ttt

10/9/2005 12:35:40 PM

i got 1.68 kN for number one, and i'm pretty sure i did it correctly

10/9/2005 12:44:49 PM

whats your final equation for number 2?

[Edited on October 9, 2005 at 2:34 PM. Reason : .]

10/9/2005 2:26:58 PM

whats your final equation for number 2?

since you have 4 control surfaces; what is the V for the two surfaces perpendicular to the water bouncing off of the circular dish?

10/9/2005 3:04:45 PM

how'd you do 4?

10/9/2005 3:35:25 PM

in number one, his answer is right

in number 2 . . . you do it exactly the same as number 1. There are just 4 control surfaces

it tells you in the diagram that V is the same as each control surface. The only thing you need to figure out for the 2 side jets is the area. Which can be figured out using the continuity equation.

number 3 is also done in the same way. But i cannot figure out y he has -825 and not 75.

10/9/2005 6:02:04 PM

i have figured out all of them but part b of number 4 and 5.

can anyone be of any help? i can help on any of the other parts

10/9/2005 6:30:01 PM

part b on 4 is just negative of your tension in the wire since the force on the vane must equal the tension force, just opposite direction

can you tell me what your equation for number 1 was?

i know i'm making a stupid mistake on that one....

[Edited on October 9, 2005 at 6:33 PM. Reason : .]

10/9/2005 6:33:02 PM

i'm getting -3.78kN for the first one.... why am i .1kN off???????

10/9/2005 6:41:41 PM

#1 i had P1a1+f=-rho(v1)^2a1+rho(v2)^2A2

10/9/2005 6:53:49 PM

if part b of number four is just negative of the tension of the wire, then wouldn't it just be -56.25N?

10/9/2005 6:55:32 PM

what was your v1 and how did you get it?

and the force of the wire is to the left and the force on the vane is to the right, so your first answer should be -56.25, and the force on the vane is opposite of that, so its +56.25

also, what were your equations for number 5?

thanks

[Edited on October 9, 2005 at 7:19 PM. Reason : 5]

10/9/2005 7:16:32 PM

my number 1 is -3.68 Kn and I showed how i got it above, or the equation.

#5 for part a it is -rho(v-u)^2A(1-costheta)
i didn't get part b.

If the force on the vane is just 56.25 N, then what is the -97.4 N he has there? what is that for?

10/9/2005 7:34:57 PM

no idea

10/9/2005 7:35:51 PM

hah . me neither.

[Edited on October 9, 2005 at 7:45 PM. Reason : ]

10/9/2005 7:45:23 PM

i know i'm being a pest, but what was your exact number for the velocity at 1 for number one and what did you do to get it?

i get the same final equation that you have, but my answer is coming out wrong

thanks again

10/9/2005 7:49:00 PM

V1=3.6 m/s
I did V1A1=A2V2
.0283=pi(100mm)^2/4 V1

10/9/2005 7:57:44 PM

hahahaha, i must need to go back to sixth grade, cause i can't find the area of a circle apparently


shitze

#5 for part a it is -rho(v-u)^2A(1-costheta)
i didn't get part b.

^^^^^^^
that gives me -251.3N for Rx, but he has his Ry as +251.3N

i dont get it?

[Edited on October 9, 2005 at 8:10 PM. Reason : ]

10/9/2005 8:06:56 PM

10/9/2005 8:10:44 PM

and ill be right back there with you in sixth grade because i didnt read your post properly.
-1000(30-10)^2(.0013)(1-cos150)
i got -938

[Edited on October 9, 2005 at 8:15 PM. Reason : ]

10/9/2005 8:12:13 PM

yep, at least i wont be the only 21 year old in a class full of 12 year olds

10/9/2005 8:25:20 PM

did you try those numbers again? it is interesting that you got the negative of your number 5, what numbers did you put in?

10/9/2005 8:46:17 PM

I am currently suffering extreme mental anguish over problem 5b. Any help would be appreciated!

10/9/2005 9:27:35 PM

i ended up getting -987 or whatever for Rx, and 251 for Ry after i looked back over my work...

10/9/2005 10:17:40 PM

can you tell me how you got 5b please?

