How much extra weight is needed for a 15lb 2x4 to withstand a 67mph drop test?

[Moose720]

I am needing a project to withstand the impact of a 15lb wood 2x4 traveling at 67mph in a horizontal fashion. I want to test my project before the final test. I want to vertically drop a 15lb wood 2x4 fixed to additional weight at a height of 10ft. The additional weight will be released just before the impact of the 15lb 2x4. How much additional weight is needed to be added to the 2x4 to achieve the 67 mph just before impact? Also what formulas are used to get this answer?

 

[berkeman]

I am needing a project to withstand the impact of a 15lb wood 2x4 traveling at 67mph in a horizontal fashion. I want to test my project before the final test. I want to vertically drop a 15lb wood 2x4 fixed to additional weight at a height of 10ft. The additional weight will be released just before the impact of the 15lb 2x4. How much additional weight is needed to be added to the 2x4 to achieve the 67 mph just before impact? Also what formulas are used to get this answer?

Remember that ignoring air resistance, different weights fall at the same acceleration (1g). So adding weight to your 2x4 will not make it fall any faster, at least not until it reaches a speed where wind resistance comes into play (which it will not from 10 feet).

So you need some sort of 2x4 accelerator, like a slingshot or other stored energy mechanism. Maybe some sort of big rubber band setup...

 

[berkeman]

Or find a private road someplace where you can get permission to drive...

 

[Moose720]

So if i drop 1000lbs and 1 lb from the same height, they will land at the same time?

 

[Moose720]

And wouldn't the additional weight be a stored engergy?

 

[berkeman]

So if i drop 1000lbs and 1 lb from the same height, they will land at the same time?

Yep. That's the classic leaning tower of Pizza thing. Or whichever astronaut on the moon dropping a feather and something thing.

And wouldn't the additional weight be a stored engergy?

Weight lifted against a gravitational field is stored potential energy, true. But that will not give you additional acceleration above 1g.

The private road is your best bet in the short term, IMO. As long as you can set up the experiment safely.

 

[Moose720]

Does anyone have any other ideas how I can simulate this test other than the private road idea. Something a little more controlled.

 

[berkeman]

How is the final test going to be conducted? Do they have a machine to launch the 2x4?

 

[Studiot]

Rather than a drop or road test why not use a spring loaded and/or rotating arm?

Incidentally it makes a heap of difference whether the 4x2 impacts end or sideways on, ie acting in a piercing or clubbing mode.

go well

 

[zts1986]

Moose, you don't need to have a slingshot and speed up the test at all. The main factor in all of this is momentum. I posted the following in your other thread, but I thought that I'd post it here as well.

This is how I would go about it:

1. Figure out how fast the object is moving when dropped from 15 feet. This can be done with the known acceleration due to gravity (32.2 ft/s^2). This is the average acceleration across the Earth's surface. If you wanted a more accurate number, you'd have to factor in your elevation. This won't be necessary or matter for this purpose, unless you're at the summit of Mt. Everest or something. Anyway, the formula for the velocity of a free-falling object on Earth given a certain height is as follows:
v = sqrt(2*g*d)
v = sqrt (2 * 32.2 ft/s^2 * 15 ft)
v = 31.08 ft/s = 21.19 mph

2. Given the velocity of the object in your experiment, the desired velocity, and the desired weight, you can set up an equation of momentum. The main principle here is that you desire a momentum produced by a 15 lb object going 67 mph. Momentum is equal to the product of mass and velocity (p = m*v). Since you're looking to recreate the momentum from the desired experiment and you cannot change the velocity, you have to add mass. That gives you the following equation:
m1v1 = m2v2
(15 lb)*(67 mph) = (X lb)*(21.19 mph)
X = 47.43 lbs

3. So, for your experiment to produce the required momentum, your total weight needs to be 47.43 ~ 47.5 pounds. You currently have a 15 lb mass so therefore, you need to add an extra 32.5 lbs of mass to your weight.

4. Just as a note, there are some inaccuracies in this experiment. The calculations I made were using point-masses, not actual objects. The difference is not very large at all and will be negligable in what you're trying to accomplish. I would not worry about those. I figure that if air resistance isn't necessary, then it isn't necessary to dive into the center of mass/inertia values for the mass.

Hope this helps!