00:01
In this problem, we're given a 4 ,000 kilogram or 4 megapilogram airplane with wings of length 5 meters and with 1 .75 meters flying into speed of 80 meters per second.
00:12
And we're asked to find the smallest angle of lift, assuming the wind is an naca 2409 section.
00:20
And we're given the density of air as i have written it.
00:24
Okay, so we're going to do a little bit of math here to figure out what we're going to find.
00:31
And we know that our sum, i already did this for another problem.
00:38
I'm going to follow this along with a highlighter.
00:48
Falling right here.
00:50
So we know that our sum of first is ma, and we can write this equation next.
00:59
That is provided in our text.
01:01
Here's a quick free body diagram, and we can figure out that our sum of our forces here will be m -a -y.
01:22
That was sum of f -y.
01:33
Okay.
01:37
Then we're going to combine a couple of equations.
01:44
Let me get over here to combine my equations.
01:48
Okay, so let's do, okay, unless rearranges, cl will be equal to 2mg over apl.
02:24
Okay.
02:27
Then we'll have okay, then we'll get cl is 2mg over 2bl density times v squared, which will equal mg times bl density v squared, density v squared...