ME274 - Basic Mechanics II: Sample Exam Problem 29

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.

The following is a reverse chronological order listing of the posts for the course blog. To add a post, click here (when adding posts, be sure to add a "label" in the box at the lower right side of the post window). To add a comment to an existing post, click on the "Comments" link below the post.

____________________________________________________

Aug 24, 2008

Sample Exam Problem 29

Right-click here to download video onto your computer.

Please add your questions and comments below.

5 comments:

Robert Shingleton said...: In step 2 (N/E), when you defined "Tg=(M/cos(theta))Agx" from equation 1 and then substituted it into equation 2, I think you forgot to multiply it by sin(theta).; November 26, 2008 at 7:34 PM
CMK said...: Oops...you are correct, I left off the sin(theta).

Thanks for pointing out the error.; November 26, 2008 at 9:03 PM
nour said...: In the kinematics part, i think there s a shorter way to find aG.
We can find the instant center C of the bar which would be on the same vertical line of A and D such as AD=AC.
That would give us aG=alpha X r G/C
Then we can solve.; November 28, 2008 at 1:45 PM
CMK said...: --to nour--
Don't forget that instant center analysis, in general, works only for velocity and NOT for acceleration. That is, aG is equal to alpha x rG_C only if aC = 0 and if omega = 0. For this problem, omega = 0 since the bar is at rest. However, do you know if aC = 0??

It is recommended that IC analysis be used only for velocity.; November 28, 2008 at 4:43 PM
nour said...: Thanks for pointing this out. I thought aC would automatically be zero. Maybe i got the same answer because aC turned out to be zero anyway.; November 29, 2008 at 4:53 PM

Subscribe to: Post Comments (Atom)