Help with a "basic" problem - force of impact?

Hello, I hope someone might be able to help with this. It's been 20++
years since high school physics, and my brain's turning around,
churning butter!

I need to know the force of impact for an object that would hit a 2x4.

The object in question weighs 145 pounds. It would fall from a height
of approximately 3 feet from the ground. It would hits the 2x4 on the
end bit of it so on a surface approximately 1.5" x 3.5".

I don't think that the calculation of 145pounds/5.25 square inch is the
simple answer, is it? Do I need to take speed into account?
Acceleration? And if so, how big a difference would it be? This
simple calculation gives me approx 27pound/square inch. Would
including speed etc into it make a huge difference?

Any assistance *greatly* appreciated.

Thx

Oh, btw, this is *not* a homework thing, haven't seen the inside of a
class as a student in way too long

On 24 May 2006 23:35:32 -0700, [Only registered users see links. ] wrote:

F = m * a
acceleration is time dependent and is influenced
by the "hardness" of the objects.
We would have to know the exact timing from
touching the bottom target to complete standstill.
Practically impossible getting closer results.

acceleration measured in multiples of "g"
aprroximately
knocking on the table (or head) 1 g
dropping my newly aquired harddisk on the floor 10 g
hammering a nail into red bricks 100g
hitting anvil with blacksmith hammer > 1000g

velocity of impact
v = SQRT (2*g*h) [m/sec]
h = height in meters
g = 9,81 m/sec/sec


to hell with imperial units.



w.

--
"Mama, can a girl of my age become pregnant?"
"Oh yes!"
"Oh Jesus!"

<[Only registered users see links. ]> wrote in message
news:1148538932.213194.199040@i39g2000cwa.googlegr oups.com...
|
|
| Hello, I hope someone might be able to help with this. It's been 20++
| years since high school physics, and my brain's turning around,
| churning butter!
|
| I need to know the force of impact for an object that would hit a 2x4.
|
| The object in question weighs 145 pounds. It would fall from a height
| of approximately 3 feet from the ground. It would hits the 2x4 on the
| end bit of it so on a surface approximately 1.5" x 3.5".
|
| I don't think that the calculation of 145pounds/5.25 square inch is the
| simple answer, is it? Do I need to take speed into account?
| Acceleration? And if so, how big a difference would it be? This
| simple calculation gives me approx 27pound/square inch. Would
| including speed etc into it make a huge difference?
|
| Any assistance *greatly* appreciated.
|
| Thx

Force = mass * acceleration.
You need to take deceleration into account.
That is, the distance over which the mass stops moving, which is the dent
in the wood. The pressure depends on area, but the net force is the same.
Try it with a 145 lb woman with stiletto heels on her shoes.
Show her a mouse, she'll climb up on the breakfast stool, then remove
the mouse and check the floor when she jumps down again.
(Stiletto heels are murder on wooden floors)
Androcles.

Hi, Your calculation is just the weight of a body at rest on the end of
the 2x4.

The force is always the same regardless of the material used. F=ma
a=gravity = 9.8 m/sec2

55.5 kg x 9.8 m/sec2 = 543 kg = 1195 lb is the force of impact. Divide
by the area of impact and you get the force applied over the entire
object. Then you have to examine the nature of the item dropped and the
nature of the 2x4. If you drop a 122 lb bowling ball on the end of a
2x4, bowling ball wins. If you drop a 122 lb watermelon on the end of
the 2x4 well...

I made a small error. 1195 is the force applied over the entire area.
Divide by by the area and get the force per sq. in. in this case.

Dear vavroom:

<[Only registered users see links. ]> wrote in message
news:1148538932.213194.199040@i39g2000cwa.googlegr oups.com...

The force would depend on the duration of the impact. Newton
discussed this with the concept of "impulse". The force is a
complicated value over the duration of contact in your example.


That is actually the pressure if the object were fluid and
uniformly distributed over that area. So no.


Consider the momentum the object has just before it strikes the
2x4. Then look at the maximum downward displacement of the 2x4,
when object and 2x4 have converted all this energy-of-motion to
energy-stored-in-the-2x4-as-a-spring. The travel between
undeformed and fully deformed requires some short period of time.

Google...
object board impulse Newton site:.edu
I get 45,000 hits.

Wrap your mind around a few of those things.

David A. Smith

[Only registered users see links. ] wrote:

Yes. One correct way to work it out would be to figure out
what the speed is after a three foot drop (sqrt(2*a*s) = sqrt(2*32*3)
ft/sec
= 14 ft/sec), then figure out how quickly it is decelerated. I often
guess a ballpark figure of 1-10 msec for small-body collisions like
this. So the acceleration would be on the order of 14 ft/sec/0.01 sec
= 1400 ft/sec^2 = 44 g or more. Thus the force would be 44 times
your estimate or more.

But another way to guesstimate the g forces is to try to google
for drop tests or shock tests, for instance of computer disk drives.
Doing that, I see testing in the 60 g range or more.

Now an object dropped on a hard surface will experience more
forces than a soft surface (because of more rapid deceleration).
Wood is softer than a hard surface used in shock testing.

All in all I think 50 g (multiply 145 by 50) would be a reasonable
estimate.

- Randy

"SAIL4EVR" <[Only registered users see links. ]> wrote in message
news:1148561644.851251.311640@i39g2000cwa.googlegr oups.com...
| Hi, Your calculation is just the weight of a body at rest on the end of
| the 2x4.
|
| The force is always the same regardless of the material used. F=ma
| a=gravity = 9.8 m/sec2
|
| 55.5 kg x 9.8 m/sec2 = 543 kg = 1195 lb is the force of impact. Divide
| by the area of impact and you get the force applied over the entire
| object. Then you have to examine the nature of the item dropped and the
| nature of the 2x4. If you drop a 122 lb bowling ball on the end of a
| 2x4, bowling ball wins. If you drop a 122 lb watermelon on the end of
| the 2x4 well...


122 lb bowling balls and watermelons... That's quite some imagination you
have.
Acceleration is rate of change of velocity, and the rate of change of
velocity
in this case is well... let's see.
Velocity due to gravity from a height of 3 feet is... hmm..
distance = 1/2 gt^2..
3 = 1/2 * 32 feet per sec per sec * t^2,
so
3 = 16 * t^2
t^2 = 3/16
t = sqrt(3/16) = 0.433 seconds

3 feet in .433 secs is err ...about 7 fps
Assuming the wood is soft enough to compress 0.05" when a bowling ball
hits it in a time of 0.001 seconds, what is the rate of change of velocity
from 7 fps to 0 fps in 0.001 seconds?
I get a = 7000 fps/s.
F = ma = 122 * 7000 = 854,000 lbf, a tad more than your 1200 lbs.
Did they teach you dimensional analysis in school?

[Only registered users see links. ]
"in imperial engineering units, F is measured in "pounds force" or "lbf", m
in "pounds mass" or "lb", and a in feet per second squared. "

Androcles.

Your are slightly out of your mind. Do they teach you common sense in
school? If you drop anything that you can lift up and let go three feet
from the ground it is certainly not going to strike with the force of
854,000 anything except mabe millipounds. Bet you never heard of that
dimension. 854,000 * 1/1000 of a lb = 854 pounds. Now you are only off
by a 1/3 instead of by a hundred. Of course maybe they did things
different in Greece. The compression of wood is not an issue here. Read
the question. How the wood responds depends on the object dropped and
that was not given. If it was a piece of foam at 5,000 mph, well that
is a different story.
By the way this is the way they taught it in the old school. I haven't
done an acceleration problem in 40 years.