me203 2008spring Spring 2008 Posted by D. Dane Quinn on Sat 16 of Aug, 2008 17:45 EDT A compilation of Spring 2008 posts

Spring 2008 Posted by D. Dane Quinn on Sat 16 of Aug, 2008 17:45 EDT A compilation of Spring 2008 posts

Posted by D. Dane Quinn on Mon 21 of Jan, 2008 20:56 EST ...and it begins. The first homework is now available on the website. Download...view...work...enjoy...and post any questions or comments here.

Posted by D. Dane Quinn on Thu 24 of Jan, 2008 12:51 EST So I had the following question via email: i am getting tripped up when its asking to write “i” and “j” in terms of e1 and e2... is there any quick advice you can give over an email yes...lots of advice. Actually, I think the simplest thing to do is to redraw the (i,j), and (e1,e2) basis so that (e1,e2) is horizontal and vertical, while (i,j) is rotated clockwise from there by an angle theta. This would be equivalent to tilting your head counter-clockwise . Then, all should be clear(er)...

Posted by Emily Shinko on Thu 24 of Jan, 2008 18:34 EST hint* you should get the same answer for problem 2 parts a and b if you did the redefinition of e1 and e2 (or i and j) correctly. if you didn’t, then you know you haven’t got the hang of the thing yet.

Posted by JenniferWelch on Tue 29 of Jan, 2008 09:20 EST I am having some trouble working through Problem 3. I am trying to use the y(x) that we defined in class on monday, and solve for Va, but I am getting too small of an answer. Is there a better way to work through this problem? Thanks, Jenn

Posted by D. Dane Quinn on Tue 29 of Jan, 2008 14:48 EST Hi Jenn, The answer should be something on the order of 100 ft/s. My advice is to check all of your variables. For example, notice that the hose is pointed down by 20 degrees, so this implies that theta = -20. Likewise the point A is below the fireman so y(final) = -30. Also, don’t forget that your units are given in terms of ft, so g = 32.2 ft/s^2. If this does not work out, you can also go back to the equations for x(t) and y(t), using them directly (instead of y(x)). Good luck...

Posted by Emily Shinko on Tue 29 of Jan, 2008 20:08 EST hint* what i did was define the (a) and (b) dimensions in terms of sines and cosines (kinda like how we did the projectile ex from monday’s class). if you solve for one value - either the initial velocity value or the time - in one eqn, you can substitute it into the other eqn.... this *should* work (because the *exact same* procedure can be used to solve #4 correctly) HOWEVER i am still getting the wrong answer in 3... according to the back of the book. i can’t seem to find my error (it’s def calculation error tho).

Posted by AngelaCoates on Wed 30 of Jan, 2008 19:48 EST I have a pretty simple question about 6b... The instructions say to assume that the nozzle is at ground level. Does that mean assume that theta = 0 or just that we’re supposed to neglect the vertical distance between the ground and the nozzle?

Posted by D. Dane Quinn on Wed 30 of Jan, 2008 21:40 EST Hi Angela, The latter, neglect the distance between the ground and the nozzle. The problem simplifies if you assume that the initial height of the water is y=0 and once it travels through the air (some unknown distance x) it again lands at zero height (y=0).

Posted by GregReinhart on Thu 31 of Jan, 2008 13:41 EST im probably skipping over something but on problem 6b dont we need to know either the angle of theta or at what time the water particle leaves the nozzle...

Posted by D. Dane Quinn on Fri 01 of Feb, 2008 00:19 EST Greg, you can solve for the range of the water as a function of theta, much like we did in class...

