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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Sep 22, 2008

Sample exam problem 12


Hi sir

In problem 12, is it accurate to say that the instant center for link CD is any point on the line CD?
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2 comments:

CMK said...

I can see why you might think so since the perpendicular to v_D intersects the perpendicular to v_C at all points along CD.

However, the IC is a specific point; a point on CD that has zero velocity. Only one point can have zero velocity [more than one would say that the link is completely stationary]. Here you need to think of a consequence of an IC: the speed of any point on that link is proportional to the distance between that point and the IC.

Therefore, the IC (point E) must exist at a point on CD such that v_D/DE = v_C/CE ( = omega_CD). That point is E shown in the figure. You can use similar triangles to find the location of E on CD.

Does this help?

September 22, 2008 at 10:42 PM
nour said...

yes it is,thank you.

September 22, 2008 at 11:06 PM

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