = energy, the obvious approach is to compute the KE and divide by the distance. That gives an average force in a sense, but it is an average over a distance, not an average over time. When we talk of average acceleration (and thus of average force) we mean the change in velocity divided by the time taken, so it's an average over time.
In practice, the force during a collision tends to increase more or less linearly from 0 to a maximum, then may stay at that maximum (as objects crumple) for a while, then quickly fall away to nothing. However, that makes it impossible to calculate the average over time without further information.
Sorry I can't be more helpful. I suggest you use the energy approach to get the answer expected, but please understand that it is wrong.
 

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