was anyone able to solve parts C and D for question 1.3? i am unsure of what to do. i was trying to use the basic formulas from physics and dynamics we learned relating constant acceleration to velocity, time, position; in this case though, there is no constant acceleration, it varies. any help?

8/24/2005 6:42:10 PM

i dont have the book yet
if you post the problem i can help

actually i can answer your question...
you have to use calculus

[Edited on August 24, 2005 at 6:48 PM. Reason : .]

8/24/2005 6:47:19 PM

well yea i know calculus is used..lol. but i dont know where to start at to get the distance. i have the terminal velocity already.

this is the problem, its not in the book:

A sky diver, with mass 70kg drops from a slow moving airplane and falls straight down. The drag force acting on the diver is F=kV^2 where k=0.25 and V is the speed of the falling diver relative to the air. find: c.) vertical distance required for the sky diver to reach 95% of the terminal speed d.) compare this distance with the distance required to reach the same speed if air resistance is neglected. Hint: the terminal velocity is the velocity when a=0.

8/24/2005 7:22:01 PM

oh 208
ok i was confused

ads=vdv

[Edited on August 24, 2005 at 7:41 PM. Reason : .]

8/24/2005 7:29:23 PM

i tried using that formula but got no where. i dunno. i'm not seein somethin here and i'm giving up for tonight. if anyone else gets it, please post some helpful advice.

8/24/2005 9:06:14 PM

I have some help for part d. Use Vf^2 - Vo^2=2g(Xf-Xo). But i have no clue how to part C. Ive tried everything.

8/25/2005 1:10:00 PM

^^^no, this is for 308. Fluid Mechanics.

adv=vds? huh

8/25/2005 3:43:46 PM

i figured out 1.3c, it starts by substituting v dv/dy for a on the right side of F = ma. Then you should have Fd - mg = m (v dv/dy). Substitute kV^2 for Fd and 70*9.81 for mg, then bring the left side to the right to group terms. basically you end up with an ugly integral dy = (70*V/0.25*V^2 -686.7) dV and integrate left from 0 to y and right from 0 to .95(1.3b answer).

8/25/2005 5:14:27 PM

my bad i didnt read the problem
i just read something falling and thought 208

8/25/2005 5:22:56 PM

how did u integrate that beastly thing? cause i integrated it and my answer isnt right.

8/25/2005 8:43:14 PM

maple, or a TI-89

8/27/2005 4:18:31 PM

yeah, when it comes to taking integrals there is no reason that you shouldn't have a comp algebra sys do it for you -- unless of course it's ultra simple and it's just like integral of x^2.

8/28/2005 7:05:57 PM

on the second homework set, i am having trouble with 2.3 and 2.4; can anyone whose started give me a hint as how to approach either one of them

thanks

for 2.3; i have my tau = mu*umax*(-8/h^2) after deferentiating du/dy where u(y)= umax(1-(2y/h)^2

not getting the right answer though, some help here please

for 2.4 i dont even know where to start, maybe force equilibrium....

help please

8/29/2005 6:39:11 PM

^ will be workin on this later today and 2morrow mornin after class in the 3rd floor study lounge.

8/30/2005 11:54:54 AM

aight, i'll stop by after class if i can't get it tonight

8/30/2005 6:12:06 PM

still having problems with 2.3.... anyone have just a hint at what i'm doing wrong, i dont want a full explanation

8/30/2005 7:45:55 PM

ttt for set 3

wtf am i doin wrong for problem 1? it seems so easy but i cant get anywhere with it. is the oil important?

[Edited on September 10, 2005 at 5:05 PM. Reason : .]

9/10/2005 4:53:27 PM

any progress?

9/11/2005 1:00:50 PM

not really, you get anywhere with these?

