Democratic Underground Latest Greatest Lobby Journals Search Options Help Login
Google

a science question

Printer-friendly format Printer-friendly format
Printer-friendly format Email this thread to a friend
Printer-friendly format Bookmark this thread
This topic is archived.
Home » Discuss » The DU Lounge Donate to DU
 
mmm Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 09:41 PM
Original message
a science question
if a 120 pound animal hits me at 60 miles an hour,
how much force is applied on impact
Printer Friendly | Permalink |  | Top
ProudGerman Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 09:44 PM
Response to Original message
1. What is the size of the impact zone?
How much "cushioning" is on the impact zone and the point of the animal that is hitting you?







Like I'd be able to answer even with this information.
Printer Friendly | Permalink |  | Top
 
Zorba607 Donating Member (157 posts) Send PM | Profile | Ignore Mon Nov-24-03 09:46 PM
Response to Original message
2. f=ma
force equals mass times acceleration.
change pounds to kilos and mph to meters per second and force is in newtons.
I think in order to determine real acceleration youd have to know if it was an inelastic or elastic collision.
Printer Friendly | Permalink |  | Top
 
mmm Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 09:52 PM
Response to Reply #2
3. you are being silly
this must be simple physics
Printer Friendly | Permalink |  | Top
 
HereSince1628 Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 09:57 PM
Response to Reply #3
5. What's not simple about that?
Printer Friendly | Permalink |  | Top
 
Kitsune Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 09:57 PM
Response to Reply #3
6. No such thing
as simple physics.

:hurts: <-- me in physics class.
Printer Friendly | Permalink |  | Top
 
WannaJumpMyScooter Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:03 PM
Response to Reply #3
11. it is simple
Edited on Mon Nov-24-03 10:38 PM by WannaJumpMyScooter
120lbs = 264kg

120mph = 132kph = 36.6m/s

m=fa
m=36.6 x 264

9662.4 would be the answer
Printer Friendly | Permalink |  | Top
 
Catshrink Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:06 PM
Response to Reply #11
13. Where are your units????
zero points for not naming your units. (But partial credit for showing your work.)

Newtons!
Printer Friendly | Permalink |  | Top
 
WannaJumpMyScooter Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:09 PM
Response to Reply #13
16. Juice Newtons? I always loved her.
Fig Newtons?
Printer Friendly | Permalink |  | Top
 
forgethell Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:12 PM
Response to Reply #11
17. except that you don't
have the accleration, you have the velocity. Units of acceleration are units of length divided by seconds squared.
Printer Friendly | Permalink |  | Top
 
WannaJumpMyScooter Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:15 PM
Response to Reply #17
19. oops. I missed that... that does make it a bit sticky
but, can we assume the car is a constant 132kph?

I assume we are talking about the car moving, not the deer, right?
Printer Friendly | Permalink |  | Top
 
WannaJumpMyScooter Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 09:55 PM
Response to Original message
4. what 120lb animal moves at 60mph?
damn, that is fast. Cheetah? Ostrich?
Printer Friendly | Permalink |  | Top
 
BiggJawn Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 09:58 PM
Response to Reply #4
8. A Deer standing in the road in front of your speeding Subaru...
Edited on Mon Nov-24-03 09:59 PM by BiggJawn
Do I get a prize? and...I'd say that the force of impact would be about $4500....More if it got into the windshield.
Printer Friendly | Permalink |  | Top
 
woofless Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:04 PM
Response to Reply #8
12. I hit a deer at 60
on my Moto Guzzi and the total was over $60,000. Thank God the deer had insurance.

Woof
Printer Friendly | Permalink |  | Top
 
mmm Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:07 PM
Response to Reply #8
15. thank you Big Jawn
This is simple physics, I shall find my own answer.
Gheesh...
Printer Friendly | Permalink |  | Top
 
HereSince1628 Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:01 PM
Response to Reply #4
10. Golden retriever jumping from oncoming pickup truck
:shrug:

I assumed that was a syntax problem, but then this alternative came to mind.

Come to think of it, it could also be a teenager or adult jumping from the back of an oncoming pickup truck.
Printer Friendly | Permalink |  | Top
 
NaMeaHou Donating Member (802 posts) Send PM | Profile | Ignore Mon Nov-24-03 09:58 PM
Response to Original message
7. what size is "me?"
critical
Printer Friendly | Permalink |  | Top
 
kixot Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:00 PM
Response to Original message
9. Ok, let me ask you a few questions first.
Is the animal's acceleration changing or constant?

If changing by what rate?