10/9/2005 10:18:36 PM

well control surface 1 has no velocity component in the y direction; so that term drops out

your second term should end up being:
rho(V-U)cos60*(V-U)A = Ry = 251

x component equation should be:
-rho(V-U)^2A + p(-(V-U)sin60*)((V-U)A) = Rx = -937.9

10/9/2005 10:30:20 PM

in number 4, his answer is like this (tension in wire, x-component of force on vane, y component of force on vane)

well i have em all right, so if any one needs help . . . ask in the next 15 or 20 min before i go to bed

[Edited on October 9, 2005 at 10:46 PM. Reason : ]

10/9/2005 10:35:22 PM

how do you get the y component of the force on the vane when it comes in from the x direction?

10/9/2005 10:43:35 PM

because the vane directs the water upwards, in order for that to happen a force must be exerted in there to change the direction

just do the same thing you did in problem 5. solve for Ry using the y-sum eqn

10/9/2005 10:48:15 PM

anyone know if solutions r posted?

10/10/2005 5:29:33 PM

yes, all the solns are on the bored for ch. 4,5,6 except for 4.1

10/10/2005 5:33:13 PM

can anyone go and take pics of the solutions for chap6 tonight and post them up?

10/10/2005 6:28:36 PM

Here are some pictures of the solutions. They didn't come out great but should get you started. I tried overlaping the pictures so if you can't read something on a try b.


http://www.tripod-productions.com/MAE308/

10/11/2005 1:29:11 PM

i took pics early this mornin cause i didnt think anyone would. anyway, i just took pics for chapter 6. so if people need em, they are there.

http://www.filefarmer.com/brianj320/solutions/

10/11/2005 3:23:18 PM

as long as he doesnt throw me a curveball, i dont think this test will be that bad

probably going to be 2 problems like he said, a bernoulli one and one like the last homework set

good luck guys

10/11/2005 7:37:14 PM

i usually get pics of his solutions, pm me or im me ... if you need em

10/11/2005 8:43:44 PM

Hey, could ne 1 give me the password to Gould's website?? I am a student in Cassidy's Fluids class and need some help and I was hoping his notes and sample tests would help me a bit....thanx a bunch...
~Di'Anthony Sweet

10/12/2005 1:47:21 AM

g'luck all

10/12/2005 7:06:01 AM

yep, threw me a good old curveball

swing and a miss

lol

10/12/2005 9:30:37 AM

damn that was kinda long, didnt get a chance to do the second one very well, but i think i got the big one correct

10/12/2005 9:30:38 AM

i can't believe he used english units, that took forever to keep converting everything

10/12/2005 9:33:17 AM

argh. i hate english units. The US needs to get the fuck with the program and convert to metric. Its soooooooo much easier.

def was a little long though. I was in the room with the long tables an i dont think anyone left early. How about the other room?

10/12/2005 9:56:58 AM

english units suck

from what I could tell most were there till the end

how can he give us problems that aren't like the hw at all?

damn and I was confident about myself going into that also.

10/12/2005 10:25:01 AM

nope, no one left early

one guy showed up 30 minutes late, i felt sorry for him, no way you can do that test in 20 minutes

10/12/2005 11:25:44 AM

i'm just pissed that i spend so much time on the first one, that i only had like 5 mins to do the second one, and i know i fucked that up BAD

10/12/2005 1:40:44 PM

Hey, could ne 1 give me the password to Gould's website so I can access his notes????????? I am a student in Cassidy's Fluids class and need some help and I was hoping his notes and sample tests would help me a bit....thanx a bunch...
~Di'Anthony Sweet

10/12/2005 6:53:20 PM

check pm

10/12/2005 8:21:46 PM

anyone start the homework set yet?

10/17/2005 1:06:10 PM

For 8.4 what are the independent variables that we need to use?

i have rho, V, mu, h, l, Ca, Fdrag which mean we have to solve 5 fucking pi groups

am overdoing it or what????

i have the other 5 this one is just fucking with my head, if i know what parameters to start with the rest is easy

10/19/2005 12:54:10 PM

I can get the Pi groups but I don't understand what to do with them to simply them. Any advice?

10/19/2005 8:03:40 PM