Welcome to the website of ME 274 for the Fall 2008 semester. On this site you can view blog posts, add your own blog posts and add comments to existing posts. In addition to the blog are links to course material: course information, information on solution videos, exams, quizzes, homeworks and other course-related material. Direct links to the homework solution videos are also available on the left side of this page.


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Nov 12, 2008

Solution to Problem on Example pg 71 in notes

The solution to the problem that came up in lecture on Monday (Example on page 71 of the notes) was that you can assume the Omega's of member AB and OB are equal but opposite in sign. (this makes the problem solvable because we eliminate one of the two variables).

The Omega's have to be equal because they are connected at point B.  If the omega's were different; the problem would "explode" or break at point B.

We cannot assume this at the exact instant that points A and O are alligned.  In the problem we are dealing with the instant right before A and O are alligned so it is still safe to assume the Omega for each bar is equal but opposite in sign.
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