anyone get how to do 12-102?

9/7/2005 5:11:48 PM

Does anyone have the problem numbers that were assigned for the homework set due on Tuesday for Buckner? I'm not able to get into the MAE websites. Thanks

9/17/2005 8:32:57 PM

bttt...ya the website is down, and I need the problem set

9/18/2005 10:27:20 AM

this is gay

somebody email buckner

9/18/2005 12:28:36 PM

I emailed him last night. He hasn't replied yet.

9/18/2005 12:35:13 PM

here ya go

9/18/2005 1:23:16 PM

Thank you so much!! is the site back up?

9/18/2005 1:25:48 PM

no, I just realized that I already downloaded it

9/18/2005 2:20:07 PM

what did you guys get for the ones that aren't in the book?

I've only finished one so far, I got F=85.7N for 13.4

9/18/2005 7:07:25 PM

^ that's what I got, although I also specified direction (against the trailer)

9/18/2005 8:31:06 PM

^ and ^^ how did you get 87.5 N for 13.4

I have worked it twice now and get exactly double 87.5 N.
Anywhere in you work you might have divided by two that I could have left out?

Thanks for the help and posting the assignment!

9/18/2005 10:16:32 PM

^
:make sure to convert km/h to m/s
:use (V)^2 = (Vo)^2 + 2as and solve for a
:apply F = ma

Even if you left out the 2 in the "2as", it wouldn't double the answer, so I don't know how you got your answer. This method works though.

BTW has anyone gotten 13.22? I keep getting 12.591 ft/s^2

[Edited on September 18, 2005 at 10:45 PM. Reason : q]

9/18/2005 10:31:14 PM

^Thanks again, I found my mistake.

9/18/2005 10:43:58 PM

I got 13.22 right

the equations of motion for B are

2T -Fg = m_Ba_B

and for A are

T - Ff = m_Aa_A

So you have two equations and two unknows, I multiplied the second one by 2 and added them together, then solved for acceleration of b and got -8.587 ft/s2

then I plugged it into a_A = -2a_AB

then I set a_A = const and used the kinematic equations to find Va final

by the way, I keep getting a really low answer (like 2.4) for 13.13, the answer in the back is 745, any help plz?

9/18/2005 11:19:13 PM

^

Add up all of the forces using vector addition. Note that there is a 4th force of weight {-2k} lb. Integrate 2 times to get the position vector. There is no initial velocity, and hence no constant of integration for the first integration, but there is an initial position. So integrate 2 times and add {3j} ft to it. Then plug in t = 3 and you'll have a vector in terms of numbers. Then find magnitude as that would be the distance of the position vector (i.e. from the origin).

[Edited on September 18, 2005 at 11:30 PM. Reason : lol n/m]

9/18/2005 11:30:11 PM

that's exactly what I did, but I keep getting 2

after integrating twice, I have s = m(3t^2 i + (3t^2 -1/12 t^4) j - 3t^2 k)

[Edited on September 18, 2005 at 11:41 PM. Reason : ahhhh nvm, you have to divide by m ]

9/18/2005 11:37:09 PM

also for 13.53 I keep getting 23.5 m/s

and on 81, 91, and 93 I'm pretty much stuck

I'm about to scan them in since it would be pretty much impossible to type

9/18/2005 11:41:25 PM

^

refer to example 13.6. It's the same, except you add in a F_f, which points into the race track and is perpendicular to N_C. From there, set up 2 EOMs with respect to the n and b direction. Take careful note of the angles as the forces will have to be trig'd to be in the proper direction. I got N_C (or F_N, depending on your preference) to be 19121.104N.

For those of you stuck on the pulley problems, make sure you use the proper sign for acceleration in your Net F equations! The proper sign will be up to you since you decide which way is positive/negative. See section 12.9

9/19/2005 2:47:53 AM

for 13.81,

They give you y(x), so find y'(x) and y''(x), and use the formula for rho. Note that these functions use radians, not degrees. I got rho = 41.434 ft. Next, you need to find the angle between the normal axis and the F_W. You need to do this in order to use the net F_n EOM. Thus, by definition, y'(x) is the tangent, or dy/dx, so theta is tan-1(dy/dx). I got that angle to be 65.765 degrees. Anyway, you end up something like this:

net F_n = ma_n
-F_n + 180cos(65.765) = 4.857
F_n = 69.0lb

[note1: a_n = v^2/rho]
[note2: Because a_n always points towards the center of curviture, I chose positive net F to be into the hill.]