Posted by ChristinaS on Thu 31 of Jan, 2008 14:58 EST any tips on Q4? i can figure out the conditions at y=-100, but don’t quite know how to find the end part (the sloping ground)

Posted by GregReinhart on Thu 31 of Jan, 2008 15:26 EST > any tips on Q4? i can figure out the conditions at y=-100, but don’t quite know how to find the end part (the sloping ground) try setting y=100 + d*sin(5.7) where 5.7 is the degree of the slope and letting the down direction being positive

Posted by JenniferWelch on Thu 31 of Jan, 2008 15:27 EST You need to look at the sloping ground like a right triangle. Because they gave you the slope of the ground, you should be able to find the other two sides of the triangle in terms of d. I got the y side to equal 1/(101.5) times d. I got the x side to equal 10/(101.5) times d. Use sine or cosine if you find that easier. Now your total change in the y direction should be -100-(1/(101.5))d and in the x direction should be (10/(101.5))d These represent your final x and final y which you should be able to use in the equations to solve for the final velocity. (I had to solve for d and t before I could find the final velocity.)

Posted by JenniferWelch on Thu 31 of Jan, 2008 15:29 EST Wow that didnt show up right. I was trying to represent the square root of 101. I hope you can read it.

Posted by IanCollins on Tue 05 of Feb, 2008 11:28 EST Problem 1: I have gotten equations for x and y values then i solved the x equation for dt and tried to plug it into the y function. Once there i am not able to solve for dt. Any hints?

Posted by D. Dane Quinn on Tue 05 of Feb, 2008 14:24 EST Basically the motion of each ball is similar, with a different initial angle for the velocity. In addition, the balls must pass through the same point in space so that they collide. So I would work backwards. For each ball to be at the same point right now, how long ago must each of them have been thrown? How’s that?

Posted by AngelaCoates on Tue 05 of Feb, 2008 13:09 EST I don’t really understand how to do these rotation problems... at all. I am coming up with equations and stuff, but I have no idea if they’re right. I don’t really feel confident with my answers. Does anyone have any suggestions of examples to look at or anything? Maybe an approach method?

Posted by D. Dane Quinn on Tue 05 of Feb, 2008 14:18 EST Well, you can go through the solutions to the last homework. In general, you want to describe the kinematics in terms of some (unknown) coordinates like theta(t) and r(t). Then identify what information is given in the problem statement, what you can determine, and what quantities you need to find. This is a pretty vague answer, but if you can identify a specific problem and/or example, then maybe I can work from there...

Posted by ChristinaS on Tue 05 of Feb, 2008 18:08 EST I think i got it except i don’t know how to find phi dot (derivative of angle btw horizontal and BP). it seems it would be zero but then the answer doesn’t make sense.

Posted by D. Dane Quinn on Tue 05 of Feb, 2008 18:14 EST Hi Christina, Sorry I missed you today. The presentation ran longer than I expected... Once you get your velocity equation, you should have two unknowns (probably phidot and xdot) and the vector equation can be broken up into two directions, giving you two equations. Then, you can solve for both phidot and xdot simultaneously...

Posted by D. Dane Quinn on Tue 05 of Feb, 2008 18:16 EST This one is pretty similar to question 06, except the velocity of the slider is in a different direction. However, just looking at the first link AB, we know that the velocity of B is perpendicular to this ink.

Posted by D. Dane Quinn on Wed 20 of Feb, 2008 08:56 EST It’s up...and please use this thread for any comments. You can only get something out of all this if you put some effort in.

Posted by VincentWeisenberger on Wed 20 of Feb, 2008 17:24 EST I have a question for number one. I have started the problem but have hit a wall when finding the angular velocity. I set three coordinate systems and then wrote an equation for velocity of point c by using the common velocity equation. Than i solved for the angular velocity of c. Does this sound like the right procedure or do i need to take a diffrent path. The answer i get does not seem right due to the fact it has an i and j coordinate component in it.

Posted by D. Dane Quinn on Thu 21 of Feb, 2008 07:22 EST Do you actually mean the velocity of point B? The velocity of C is zero, as this point is fixed in the corner. As I discussed in class, you should express the velocity of point B in terms of link BC as well as in terms of link AB. The contact point is fixed relative to link BC while link AB does not rotate (zero angular velocity) but the contact point moves relative to this link. In general, there is no problem with having i and j components in your velocity equation, but the angular velocity should only be in the k direction. Have you checked that your cross-product has be taken correctly?