What units do you want the answer in?
(*Valid units only, please, don't come back with you want it in "liters")
Printer Friendly | Permalink |  | Top
 
GreenCommie Donating Member (320 posts) Send PM | Profile | Ignore Mon Nov-24-03 10:07 PM
Response to Original message
14. From an engineer...
Assuming that the vehicle does not deccelerate, acceleration is constant with respect to time, the velocity vectors are colinear, the velocity of the animal is zero in the direction of the car's velocity, no friction forces, no damping forces, inellastic collision, and a bunch of other things....

the change in velocity equals the intergral of the acceleration with respect to time from time 1 to time 2. The only unknown is the acceleration constant (a). Pick an amount of time that this occurs over (say 0.4 seconds) and solve for a. (a will grow larger as time grows shorter.) Then Force = mass times acceleration (F=ma). The units are either in pounds or Newtons. The calculations are done on the animal, but Newtons third law (action, reaction) allows us to use it to get the force on the car too.

So, for a 120 lb (wieght, not mass) animal at 60 mph in 0.4 seconds, it comes out to be (120 lb x 220 ft/sec^2) 26400 lbs of force. Of course this is a rough estimate. It probably takes longer than that, and there are a lot of assumptions made (I'm sure most people would brake). Still, it gives you an idea.


Hope this helps.

Printer Friendly | Permalink |  | Top
 
kixot Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:17 PM
Response to Reply #14
20. Wayyyyyyyy overcomplicated.
F may equal ma but if you have no change in acceleration then it doesn't really matter how long of a domain you integrate over because the force (or should I say 'momentum') will be same at x as it is at any other x'. The equation to have in mind here is p = mv and since we don't have distance to deal with or anything else for that matter we can assume the two obejects are points for all intensive purposes.
Printer Friendly | Permalink |  | Top
 
GreenCommie Donating Member (320 posts) Send PM | Profile | Ignore Mon Nov-24-03 10:25 PM
Response to Reply #20
25. Yeah, it proabably is
But it gets at where this stuff is coming from, and it also allows time to play a factor.
Plus, I like to say "integral." :)

Distance doesn't come into play however, because two velocities are known. Whether using (delta v) = a * (delta t) or p=mv, all you need are velocities and times. Time is usually arrived at through experience (poor deer).
Printer Friendly | Permalink |  | Top
 
Endangered Specie Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:28 PM
Response to Reply #25
27. if acceleration is assumed constant
you can find the acceleration if you know the distance over which said deceleration took place v(f)^2-v(i)^2=(2a)(distance)
Printer Friendly | Permalink |  | Top
 
GreenCommie Donating Member (320 posts) Send PM | Profile | Ignore Mon Nov-24-03 10:35 PM
Response to Reply #27
31. True...
but distance is a by product of the time of the action. So it's just extra work to calculate distance (unless it's given).
Printer Friendly | Permalink |  | Top
 
BrewerJohn Donating Member (499 posts) Send PM | Profile | Ignore Mon Nov-24-03 10:13 PM
Response to Original message
18. It's all in the details, unfortunately
Force is the rate of change of momentum (of the impacting animal), so the force vs. time curve during the collision depends on the details of the collision, which are probably unpredictable. You can estimate an average force over the collision time, though, by estimating the total time the animal is in contact with you, and knowing whether it bounces off (and how fast) or just falls at your feet.

For example, if it hits you and then falls flat, the total momentum change is about (54.5 kg) X (26.8 meters/sec) (those being the metric equivalents of 120 pounds and 60 mph) = 1461 kg-m/s. If this happens over 0.1 second the average force is (1461 kg-m/s)/(0.1 s) = 14,610 kg-m/s^2 (or newtons), which converts to around 3285 pounds of force.
Printer Friendly | Permalink |  | Top
 
kixot Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:20 PM
Response to Reply #18
22. hey, Oppenheimer, stick to the details given.
Object of given mass m exerts force x when traveling at velocity v. First year stuff here.
Printer Friendly | Permalink |  | Top
 
Waistdeep Donating Member (469 posts) Send PM | Profile | Ignore Mon Nov-24-03 10:23 PM
Response to Reply #22
24. What?
"Object of given mass m exerts force x when traveling at velocity v. First year stuff here."

Your statement has nothing to do with any first year physics I ever taught.
Printer Friendly | Permalink |  | Top
 
Endangered Specie Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:26 PM
Response to Reply #24
26. Re:
"Object of given mass m exerts force x when traveling at velocity v. First year stuff here."

No forces acting on object unless velocity changes!
Printer Friendly | Permalink |  | Top
 
GreenCommie Donating Member (320 posts) Send PM | Profile | Ignore Mon Nov-24-03 10:30 PM
Response to Reply #26
29. The velocity of the animal changes....
producing a force. Then Newton's Third Law of Motion says that the force on the deer applied by the car is the same as the force on the car applied by the deer.

So there is a force on the car, even if the car's velocity never changes.

Printer Friendly | Permalink |  | Top
 
kixot Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:38 PM
Response to Reply #26
34. acceleration due to gravity is inherent in the unit of weight given,.
thus, we have force, not momentum.
Printer Friendly | Permalink |  | Top
 
kixot Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:36 PM
Response to Reply #24
32. You better have taught it.
*Velocity equals change in position over change in time - given
*Mass of object - given
*Force - defined as the product of mass and acceleration


NOTE - the acceleration is implied by the units of mass given, LBS!!!



