Printable View
A truck weighing 14000 lbs. Moving 40mph down a 15 degree slope. How much energy? Hits a car sitting still into two other cars. How far would the original car have to be away from the other car to not make contact? The car the truck hits weighs 3052 lbs. If it matters the tires have a contact patch of 8x8.
Just divide by zero.
Its an energy balace. ( Kinetic energy + Potential ) initial = ( kinetic + potential ) final + losses. the distance term is in your losses. Have to figure out how much energy to overcome static friction and then distance covered with dynamic friction.
Thats where I would start at least.
I cant remember off the top of the my head how to calculate energy to overcome static friction. But if you make an assumption that it is negligible compared to the total loss your loss term is mew*normal force*distance sliding.
Disclaimer: Please dont reference me in a court case where you totalled 3 cars.
What the deceleration force? Friction? Uphill roll?
Sent from my Galaxy Nexus using Tapatalk 2
You need to give us some more information and a pretty picture would be nice.Quote:
Originally Posted by EvanH
If it is a simple particle collision problem, this should help.
http://i601.photobucket.com/albums/t...tin/img001.jpg
Edit: Teach said to ignore Us for my example. Oh, and you mentioned energy.
Your question is too vague. You need more information.
1) What was the distance of the truck from the car when the 40 mph speed of the truck was measured? Or was 40 mph the speed of the truck when it struck the car?
2) Was the collision between the truck and the car perfectly inelastic?
3) Are we assuming a one-dimensional or multi-dimensional collision?
4) What is the friction coefficient of the car (or car-truck body if the collision is perfectly inelastic)?
A ford F-350 1ton commercial truck with an auto crane, and air compressor plus what ever tools, plowed into my GF's caliber from behind forcing her into another car, into another. Her rear trunk is now against the back seats, and the front is crumpled and a broken radiator.
So three cars, one truck. Im just trying to figure out how far away she would have had to be from the car in front of the caliber to never have contacted the other car then pushed into the other car.
What other info? I can provide, if specified, more. Its not my strong point.
Truck, 14000 lbs gross, moving 40 mph. Caliber, brand new tires, 3052 lbs sitting still breaks applied because stopped at light. Everything took place on a 15 degree downward slope, 72 degree air temp, dry, fresh paved asphalt road.
truck at 40 when striking car.
what is meant by inelastic? and the dimension?
not sure the friction coefficient, air or tire contact.
No 123Quote:
Originally Posted by EvanH
I was operating under the assumption that this was an academic question. Since it's a real-world problem, we can do a little more hand-waving.
Perfectly inelastic means that the truck stuck with the car throughout the collision, ie the car didn't bounce off the truck.
Let's assume one-dimensional since it was likely that the car was rear-ended, and we will assume that the collision was square enough with the rear to cancel out effects from the two other dimensions.
Was the car also at a 15 degree decline?
Momentum of the truck is easy. p = mv = 14000 lb x 40 mi/hr = 560000 lb-mi/hr
Momentum must be conserved, and hence transfered to the car assuming an elastic collision. But we still need the speed of the truck after the collision to find the velocity of the car after impact, since:
mass of truck * velocity of truck (before collision) = mass of truck * velocity of truck + mass of car * velocity of car (after collision)
Once finding the velocity of the car, you would need to assume a reasonable number for the friction coefficient of the car with brakes applied. Then you would have to use the energy equation to determine the distance the car would travel with these assumptions, taking into account the 15 degree decline by factoring in potential energy.
In other words, this isn't as easy as you think.
Found an estimate COR of .15 and COF 0.9 static, 0.7 kinetic if you feel like doing the math.Quote:
Originally Posted by BigTyla
I knew this was going to be the case. There are crazy complex simulation codes to do what you want us to do on an internet forum for free.Quote:
Originally Posted by vroom4
I laughed when I read it was IRL. Think of some poor soul trying to account for all the shit in the back of a fully loaded utility truck during an impact :lo5l:
I think its safe to say your gf would've needed 4-5 car lengths between her and the next vehicle.
I figured it would have to be something ridiculous like that. Something no one would do unless they knew it was going to happen. I am just doing this for kind of my own knowledge gain, and semi help with math. Thank you for the help.
There are so many variables to this problem in the real world it'd take forever to figure out. You have to know how the car crumples to find the energy absorbed from that, the friction forces, even the amount of energy absorbed by the springs in every vehicle as they get hit.
My dynamics professor was asked to do something very similar to this to prove the client couldn't have gotten whiplash from a rear ending and he declined it.
lol at this whole thread
force = M*A
that's the first thing you have to figure out. 15 degree slope doesn't matter if the speed is constant at 40mph. its just a function of the speed and the mass behind it. everything else will follow that.