9/11/2005 1:08:07 PM

at least now i know other people are having trouble with these, they dont seem hard initially, but a couple are confusing the shit outta me, can't seem to get the right answer

1st one; the oil has to factor in somehow with an extra pressure element

the underwater gate with water on both sides isnt complicated but i only get the right answer if i take the pressure on one side??

i'm taking a break, but i'll hop back on here later and see if i can help you guys

if you can give hints on any of the problems, i'd appreciate it

9/11/2005 2:44:59 PM

exactly..they dont seem hard at 1st but i'm so damn confused right now with this stuff. i've only been able to get 2 and 3 (well 3 for the most part). the oil has to figure in somehow i would think for 1 but then again i dont know.

9/11/2005 3:18:45 PM

Does anyone have a clue about 3.5 or 3.6, got all the other ones but just cant seem to get close to anything on 5 or 6.

9/11/2005 3:26:54 PM

can u give some hints on 1 and 4 since u got them? i'm goin insane from this stuff cause i'm missing somethin and i dont know what.

9/11/2005 3:53:52 PM

yeah, some hints would be appreciated...

i'm mostly confused on 1, 4,5,6,7

hah, thats not good is it?!

i just hope for the ones i still cant get for tomorrow, he has the solutions up quick so i can see what i'm not understanding

9/11/2005 4:00:32 PM

how'd you go about to get 3?

9/11/2005 4:31:35 PM

find hc by using the centroid calculation in that table, calculate area of the triangle, fing roe*g from the specific gravity, fr is roe*g*hc*area

yp is the second moment equation over (hc times area) + yc

9/11/2005 4:36:32 PM

have you gotten #7?

9/11/2005 4:37:18 PM

4 is easy, just follow either one of those examples from class about force on a rectangular gate and do a force on each side seperately so you have 2 resultant forces, one on each side. the trick is to measure your Hc,Yc,Yp on the right side from the stopper since the free surface is lower on the right side. 3 is the same thing, just look up the centroid of a triangle in that table in chapter 3, and thats your hc, do Fr = gamma*Hc*Area(triangle), then remember Yc = Hc when youre finding Yp.

9/11/2005 4:39:32 PM

^^^^^^^^^^^^^

have you gotten #7?

9/11/2005 4:44:03 PM

wtjessup: i've worked through 4, using FOR THE RIGHT SIDE: hc=1, yc=1.4142, ending up with yp=1.8856m and on the LEFT SIDE i'm using hc=3m; yc=4.2426m getting 4.3997 m

are these the same values you got/used or am i doing my geometry wrong?

thanks

9/11/2005 4:57:49 PM

nevermind, i fucking typed my moment equation into my calculator wrong

9/11/2005 5:00:16 PM

Quote :

"4 is easy, just follow either one of those examples from class about force on a rectangular gate and do a force on each side seperately so you have 2 resultant forces, one on each side."

so u get 2 resultant forces then what? subtract 1 from the other?

9/11/2005 5:09:01 PM

draw up a FBD with your resultants on each side, and find the moment arms from the yp values for each, sum around b to get Fa, then divide by width of gate for force per unit length

9/11/2005 5:12:52 PM

i'm still stuck on 1, and 7 mainly; havent worked through 5,6 yet

9/11/2005 5:14:02 PM

where r u guys puttin the y axis for 4 to measure from?

9/11/2005 5:45:03 PM

draw it up parallel to the gate up to the left side's free surface, but remember to take your yc, hc, and yp on the right side from point a

9/11/2005 6:24:38 PM

how is his answer for 1 right? you have to add Patm and the pressure from the fluid, and his answer is like half of Patm.

9/11/2005 8:10:30 PM

i've given up, i've spent 4 hours on three problems today, i'm just putting down some bullshit

9/11/2005 8:31:26 PM

The answer to 1 is right, it's just you have to use an obscure conversion factor.

lb/(ft*s^2) = 0.0310809502 lb/ft^2

And yes you need that many decimal places else it won't round right in your calculations. That and make sure that the area you multiply by is 0.4444ft^2, again because of rounding issues. I was using only 0.44 at first to no evail, plugged in an extra two decimals and boom.