Printer Friendly | Permalink |  | Top
 
BrewerJohn Donating Member (499 posts) Send PM | Profile | Ignore Mon Nov-24-03 10:42 PM
Response to Reply #22
37. Uh, I don't think so
Mass times velocity is momentum, not force.
First year stuff, yes.
Printer Friendly | Permalink |  | Top
 
Waistdeep Donating Member (469 posts) Send PM | Profile | Ignore Mon Nov-24-03 10:20 PM
Response to Reply #18
23. Even so 0.1 sec is too much time
Consider that in 0.1 sec the car has moved almost 9 feet. You can be sure that most of the deer (assuming it stuck on the car) was accelerated to 60 mph before the car moved a few feet -- in much less than 0.1 sec. I'm thinking that 0.01 - 0.02 sec is a much better estimate. That makes the force much bigger, as you might expect.
Printer Friendly | Permalink |  | Top
 
kixot Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 11:06 PM
Response to Reply #18
39. ok, I thought about it and you're right with the point you make.
i was assuming there would be a dead stop and the entire force of the impact would be absorbed, which of course is absurd. A bounce or deflect would make a significant difference as you point out.
Printer Friendly | Permalink |  | Top
 
Endangered Specie Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:19 PM
Response to Original message
21. youve got it dead wrong people!
Edited on Mon Nov-24-03 10:23 PM by Endangered Specie
mph is not acceleration, it is velocity. Velocity is distance/time, acceleration is distance/(time^2) or change in velocity!!!! COMMON MISTAKE!!!


In order to figure out the force applied, you need to know the deceleration period, velocity wont help unless you know either how long in length or time it took you to decelerate to 0mph(assuming at the end, you end up at 0mph...).

Physics is never simple.

The problem cant be solved with the info given.

Some DU'ers need to take physice again!
Printer Friendly | Permalink |  | Top
 
kixot Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:29 PM
Response to Reply #21
28. Yes it can be answered with the information given.
Since no mention is given to horozontal acceleration we can assume the only accelleration acting on this model is due to gravity and so it is negligible but this still applies and so we still have an acceleration, so we still hav ea force.
Printer Friendly | Permalink |  | Top
 
Endangered Specie Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:34 PM
Response to Reply #28
30. ok, then answer it
Edited on Mon Nov-24-03 10:38 PM by Endangered Specie
The acceleration (weight) you are talking about is perpendiculer, and wont have any effect. F=m*a*cos(theta). if perpendicular cos(90)=0. F=0
Printer Friendly | Permalink |  | Top
 
WannaJumpMyScooter Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:37 PM
Response to Reply #30
33. I did, above...
120lbs = 264kg

120mph = 132kph = 36.6m/s

m=fa
m=36.6 x 264

9662.4 Newtons

would be the answer

Printer Friendly | Permalink |  | Top
 
Endangered Specie Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:39 PM
Response to Reply #33
35. I REPEAT:
Edited on Mon Nov-24-03 10:40 PM by Endangered Specie
Velocity (m/s) CANNOT substitute for 'a' (acceleration). THe two are completly different!

That is one of the most common mistakes in a physics class, confusing velocity and accel. Accel is a "change in velocity"

That is not a solution
Printer Friendly | Permalink |  | Top
 
kixot Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:41 PM
Response to Reply #30
36. I know the perpendicular application of the weight nulls it in the sum
but the unit remains and that's what gives the (kg*m)/(s^2) needed to make it force, not momentum.
Printer Friendly | Permalink |  | Top
 
Endangered Specie Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 10:44 PM
Response to Reply #36
38. Keep in mind
That even though you are given pounds (a weight), The formula calls for a mass (kg),which has nothing to do with weight. lbs (weight 'mass*gravity')and kg (mass) are NOT the same, the translation 2.2lbs=1kg only works on earth.
Printer Friendly | Permalink |  | Top
 
WannaJumpMyScooter Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-24-03 11:06 PM
Response to Reply #38
40. Hello, we ARE on Earth, no?
Printer Friendly | Permalink |  | Top
 
DU AdBot (1000+ posts) Click to send private message to this author Click to view 
this author's profile Click to add 
this author to your buddy list Click to add 
this author to your Ignore list Thu Oct 30th 2014, 02:27 PM
Response to Original message
Advertisements [?]
 Top

Home » Discuss » The DU Lounge Donate to DU

Powered by DCForum+ Version 1.1 Copyright 1997-2002 DCScripts.com
Software has been extensively modified by the DU administrators


Important Notices: By participating on this discussion board, visitors agree to abide by the rules outlined on our Rules page. Messages posted on the Democratic Underground Discussion Forums are the opinions of the individuals who post them, and do not necessarily represent the opinions of Democratic Underground, LLC.

Home  |  Discussion Forums  |  Journals |  Store  |  Donate

About DU  |  Contact Us  |  Privacy Policy

Got a message for Democratic Underground? Click here to send us a message.

© 2001 - 2011 Democratic Underground, LLC