Use the angle to solve for the net F_t EOM, which will give the proper a_t they are looking for in the second part.

[Edited on September 19, 2005 at 4:21 AM. Reason : re-clarified]

9/19/2005 4:09:30 AM

thanks man

9/19/2005 9:19:35 AM

I have all the homework problems done if anyone needs help.

I'm looking to study this dynamics shit tonight in the library, anyone interested in joining?

aim: britishent

9/19/2005 2:21:33 PM

quick question:

in 13-91, are we looking down on it or at its side? That is, is there a force of gravity in the EOMs?

[Edited on September 19, 2005 at 8:52 PM. Reason : claire]

9/19/2005 8:50:26 PM

^

n/m, I got noobed by the chain rule

9/19/2005 9:35:13 PM

yeah 91 is crazy, there's all kinds of secants and nonsense

I still can't come up with the right answer though

hey S, have you gotten 93? I solved for ar and atheta

then did

mar = Pcos(phi) + F_Nsin(phi)

matheta = P sin(phi) - F_Ncos(phi)

where phi is the angle btwn ut and ur, and tan (phi) = r / (dr/dtheta)

then solved for P, but I keep getting the wrong answer for P

am I on the right track?

9/19/2005 10:00:20 PM

^

I did the same except my orientation of FN was into the curve (see last sentence on p. 123). I get whacky answers.

9/19/2005 10:07:49 PM

well my answers aren't way off, but I still don't know what I'm doing wrong

9/19/2005 10:20:28 PM

I got it; we pretty much have the answer right in front of us; make sure to use radians in the r = f(theta) and phi calculation

at 45 deg or 0.785 rad...
r = 0.314m
r' = 2.4 m/s
r'' = 0 m/s^2
theta = 0.785 rad
theta' = 6 rad/s
theta'' = 0 rad/s
phi = 38.146 deg

mar = Pcos(phi) - FNsin(phi)
matheta = P sin(phi) + FNcos(phi)

should evaluate into something along the lines of...

Fpcos(38.146) - Fnsin(38.146) = -22.619
Fpsin(38.146) + Fncos(38.146) = 57.6

make sure you use the right formulas for ar and atheta

[Edited on September 19, 2005 at 10:32 PM. Reason : wow, I think I actually feel LESS confident about the test after doing these problems. this is suck]

9/19/2005 10:26:40 PM

I went to the review session, he said there will be nothing along the lines of number 91, with all the crazy trig derivatives

^ I get 75.7 degrees for phi

tan phi = .4 (45 pi/180) / (.4*.2)

is that right?



[Edited on September 19, 2005 at 10:40 PM. Reason : .]

9/19/2005 10:36:47 PM

degree to radian conversion: (2*pi/360), so

r = (0.4)(45)(2*pi/360)
dr/dtheta = 0.4

9/19/2005 10:44:09 PM

well 2* pi / 360 is the same as pi / 180

and dr/dtheta is r' right, which is 2.4 at theta = 45 deg

9/19/2005 10:46:14 PM

^

oops lol

no, dr/dtheta is NOT dr/dt. dr/dtheta differentiates w/ respect to theta, whereas the latter is with respect to t.

r = 0.4theta

dr/dtheta = r'(theta) = 0.4
dr/dt = rdot = r'(t) = 0.4*theta'(t)

9/19/2005 10:51:29 PM

oh thanks a lot!!

Only one left, number 91

all those crazy secants, there's so many places I could be making a mistake

9/19/2005 11:06:12 PM

for 91, at theta = 30 deg,

r = 0.577
r'(t) = 0.667
r''(t) = 3.849

good luck on the exam tomorrow

9/19/2005 11:09:32 PM

you too

9/19/2005 11:13:15 PM

What did you guys get for the exam questions? These are the important ones, as getting them right will pretty much mean you got the other parts right. I got:

1-2c. -5000*sqrt(2) N
1-3a. 18/5 N

9/26/2005 9:07:42 PM

^ I got the same answer for 1-2c. I keep getting 26/5 N for 1-3a, but I am not sure whether it is right.