Also in part b of #1, there are 4 forces on that block you have to deal with to solve the problem. I was assuming three and then went into see him. He corrected me on that front, however the answer you end up getting is off by a few decimal places, but whatever.


My question:
Anyone have a clue how to get the area of the block in #5 when you don't know the horizontal distance from the edge of the parabola to the right of the block edge?

9/11/2005 8:34:40 PM

oh wait, that was just for pressure, then gotta multiply by area to get force.

hmm . . .still off by about a hundred.

ah . . . obscure conversion factor.



[Edited on September 11, 2005 at 8:40 PM. Reason : ]

9/11/2005 8:35:00 PM

DPK, can you be a little more specific on how you found the pressure? i'm still not getting the right answer

thanks

9/11/2005 8:39:59 PM

I believe what you do is this: you know that the pressure is rho*g*h at a given depth in a fluid. So you have Patm + (rho)oil*g*(h)oil + (rho)water*g*(h)water

that gives you the pressure at the bottom of the block. Then F=PA

and to find the tension, I'm not sure about the fourth force. One is W, acting down. One is tension, acting down. And one is the bouyant force acting up.

I guess there might also be a difference in the force of pushing down on the top of the block and the force pushing up on the bottom of hte block. But I would think that would be negligible.

Either way, I'm still getting a slightly wrong answer.

[Edited on September 11, 2005 at 8:49 PM. Reason : on to # 2]

9/11/2005 8:47:58 PM

Well you are trying to find the pressure at the bottom of the block right? So that's the sum of the atmospheric bearing down, plus the oil pressure, and then the pressure from the top of the water and the distance through the block (remember to convert to feet from inches).

My units for pressure ended up simplifying down to lbf/ft^2 making note of the conversion factor in my above post. That darn question took me about 2 days to figure out.

9/11/2005 8:52:08 PM

Has ANYONE figured our number 6?

[Edited on September 11, 2005 at 9:01 PM. Reason : #6]

[Edited on September 11, 2005 at 9:02 PM. Reason : #6]

9/11/2005 8:55:07 PM

In part b, you have a force going down on to the block from everything above (you have to find the pressure at the top of the block to get this), then you have the weight of the block going down, the tnesion of the wire, and then the force going up onto the bottom of the block.

And if y'all get to #5 and can figure something out, please give me a holla I'm going crazy.

[Edited on September 11, 2005 at 8:58 PM. Reason : #5]

9/11/2005 8:56:24 PM

grrrrrrrrrrr, can you give me some numbers, i'm still way fucking off, and i either have my rho calculated wrong or something

9/11/2005 9:02:49 PM

The density you should use for atmospheric is 2116.2 lbf/ft^2 (front of book) and your rho for water is 62.30 lb/ft^3 (also in front). For oil you are multiplying the rho of water by the S.G. of oil to get your density of oil.

Then you pretty much just plug everything in to the equation honda posted. My advice is DO NOT add the atmospheric to the rest of pressure until the last step so you can see all the units. Do the pressure of oil and water, and add them together. Your units will end up being lb/(ft*s^2), convert that with the factor up a few posts to the units for atmospheric, add together, then that's your pressure.

Multiply that by 0.4444ft^2 to cancel out the ft^2 and you get your force.

Also note that your water height is the 2ft + the height of the block.

[Edited on September 11, 2005 at 9:10 PM. Reason : water height]

9/11/2005 9:09:15 PM

god i'm retarded:

alright i have 49.84 lbf/ft^3 for the rho-oil
hoil - 3 ft
hwater - 2.66666 ft
rho water - 62.3 lbf/ft^3

are these right?

9/11/2005 9:17:41 PM

yes

finally . . . i've finished half of the first problem

now just 6 more to go (cause i don't think i'm gonna finish the other half)

so how do u do the second problem?

[Edited on September 11, 2005 at 9:21 PM. Reason : ]

9/11/2005 9:19:37 PM

i'm getting: 12272 for my pressure....

not right huh?

9/11/2005 9:20:52 PM