Posted by AngelaCoates on Wed 20 of Feb, 2008 17:41 EST I’m stuck on problem one, too. I defined 3 coordinate systems, but I solved for the velocity of P instead of C. My answer is really close to what is in the back of the book, but I’m messing up somewhere with the sines and cosines. The only thing that I wasn’t really sure of when I was working through it was the velocity of P that I found with respect to the bar AB. I defined coordinates fixed in the bar and said that the velocity of P is the same in magnitude as the translating velocity of the entire bar, but in a different direction. I don’t know if that makes sense, but if it does, is that a bad assumption?

Posted by D. Dane Quinn on Thu 21 of Feb, 2008 07:26 EST That is it...the speed of the contact point is not simply the speed of the translating bar (they call the contact point B in the text but I called it P yesterday in class...I’ll stick with B here so everyone can follow). If you apply the velocity equation for link AB, keeping in mind that this link has zero angular velocity because its not rotating, you should be able to find the correct velocity of B.

Posted by MichaelReith on Wed 20 of Feb, 2008 20:11 EST Is there any hint or direction you can direct me in on question 2? The pulleys seem to be giving me trouble. Is this like a pulley problem from last week?

Posted by D. Dane Quinn on Thu 21 of Feb, 2008 07:31 EST Since the angular velocity and acceleration of the bar is given, you can find the velocity and acceleration at B. Then, this becomes very similar to problem 3 (12-195 in the text) from last week, with the velocity of B known (but not simply in the vertical direction as in the previous problem.

Posted by VinceGreczanik on Wed 20 of Feb, 2008 20:57 EST Doesn’t seem like we are given much information for prob 4, i hate these problems that dont have numerical values... have no clue where to start??????? Also any suggestions on problems that have no “Givens”?

Posted by D. Dane Quinn on Thu 21 of Feb, 2008 07:35 EST Since the wheel does not slip at point B the instantaneous velocity of this point must be zero, since the ground has zero velocity. The you can use the velocity equation to find the velocity of A, as both of these points are fixed in the disk. If it helps, assume some numbers for the radii and angular speed of the disk. Then once you are comfortable with your approach, go back and use the original variables.

Posted by JeremyCessna on Thu 21 of Feb, 2008 14:15 EST I have a beginning question for 2. When it says that the cord is 6m long does that mean the length of Sa or Sa + section DB. Also on question 6 what is the path of the velocity of the person with respect to the body?

Posted by D. Dane Quinn on Thu 21 of Feb, 2008 16:19 EST The length of the rope between the end of the bar, the pulley, and the mass is constant, its S_a + DB. In question 6, the path of the girl in part a is straight across the disk (straight line relative to the disk). For part b, she moves along a path with constant radius in the disk.

Posted by ryanfearon on Thu 21 of Feb, 2008 16:30 EST Dr. Quinn, I am struggling with the relative motion equations for problem six, especially with relating the girls horizontal velocity to the angular velocity/acceleration of the disk. I know that the acceleration is equal to the cross product of the angular acceleration and radius plus the cross product of the angular velocity and the cross of the angular velocity and the radius, but i’m not sure how to figure the girls velocity in

Posted by D. Dane Quinn on Thu 21 of Feb, 2008 17:35 EST Ryan (and everybody else), The acceleration (with respect to the ground) is not just the two terms that you describe below, but contains the acceleration of the girl with respect to the disk, the Coriolis acceleration (2 omega cross v), and the acceleration of the center of the disk, which vanishes in this problem. So the motion of the girl relative to the disk (drawn in the horizontal direction but in general rotating with the disk) enters into both of these terms.

Posted by AngelaCoates on Wed 05 of Mar, 2008 16:44 EST So I’m having trouble relating the acceleration in the truck to the acceleration at B.... any ideas?

Posted by AngelaCoates on Wed 05 of Mar, 2008 16:54 EST Also, this is a question about the homework in general. When we’re doing our homework, do you want us to leave our answers in terms of variables? The homework problems posted online a lot of times don’t have the actual values written in the problems, but then the problems in the book do.

