Recent content by vettett15

Hmm, here is where I got my data from: http://web.mit.edu/nnf/education/wettability/wetting.html Which states: When the solid has a high affinity for water - in which case it is called hydrophilic (high energy e.g. glass)- water spreads. In the opposite case of hydrophobic (low energy e.g...

Andy, My understand is, and perhaps I'm wrong, a higher energy surface will allow water to wet more easily and vice versa. I suppose the question is, if water is building up in an interior corner would it be better to have a higher or lower surface energy material? My thought is a lower...

Hey guys, I'm working on a project which demonstrates the venturi effect, by flowing gas through a nozzle (which is housed in a 2 piece housing) and it flows out the exit, the air is picking up water along the way. We have noticed that water droplets are forming in the corners of the...

Guys, Maybe someone can help, as I am having a brain fart here trying to think of how this all works. Say for instance I have an electrical part that dissipates 10 W of power and I am removing this power through the use of a thermal strap to a heatsink. I think the...

Yeah by linearly increasing load I mean a triangular distributed load that is increasing the further away from the support you get. So in the first part I have the load required for the triangular distributed load and then I found the resultant force of that triangular distributed load...

Yes both ends are inclined. They are 0.005" higher than where the support it

Hey Guys, I am working on a situation where I have a beam that is simply supported in the middle, and the two ends of the beam are .005" higher than the middle where the support sits. I am trying to figure out with a linearly increasing load what the force is to make the two ends...

kamerling, Thanks for the reply, I see what you are saying if you have a force in line with the direction of motion than that force is F. The components are Fcos(theta) and Fsin(theta). I am thinking along this line of thought. The horizontal and vertical motions are independent, so...

question kamerling, I was looking at your formula again and had a question. making Cd*A*p/2 equal to C to make this more simple. You have total drag force as: C*v^2 which is the same as C*(vx^2+vy^2) Then you say the horizontal component is: C*vx*(vx^2+vy^2)/sqrt(vx^2_vy2)...

ok that makes since, going from what you have as total drag to the x and y drag components it looks like you just put a ratio of vx/v and vy/v in there. v equaling sqrt(vx^2+vy^2). That makes sense. Thanks

another question I also had another question, on the same website I sent you, they are getting equations of motion in the horizontal and vertical directions. My question is, for both the horizontal and vertical directions they are using the same drag. If you thought of the horizontal and...

retarded I am Thanks so much for the reply. I can't believe I missed that, I just assumed that everything was multiplied by t so the zero would wipe it all out. Thanks again.

Hey guys, I am working on deriving some equations of motion for an object with air drag. I am using this site to help me: http://www.grc.nasa.gov/WWW/K-12/airplane/flteqs.html Here is my problem, at the bottom of that page they have this equation for horizontal velocity: 1. u =...