g = 32.17 ft/s2
mkid = 80 lb.
AS = 6 ft2
CD = 1.0
p = 0.074887 lbm/ft2
∆t = 0.5s
F = 17132.83 lb
I think... But I couldn't remember if you had to multiply by g inside the radical or not. If not you get F = 3020 lb
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Fall 2008 -- Purdue University -- West Lafayette, IN
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9 comments:
g = 9.81 m/s2
mkid = 45.36 kg
A = 0.557 square meters
Cd = 1
p = 1.2041 kg/m3
∆t = 0.5s
I get F as 3304.69 Newtons.
80 pounds is the weight of the kid, mg. So you don't need to multiply again by g.
I'm pretty sure my answer is right, but it might not be.
I got 3304 N as well. I converted from English units too because they're more confusing to work with.
tim, I got 36.29 for the kg, but otherwise I think your method looks alright.
As for the the english answer I got 1771.05 lbs. when I used (80/32.17) for the mass rather than 80, which i thought was the weight.
well, yeah, you can choose the mass as anything you want...
This is great stuff! I got a little different number for the mass of the kid. Although, after watching it a couple more times, I think it should probably be a little higher.
Anyway, there are a couple of things to notice here. First is that the terminal velocity depends a lot on the surface area. That can be a tough one to estimate.
The other parameter that is tough to estimate is the "contact time" between Hancock's arm and the kid. If we let it range from 0.2 s to 0.5 s, the range of answers should be between 2365 - 5912N (532-1329 lbs.). Those are big numbers. If if his spine didn't break, he wouldn't have many internal organs left.
While this problem was an obvious pitch for ME 577, it also showed us that we could do some interesting things with the impulse momentum equations.
I think the mass of the kid should be AT LEAST 100 lbs. My brother is the same age/size and I think he's about 115 lbs (aka 52.2 kg)
Also if you watch the video on youtube (http://youtube.com/watch?v=K588VwOZPbw), you'll see the acctual throw is pretty quick - I'd be apt to go with Nauman's ~.2 sec.
g = 9.81 m/s2
mkid = 52.2 kg
A = 0.557 square meters
Cd = 1
p = 1.2041 kg/m3
∆t = 0.2s
So I got 5680.958 N
Also, after watching the video again, it seems like his catch would relieve some of that force.
Any idea what force would be needed to rip him in half? I felt really creepy trying to look that up "forces needed to rip a person in half" on the internet...
Good point. I've looked up enough of those kinds of things that I'm probably on a couple of lists. It turns out that those numbers depend on how it happens. We'll leave it at that for now.
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