9/26/2005 10:39:18 PM

I keep getting 36/7 for the 1-3a. I have two tensions on my block B FBD, is it supposed to be 3 tensions?

9/27/2005 10:48:07 AM

Has anybody got 16-44 or 16-64 for the homework that is due Tuesday?

I have no idea how to even start 16-44 so any on help of that would be great.
Thanks in advance!

11/6/2005 11:37:22 PM

for 44: Think about what common angles the triangles d-y-A and h-x-B share. what trigonometric relationship can you use to get a symbolic answer?

for 64: This is just like 54 except the side with E has one more link. What had me stuck here for a while was that I didn't notice that the angular velocity of BC I solved for in the first part can be used for the BD in the second part of the problem to avoid having two unknowns at the end.

[Edited on November 7, 2005 at 5:29 PM. Reason : numbers]

11/7/2005 5:25:04 PM

^ Thanks

Has anyone gotten 16-22? I keep getting 931 rad/s which is exactly double the right answer. My time is 0.04469 s and my Wa is 2791.65 rad/s. Are both of these numbers right?

[Edited on November 7, 2005 at 9:34 PM. Reason : time & Wa]

11/7/2005 9:32:46 PM

^You can skip the use of time

ads = vdv
alpha * dtheta = w * dw

they give you alpha as a function of theta, so plug that into the above equation. Integrate with theta-bounds of 0 to theta and w-bounds of 20 to w. Lower bound is 20 since that was initial angular velocity. Integrating results an equation with theta and w. Plug in the theta they give you (10 rev -> 20pi radians) and get w. That is wA (I got 1395.939 rad/s).

use: wB = wA * (rA/rB)

So, multiply that by the gear ratio (rA/rB). I get 465.3 rad/s.

Can anyone please point me into the right direction for 16.38? I tried using law of cosines, but it didn't work (I got 8.28). These problems are like related rate problems from Calc I, argh.

[Edited on November 7, 2005 at 10:00 PM. Reason : m]

11/7/2005 9:58:47 PM

for 22: did you use w dw = a d(theta) ? shouldn't really need a time.

for 38: I think he did this one in class if you have it in your notes, but if not, just need to differentiate an equation that gives x based on theta. just break it into two right triangles and go from there, first part of x should be .2cos(theta) and other part would be sqrt(.75^2-(.2sin(theta))^2).

[Edited on November 7, 2005 at 10:00 PM. Reason : like he said ^]

[Edited on November 7, 2005 at 10:02 PM. Reason : 38]

11/7/2005 9:59:42 PM

^ and ^^ thanks, I don't know what I was thinking trying to use constant acceleration.

For 16-38 start with this: Vp=Vb+Vp/b=Vb+(Wbp)X(Rp/b)
Also assume Vp in the y-direction is zero. Just work the rest of it out like the examples he showed us in class on pg. VI.27.

[Edited on November 7, 2005 at 10:10 PM. Reason : constant acceleration]

11/7/2005 10:08:18 PM

16-60 is fun.

11/7/2005 11:33:58 PM

16-74 help~? thx

11/8/2005 12:54:49 AM

I don't know how to start 16-91 using the instant center method.

I am also stuck on 16-137 part b. I keep getting -1.5i for the answer. The book shows -1.69i. Does anything other than the radius change from part a?

Thanks for any help you can give.

11/14/2005 8:41:52 PM

For 16-137: did you also consider acceleration of the girl with respect to the origin?

11/14/2005 8:58:59 PM

^Thanks, I would have never thought of using v^2/rho in this problem.

I still need some help with 16-91 if anyone has a clue how to do it.

11/14/2005 9:06:10 PM

For 16-147 does anyone else get
Vc=-11i+17.3j
Ac=-34.6i-19.5j

Is there anything different between 16-147 and 16-142 about what we have to take into account or procedures? I got 16-142 right and followed the exact same steps, but can't get 16-147 right.

11/14/2005 9:43:51 PM

for 16-91: was your IC above the center bar? a 30-60-90 triangle with .3 as hypotineuse. Once you get the sides for this triangle you just need to find Vb, (Vb) / (r b/ic) = omega of BC. With this you can find Vc = omega BC * (r c/ic). once you have Vc plug it into the equation Vc = omega CD *(r c/d)

11/14/2005 10:21:43 PM