Wednesday, October 22, 2008

HW: New homework, new experiment

Hi, students and TAs of 6A, I created this blog to facilitate useful discussions pertaining to our on-going class. I have been receiving emails from students regarding homework problems or technical issues regarding the homework site. I've been thinking that an open forum like this may benefit more students.

I have never created any blog. So, this is a new experiment for me.

Please leave comments on this blog site if you have questions about homework or any topics worth discussing in relation to our class. Any students or TAs can leave comments and reply to other comments. Comments may be moderated. Any academic or technical topics related to class are welcome.

I would like very much to see interactions between students. If students interact within themselves and come up with solutions, that would be quite ideal and would benefit you the most in the long run. I will facilitate discussions, of course, and TAs are also welcomed to make comments as well.

OK, so much for the new experiment... The new homework (#4) is open for you. There are some extra credit problems. Good luck!

8 comments:

Alex said...

I have a question about the elevator ranking problem.

I just calculated the force on the
contacts (eg (m1 + m2) * g in order to look at weights 2 on 3 and 1 on 2
. 1 and 3 have no contact which makes them 0. THe floor is special because it provides the contact of all the weights and g so its (m1+m2+m3)g.

Although when I rank these weights I suspect my calculations are somewhat haywire. Any adivce is welcome. Thank you!

Zak said...

On homework problem 5.56, the one about the space station, I dont even know where to begin. Could i have a push in the right direction?

Sam said...

I have the following comment from an anonymous student for problem 4.51. Can any student please shed some light on this issue, to help out other students who might have a similar opinion/difficulty with this problem?
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how do we get the distance of stretch? the book doesn't really explain much about any equations or techniques to finding the distance in terms of horizontal movement...all there is scenarios when the spring is in the vertical direction.
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Sam said...

Zak and Alex, those are good questions... As a rule, I will wait about 1 day before I inject some comments myself. So, good students of 6A, please help out here, by sharing your wisdom. Give some hints about how to think about these problems, without saying explicitly about the final solution. Sharing doesn't hurt nobody, and actually it will help you a lot if you explain your knowledge, since the process of explaining will award you a better understanding of your knowledge.

Alex said...

About the air plane turn problem.

I made a reflection and realigned the vectors fg and fw. then I solved for fw without really plugging in any numbers. Then I sloved for fx as the missing component towards center, and came up with an equation - incorporating v^2/r for the acceleration. Thus I have an equation in terms of theta, v, r, because mass cancels out.
So I solved this and my answer was 5837 using a 40 degree angle an 261.1 m/s as velocity. Masteringphysics says im wrong, at 5800 5900 and 6000. What the heck am I doing?

Zak said...

Alexander,

Try re-reading the problem. There is a piece of information that you are missing. The angle cannot EXCEED 45 degrees.

Hope that helps

Alex said...

In my case, the given angle limit was 40 degrees. Therefore, taking into consideration the circular motion of the plane at given velocity the angle 40 degrees should yield the minimum radius. The more the degrees the sharper the turn - meaning a smaller radius. (So limit of angle is smallest radius) Do you agree?

Sam said...

Elevator Ranking Prob: Please start from the top object (3) and work out the free body diagram for each object. In so doing, keep in mind that normal forces act in both ways, and they are Newton's 3rd law pair. First, you will be able to figure out the normal force by 2 on 3, by considering the free body diagram for 3. Now that gives you the normal force by 3 on 2, by Newton's 3rd law. Then, when you can consider the free body diagram of 2, the only unknown is the normal force by 1 on 2, and this can be easily obtained. And on and on.

5.56 : Please think about the free body diagram for a person who is standing inside that station, as pictured. What is the force that keeps the person in a uniform circular motion?

5.40: The numerical values may be different for different individuals ("random number" problem). I don't see anything really wrong with what you guys are saying, but perhaps the free body diagram is incorrect? This problem is almost identical to the "banking the road" example (of Chap. 5) that I did in class. Maybe that helps...