Posted by D. Dane Quinn on Sat 08 of Mar, 2008 12:31 EST Hi all, In general I like to work out problem solutions in terms of variables. I find that I can often simplify the expressions when I work in terms of variables. However, I also substitute in the given variables in the homework solutions. I don’t really mind if you don’t work in terms of variables. Ultimately for problems in which the numeric values are given, the numerical answer is what is evaluated.

Posted by ZachRomanin on Wed 05 of Mar, 2008 17:46 EST Hey Dr. Quinn, on problem 2 of the homework, i have my equations for the safe and boy complete, but i cant figure out how to determine the actual value for acceleration, thru the pulley equation i did figure out that 2Vs+Vb=0, so this corresponds to acceleration, such that 2As+Ab=0, but what do i do now? they don’t give us the acceleration the boy or the safe? any help would be sweet

Posted by AngelaCoates on Wed 05 of Mar, 2008 17:53 EST ...I’m not Dr. Quinn but I can tell you what I did... I solved for As in terms of Ab, and I used Newton’s second law to come up with an equation for the force the boy exerts on the rope in terms of his weight and his acceleration... Then you find that the tension applied to the safe is equal to the force the boy applies since the pullies are massless. So when you use F=ma on the safe, you can substitute those so that the only unknown is the acceleration of the boy. I can’t guarantee that this is the right way to do it, but I got the right answer

Posted by EugeneLeson on Thu 06 of Mar, 2008 15:53 EST I’m am also getting stuck on how to the acceleration of block “b” in problem #5. Do you use trig to find it or is it another way i am not thinking of??

Posted by ZachRomanin on Thu 06 of Mar, 2008 22:45 EST > I’m am also getting stuck on how to the acceleration of block “b” in problem #5. Do you use trig to find it or is it another way i am not thinking of?? I had the same question about problem 5, so i went and talked to Dr. Quinn, what i found out was that if you use the kinematics equations for velocity and acceleration with respect to the pulley you can find the velocity and acceleration of a point on the rope, now i haven’t actually finished the problem yet, so i can’t say i got it to work, but hopefully we can both figure it out with this lil tip he gave me.

Posted by D. Dane Quinn on Wed 12 of Mar, 2008 11:28 EDT The homework, which was originally due on Friday March 14, will now be due after the break...

Posted by ryannaymik on Fri 14 of Mar, 2008 21:45 EDT Dr. Quinn, I am having trouble getting started on problems 6 & 7. I have defined coordinates systems and labeled the forces. Can you give me some advice?

Posted by D. Dane Quinn on Thu 20 of Mar, 2008 20:54 EDT Advice? Buy low, sell high...but that’s probably not what you were looking for. How about this: once you have defined an appropriate set of coordinates, identify what the acceleration is in terms of these. I would probably use polar coordinates for each problem. Then, try to identify the forces acting on the particle using a free body diagram, and put it all together, with the sum of the forces equaling the mass times acceleration. If you have done the first two steps correctly, then everything should already be there. Finally, you need to express momentum balance in terms of a single set of directions, and I would go with the directions used with the polar coordinates.

Posted by JeremyCessna on Thu 20 of Mar, 2008 20:27 EDT Dr. Quinn, On the first problem is there a normal force at the pully A. I think there isn’t becuase we are assuming the pullies are massless but if there isn’t a normal force then the freebody diagram for the pully at A doesn’t make sense. P.S. What date will the HW be due? Thanks,

Posted by D. Dane Quinn on Thu 20 of Mar, 2008 21:05 EDT The pulley at A has two forces that can be easily identified, as well as one moment. The moment is applied by the motor and balances out the force that arises from the cable. The second force serves to anchor the pulley to the wall. I assume that it is this latter force that you are referring to as the normal force...so it certainly exists. A good thing to remember: momentum balance must always hold on every part of the system. If it has no mass then the sum of the forces vanishes, but this still satisfies momentum balance. Having said all of this, you should be able to work through this problem without really considering what happens at A. The kinematics is given directly in terms of the fuel assembly, instead of the motion of the motor. So knowing the acceleration of the assembly, we should be able to find the tension in the cable. The homework is due March 24...

