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Fall 2008 -- Purdue University -- West Lafayette, IN
Blog Post LABELS
- Blog Challenge Questions (9)
- Chapter 2 (2)
- Chapter 3 (19)
- Chapter 5 (11)
- Chapter 6 (23)
- Chapter 7 (7)
- Chapter 8 (9)
- Exam 1 (5)
- Exam 2 (4)
- Exam 3 (17)
- Exams (2)
- Final Exam (7)
- General (6)
- Gravity (2)
- Homework (2)
- in class (1)
- Lectures (1)
- Quizzes (2)
- Spring 2008 Archive (87)
- Spring-Mass Simulator (1)
- Student-Generated Solutions (1)
- Summer 2008 Archive (72)
10 comments:
When you summed the moments about point D, I think I am having a moment about myself and not understanding why this is in the k direction. Can you please explain this reasoning? Thanks a lot.
Actually after looking at it for another 2 minutes, I think I understand but can you tell me if I am correct or not.
Basically on all of these problems when you sum the moment it would be in the k-direction, or atleast on Sample problem 29. However non of those problems ask you to write it in vector form. When you do the right hand rule, my thumb is pointing "into" the page or in the k-direction. The right hand rule also gives the sign as well.
You really need to always write down the moments in vector form to get the correct direction (i.e., the correct sign) on the moments.
For this question, the total moment about D on the wheel is a sum of that due to F and that due to FBC. The moment about D due to F is CCW (and therefore, positive by the right hand rule). The moment about D due to FBC is CW (and therefore, negative by the right hand rule).
The moment about A on the bar due to FBC is CCW (and therefore, positive).
The right hand rule used above depends on the choice of coordinate axes. The axes that I drew (with x to the right and y up) has positive z OUTWARD (which corresponds to CCW moment). Had you used a different set of coordinate axis, you might end up with positive z inward (CW). To be safe, I recommend using coordinate axes that have z outward. This is easier to keep straight when working a problem under the pressure of time (exam).
I hope that this helps. If not, let us know.
Good question.
I'm confused, the problem says that the parts move in a horizontal plane, doesn't this mean that there would be no weight forces in this problem because they would be into the page?
--to john--
You are correct; horizontal plane does mean that you should not include weight.
I need to read my own questions a bit more closely! Thanks for pointing out this. Apologies for the confusion.
When using the moment for the bar AB, using the parallel axis theorem, why is L_ab/2 used instead of L_ab? Thank you.
In general, the parallel axis theorem says that IA = IG + m*d^2, where d is the distance between points A and G. In this case, IG = (1/12)*m*LAB^2 and d = LAB/2. Therefore: IA = (1/12)*m*LAB^2 + m*(LAB/2)^2 = (1/3)*m*LAB^2.
Does this help?
I do not understand why you do not need to know alpha_BC when it is the equation. I have it in both my i and j vector equations, so unless you assume a_D is 0, which should make alpha_BC 0, or if it is not supposed to be in the i_vector equation, how do you get an final answer with it in there since it is an unknown?
If you consider the "j" component of the kinematics equation for the wheel, you will see that aD = 0. The "i" component of that equation gives aC = - R*alpha_w.
When you combine the three vector kinematics equations, you will get two scalar equations ("i" and "j" components). These two equations involve alpha_w, alpha_BC and alpha_AB. You can eliminate alpha_BC from these two equations resulting in one equation relating alpha_w and alpha_AB. [Since the "j" component equation involves only an alpha_BC and nothing else, then alpha_BC = 0 as you have observed.]
Apologies if this was not clear from my discussion on the video.
I'm actually glad that you left this unsolved because personally the hardest part of these problems is keeping signs correct. I worked it through and found the correct answer.
Thanks
Alex
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