Solving for Impact Force and Height: Equations and Attempt at a Solution

[Moose720]

Homework Statement

I have a problem where I need to know what the impact force of an object with a mass of 15lbs traveling at 67 mph would be? I also need to know at what height the same object would need to be dropped from to achieve this same impact force .

Homework Equations

Not sure which equations to use and in what order.

The Attempt at a Solution

It has been a number of years since I have had to do a problem like this. The old wise tale of "you don't use it, you lose it" strikes again

 

[PhanthomJay]

You can find the the height that will produce the same impact force, but you can't determine what that impact force is, unless you have more information such as the time elapsed during impact or the displacemnt during the impact. It will make a big difference in the impact force for example if the object slammed into dense soil versus into a pile of hay.

Incidentally, an object that has a weight of 15 pounds has a mass of 15/32.2 = 0.47 slugs, per W=mg.

To find the equivalent height, you can use the basic kinematic equation for free fall that will result in a speed of 67 mph just as it hits the ground, the one that relates speed with the gravitational acceleration and distance fallen. Or initial PE equals final KE.

 

[gneill]

"Impact force" is a pretty loose statement. It could vary greatly depending upon how the object comes to rest, bounces, sticks, shatters, etc., etc. The forces involved may vary over the time it takes for the object to finish doing what it's doing when it "hits".

Is there a more precise statement of the problem?

Or, perhaps what you're looking for is the impulse or momentum delivered?

 

[Aviation]

"

Or, perhaps what you're looking for is the impulse or momentum delivered?

I think that is what he wants to know. I remember reading in a physics book once that there are situations where you can say the force is equal to the mass times the velocity.

 

[Moose720]

I am researching a product that has to withstand a 15lb 2x4 that is shot from a pressurized air cannon traveling at 67mph at moment of impact . I want to mimick this test but I want to drop the object instead of using an air cannon. The height can vary as long as it is under 25ft and also the mass of the object being dropped can vary in order to achieve the same moment of impact as the 2x4 being shot from the air cannon. Hopefully this clears things up about what I am trying to achieve.

 

[Aviation]

I am researching a product that has to withstand a 15lb 2x4 that is shot from a pressurized air cannon traveling at 67mph at moment of impact . I want to mimick this test but I want to drop the object instead of using an air cannon. The height can vary as long as it is under 25ft and also the mass of the object being dropped can vary in order to achieve the same moment of impact as the 2x4 being shot from the air cannon. Hopefully this clears things up about what I am trying to achieve.

There are two ways to solve this problem(for finding your drop hight). As you know v = g*t in this case so that is the first way to determine your hight. If you want to incalculate the airresistance too you have to get some more badass equations which I do not suggest. Still, to calculate the impact in your case it's going to bring some problems because the exact force is relative to the materials etc. You could try to visit this page: http://en.wikipedia.org/wiki/Impulse_(physics )

 

[gneill]

I am researching a product that has to withstand a 15lb 2x4 that is shot from a pressurized air cannon traveling at 67mph at moment of impact . I want to mimick this test but I want to drop the object instead of using an air cannon. The height can vary as long as it is under 25ft and also the mass of the object being dropped can vary in order to achieve the same moment of impact as the 2x4 being shot from the air cannon. Hopefully this clears things up about what I am trying to achieve.

To achieve the same velocity (67 mph) an object would need to be dropped from at least 150 feet. So you're looking at either alternative testing methodology or finding larger premises!

Without knowing what the dynamics of the product are it is difficult to advise. Is the 2 x 4 expected to bounce? Be trapped? Deflected in some way? does the product have crumple zones, special padding, etc.? What is the mass of the product? Will the product move, or is it fixed in place?

It could be that you want to mimic the total energy delivered over a given cross sectional area, over a reasonably similar length of impact time. In that case you might calculate the kinetic energy of the 2 x 4 moving as you say, and replace it with a slug of iron dropped from an appropriate (comparatively small) height.

Note that I take no responsibility for any accidents or product liability issues that may arise from suggestions made during these discussions. This is free advice, and you get what you pay for. Consult with a licensed (paid!) engineer for actionable advice!

 

[Aviation]

Note that I take no responsibility for any accidents or product liability issues that may arise from suggestions made during these discussions. This is free advice, and you get what you pay for. Consult with a licensed (paid!) engineer for actionable advice!

I don't think somebody on this forum wil drop a 2x4 from a distance of 150 feet and be anywhere near:P

 

[Moose720]

Ok let me ask this then, if I were to drop 100lbs from 20ft, what would the speed(mph) be just before it hits the ground? What equation or equations are used to find this out?

 

[PhanthomJay]

In the absence of air resistance, it's speed would be found using v² =2gs, where v is the speed just before it hits the ground, g is the acceleraation of gravity equal to 32.2ft/sec/sec, and s is the height. So v = 36 ft/sec or about 25 mph. It's momentum just before impact would be mv = (100/32.2)(36) =112 slug-ft/sec. The momentum of the air cannon driven 15 pound projectile at 67 mph is about 46 slug-ft/sec.

In actuality, you'd have to consider air resistance in the vertical drop, which will slow down the object to a smaller speed at impact, which depends on its size and shape and whether its dropped on edge or on end, etc.

 

[gneill]

Ok let me ask this then, if I were to drop 100lbs from 20ft, what would the speed(mph) be just before it hits the ground? What equation or equations are used to find this out?

About 24 mph. The formula is

v = \sqrt{2 g d}

where g is the acceleration due to gravity, and d is the height from which the object is dropped. Note that the speed is independent of the weight (mass) of the object. This is true so long as forces due to air resistance are much smaller than those due to gravity.

Working in feet and miles per hour, you'll have your hands full with unit conversions. Here's a short cut.

v = 5.469 \times \sqrt{feet}

Here the result will be in mph if "feet" is the number of feet for the drop.

 

[Moose720]

To double check my work, a 41.25lb dropped from 20ft would equal 46 slug ft/sec. Correct?

 

[PhanthomJay]

Yes, correct. In the absence of air resistance, a 41 pound object dropped 20 feet will have the same momentum at impact as a 15 pound object fired horizontally at 67 mph.

 

[Moose720]

Thank you very much!

 

[PhanthomJay]

To simulate the test, you'll have to use about a 25 foot long 2 x4 to get a weight of about 40 pounds. You can't just use any object weighing 40 pounds. And you'll have to drop it straight down on it's end so that it hits squarely. Note that I take no responsibility for any accidents or product liability issues that may arise from suggestions made during these discussions.