Posted by D. Dane Quinn on Wed 26 of Mar, 2008 18:39 EDT Homework...just like the Energizer Bunny it keeps going, and going, and going...

Posted by ChristinaS on Wed 26 of Mar, 2008 21:24 EDT Where is the force supposed to be applied to the system at?

Posted by JenniferWelch on Thu 27 of Mar, 2008 10:37 EDT I’m having trouble with this problem. What is the mass of block A? How do I relate the force F to the veritcal motion of Block B?

Posted by D. Dane Quinn on Sun 30 of Mar, 2008 17:00 EDT I’m running out of clever things to say...just go download the homework. Find enough clever things to say, and you’re a Prime Minister; write them down and you’re a Shakespeare. George Bernard Shaw

Posted by D. Dane Quinn on Mon 07 of Apr, 2008 22:25 EDT Yep...its that time again... If life was fair, Elvis would be alive and all the impersonators would be dead. - Johnny Carson

Posted by AngelaCoates on Wed 23 of Apr, 2008 18:05 EDT For problem 7, any ideas on how to find the acceleration of the mass center of the spool?

Posted by JustinButterfield on Thu 01 of May, 2008 10:28 EDT I was wondering if you had any suggested problems out of the book for chapters 18 and 19 since we did not have any homework over that material. If you could, preferably problems that relate to the final or that are similar.

Posted by D. Dane Quinn on Thu 01 of May, 2008 17:12 EDT I just posted a number of suggested problems from the last chapters to help in your preparation for the exam. Cheers... p.s. Here’s a link to the Tacoma Narrows bridge collapse... Oh, this is cool...

Posted by NoahKlinksiek on Tue 06 of May, 2008 19:38 EDT > I just posted a number of suggested problems from the last chapters to help in your preparation for the exam. Cheers... > > p.s. Here’s a link to the Tacoma Narrows bridge collapse... > > Oh, this is cool... Will the questions on the final be more geared towards the last chapters we did or a mix of everything?

Posted by D. Dane Quinn on Tue 06 of May, 2008 20:45 EDT The exam tomorrow will be comprehensive, although it will be geared more toward the material in Chapters 17, 18, and 19...

Posted by JustinButterfield on Mon 05 of May, 2008 18:52 EDT Dr. Quinn, I’m having trouble on 17-103, basically my problem is relating the acceleration’s of the system. If you could possibly do a video solution of this problem that would be great, thanks.

Posted by D. Dane Quinn on Tue 06 of May, 2008 13:50 EDT I generated the solution for 17-103, and it should be posted by 2:30pm...

Posted by ryanfearon on Tue 06 of May, 2008 14:18 EDT Hey, wondering if the video solutions are working for everyone else? For me they all play for like 15 seconds and then just freeze up...

Posted by RobMilk on Tue 06 of May, 2008 15:02 EDT > Hey, wondering if the video solutions are working for everyone else? For me they all play for like 15 seconds and then just freeze up... Yes, Im having that issue as well. This is bad timing.

Posted by ZachRomanin on Tue 06 of May, 2008 16:48 EDT > Hey, wondering if the video solutions are working for everyone else? For me they all play for like 15 seconds and then just freeze up... yep same thing happening for me too

Posted by JustinButterfield on Tue 06 of May, 2008 17:43 EDT > Hey, wondering if the video solutions are working for everyone else? For me they all play for like 15 seconds and then just freeze up... Yeah same thing for me too and I’m using Mozilla.

Posted by D. Dane Quinn on Tue 06 of May, 2008 20:47 EDT It is 8:45pm and they seem to be working fine for me. Is anyone else still having this problem?

Posted by NoahKlinksiek on Tue 06 of May, 2008 21:53 EDT > It is 8:45pm and they seem to be working fine for me. Is anyone else still having this problem? not working for me

Posted by ryanfearon on Tue 06 of May, 2008 21:45 EDT > the videos still arent working, i tried on firefox and explorer..