ok for #4:

i found Fr on the angled portion to be 1470kN at yp=4.305m; found Fr on vertical portion=490kN at yp=0.5333m

I resolved the Fr on the angled portion into x-y components; Frx=882kN and Fry=1176kN

weight of angled portion of gate is 100kN acting 1.5m from hinge, weight of vertical portion is 20kN acting 3m away

moment equation around hinge was:
490kN(4.46666m) + 882kN(1.5552m) + 1176kN(1.1664m) +100kN(1.5m) + 20kN(3m) - W(3m) =0

W = 1714.01kN was what I got

9/13/2005 5:59:45 PM

Quote :

"Pc = Pb + rho*g*h Pb = 101.754 kPa rho = 998 kg/m^3 g = 9.807 m/s^2 h = .36 m Pc = 105.277 kPa"

how the hell did you get Pb of that value?

9/13/2005 6:02:03 PM

Quote :

"I came up with an equation: V= (hmgsin(2.15))/(uWL) u=equals the dynamic viscosity of H2O at 20C"

how did you come up with that equation:

i did:
tau = mu*(du/dy)
(du/dy) = V/h
tau=mu*v/h

F=tau*A
F=mu*V*A/h
V=Fh/muA

F=ma=mgcos2.15

and got 113.075m/s

9/13/2005 6:06:10 PM

tau = u(du/dy)

F = (tau)A

mgsin(theta) = (u(du/dy))(W*L) (du/dy) = V/h

9/13/2005 6:20:23 PM

gotcha; was doing my algebra wrong, bahaha

9/13/2005 6:23:04 PM

#1 is kicking my ass, does anybody have any clue of how to do it....any hints or tips would be appreciated

9/13/2005 6:38:15 PM

i'm still having trouble with number one as well, also has anyone worked through 4 all the way, i just want to see if my solution matches (posted above)

9/13/2005 6:47:11 PM

I think I could figure out #1, I am just having trouble with using the gas constant to find P@a

9/13/2005 6:51:43 PM

what i was trying was, you know Patm, so if you go down in the water to the level that a is at and find it using rho*g*h thats Pa i think, but i'm not sure how to find Pb and then add that to the depth over at C

9/13/2005 7:05:05 PM

ok, well i think for b, you use the gas constant to get the density of air and times that by the heigh (25cm)...but i could be wrong

9/13/2005 7:06:36 PM

ok heres what i just did:

Pa = Patm +rho*g*.2m

Pb = rho*R*T

Pc = Pb + rho*G*.36m

got:
Pa=101.878kPa
Pb=101.246kPa
Pc=104.769kPa

which makes sense, cause Pb should be a decent amount less than Pa since its higher in the same fluid, and Pc > Pa since its lower in the same fluid

9/13/2005 7:11:46 PM

OK THANKS, that seems to make since

9/13/2005 7:17:39 PM

for people that are having problems with 4:


break it up into two problems. One with the 1m of water on the vertical section. And one with the 4m of water on the angled section. Find your yp and the Fr in each case. Then just take the moment about the hinge using those two forces and then the weight acting up

9/13/2005 7:27:28 PM

Quote :

"Pb = rho*R*T"

That seems to make sense, but what keeps us from saying that Pa=rho*R*T? Doesn't the pressure of a gas in an enclosed container equal throughout? Even though we know that due to the liquid Pa = Patm +rho*g*.2m

Maybe I'm reading too much into this problem. I don't know.

9/13/2005 7:28:30 PM

Quote :

"for people that are having problems with 4: break it up into two problems. One with the 1m of water on the vertical section. And one with the 4m of water on the angled section. Find your yp and the Fr in each case. Then just take the moment about the hinge using those two forces and then the weight acting up"

dont forget the weight of the gate itself acting down, just take that in two parts as well

9/13/2005 7:31:07 PM

^yeah, just noticed that . . . maybe i should have actually read the problem before saying how to do it

how did u break the weight up into two parts?

nevermind . . . the weight is in N/m^2 . . . just find the area of the vertical, and the area of the angled


did he say that the test will be just open book? or will it be open book / open notebook?

[Edited on September 13, 2005 at 7:40 PM. Reason : ]

9/13/2005 7:38:39 PM

Just open book. No notes.

9/13/2005 7:42:14 PM

i noticed for 1 that it asks for absolute pressure. do we have to find that using Pgage=Pabs-Patm? or is the value we find the absolute value?

9/13/2005 7:50:21 PM

i can honestly say

i'm terrified of this test

9/13/2005 8:02:53 PM

i think we all are, so lets all do bad together

hear that? no one do good. k?

9/13/2005 8:04:10 PM

I just don't like the fact that the solutions are not online, and i'm having to go on my previous work to study instead of how he wants us to do it.

I realize now I'm going to have to go and copy down the solutions to every hw set outside his office.

9/13/2005 8:07:29 PM

i'm going to start doing what brianj320 did and take pics with my cellphone or bring my digitalcam to campus, saves a shitload of time and i can just look at them on my pc

9/13/2005 8:11:11 PM

I really wish I hadn't been forced to miss two classes where he went over some of those examples in his class notes

is broughton open right now? I think i'm going to go over there cause i wanted to look at the solutions on his board


[Edited on September 13, 2005 at 8:18 PM. Reason : .]

9/13/2005 8:13:51 PM

the set 3 ones are up on the previous page... i think its still open till at least 1030 cause of labs and shit

9/13/2005 8:29:50 PM

Quote :

"i noticed for 1 that it asks for absolute pressure. do we have to find that using Pgage=Pabs-Patm? or is the value we find the absolute value?"

also what'd people get for 2?

9/13/2005 8:33:35 PM

do u think that on the test we can just do the pressure in a fluid is

P= Patm + rho*g*h


or do u think we need to show the integration like he did in Ex. 3.2?

9/13/2005 8:51:05 PM

DiscProdiG-
how did you find a yp on the vertical portion of the wall for #4? i didnt think you could because the y-axis is diagonal to the vertical part.
with that in mind, i got W=648.69 kN with the sum of the moments being:
3W-3Fr(vert wall)-1.5(weight of wall)-(5-yp)Fr(angled wall)=0
am i forgetting something here?

9/13/2005 8:52:05 PM

u find the yp for the vert. the same way as u normally do

yp = yc + I / (yc*A)

and u know yc is just half on the height of a vertical wall

for your moment equation, u should have:

3*(weight of the vertical section) + 1.5*(weight of angled section) + Frvert*(5-yp) + Frangle*(distance of hinge from the free surface (6.25) - yp) - 3*W = 0

that 6.25 is measured along the line of the gate



[Edited on September 13, 2005 at 9:01 PM. Reason : ]

9/13/2005 8:59:54 PM

for # 1. i don't see how we can use the rho given in the front of the book.

That is for stp, but in our case the pressure is being increased by that 20cm column of water.

Is it that the pressure difference is so negligible that we can use the same rho?


and can BEAVERCHEESE or DiscProdiG explain # 2?

is ur x-axis parallel with the incline that the block is on and u are saying that the block's height will only change by the thickness of the film of water?

9/13/2005 10:16:34 PM

i need some help on 3. anyone?

9/13/2005 10:18:31 PM

Just fyi to anyone who wants, the door to Broughton is open on the corner of Stinson across from Polk hall. Just go up the mini flight of stairs, the right door should be ajar (it was ajar when I got there, left it that way when I left).

Took me some pics of all the problems he had up there.

9/13/2005 10:24:09 PM

x-axis is parallel to the incline, and the value of h is small enough to negate any changes because of it

so just follow the homework that is like it (not on incline but similar) and find du/dy, use F=ma, find a converted for the incline using gravity, set F=tau*A and F=ma equal and solve for V, plug and chug

number 3:
follow homework problem similar as well,
P1-P4 = (P1-P2)-(P2-P3)-(P4-P3)

where (P1-P2)=(gammaH2O*(elevationA-h)
(P2-P3)=(gammaHg*h)
(P4-P3)=(gammaH2O*(elevationB+h)
plug and chug

thats what i did at least, seems correct to me; who knows since we dont have solutions, but me and cutiegilr ended up with that

[Edited on September 13, 2005 at 10:27 PM. Reason : ..]

9/13/2005 10:26:12 PM

y do u add B and h for #3?

9/13/2005 10:38:00 PM

for #2 BeaverCheese you said mgsin(theta)=(mu du/dy) (Lw du/dy)=V/h

where did the mu go?

9/13/2005 10:49:06 PM

wow, he really should have had at least one office hour period between the time this was posted and the test.


and between the time the HW soln's were posted and the test for that matter. I mean shit, he didn't post it til prolly around when he was leaving yesterday. So that left us just one day, a day when he doesn't have hours to ask questions.


HMM . . . I guess the TA did have office hours today. I wish I knew that earlier.

[Edited on September 13, 2005 at 10:52 PM. Reason : ]

9/13/2005 10:50:24 PM

He really took his time posting that test, I was waiting for that thing to go up for a few days.

9/13/2005 11:13:51 PM

what did anyone get for #3 answer?

I got 6.089 e4 lbf....this doesnt seem right

9/13/2005 11:26:44 PM

my answer for 3 was huge. and like no 1 has been able to help with it.

9/13/2005 11:50:10 PM

did you do what discprodiGi said to do about the P1-P4 then break it up?


Maybe it gave it to us on such short notice and didn't give us the solutions because the test will be very similar???
Not that it would necessarily help because I am already struggling, hah.

9/14/2005 12:02:25 AM

what i'm confused about with 3 is where the origin is at to measure from. like for a and b they're both labeled as 8 and 12 ft. but where is that bein measured from, bottom of h or top of h?

9/14/2005 12:06:40 AM

i think it is at the bottom of h

That is why they would subtract h from pt A and add it to pt B

9/14/2005 12:11:05 AM

if thats the case then y would u add B and h. wouldnt the distance to B just be B=12ft?

9/14/2005 12:14:32 AM

hmmm...well, now that i look at it, yeah that seems right. it seems like you would either subtract h from part one and use 12 for the last, or use 8 for part one and add h in the last part

9/14/2005 12:22:38 AM

yeah, i was wondering the same thing earlier. i came to the conclusion that it was from the bottom of h

9/14/2005 12:39:48 AM

have you done number 2?

9/14/2005 12:53:33 AM

good luck everyone

9/14/2005 7:23:34 AM

dont worry i think we are all going to do bad, maybe a curve will be in order...

probably not

9/14/2005 7:28:55 AM

Haha, oh this is going to be interesting. I haven't slept since yesterday and am running on 5 hours from then.

9/14/2005 7:30:27 AM

I just need to study more next time

It seemed like that test should have been cake, but I kept getting tripped up on a few steps

I'm hoping for a healthy chunk of partial credit, because I had all the basic ideas down there, just little things that a few more hours of studying would have cleared up

9/14/2005 9:43:58 AM

Yeah, it def could have been a lot worse.

But that doesn't mean that it was a walk in the park.

9/14/2005 1:09:13 PM