Following is a discussion on the Reviewer item titled: Problem 531 | Friction. Feel free to add your own comments!
Home • Forums • Blogs • Glossary • Recent
About • Contact us • Terms of Use • Privacy Policy • Hosted by Linode • Powered by Drupal
About • Contact us • Terms of Use • Privacy Policy • Hosted by Linode • Powered by Drupal
Forum posts (unless otherwise specified) licensed under a Creative Commons Licence.All trademarks and copyrights on this page are owned by their respective owners. Forum posts are owned by the individual posters.
i wonder how would your Friction Resultant at point A in that direction is correct? there is any technique how to assured my friction assumption is correct?
and why your Rbcos30(2Lcos(alpha)) is rotated in different direction?
sorry for my bad english.
There is a "minimum" value of angle α that will cause the bar to impend to the left.
The best way is to calculate for two directions of impending motion (to the left and to the right), then compare the results. Pick the one which satisfies the required.
That equation is rotation for moment about the left end, not for friction.
I can feel you, hahaha, we are on the same boat, hehehe. The important thing here is that we're able to communicate and we understand each other.
thank you very much sir...hahaha.^-^ but i have one thing to clarify
i wonder how would your Friction Resultant at point A in that direction is correct?
-There is a "minimum" value of angle α that will cause the bar to impend to the left. you answered.
now, it always happen on inclined bar or any object has the same problem as that?
because when i try things and in my logical thinking it always impend to the right based on the gravity and the weight.
sorry if i'm annoying.T-T
The bar may actually move to the left or to the right depending on the value of alpha.
Imagine alpha to be very small, almost zero degree, if friction cannot hold it in equilibrium, the possible movement is that the left end will go down and the right end will go up, this is the direction of movement you saw.
Now think for very large alpha, say 75 degrees or more, the possible movement is that the right end will slide downward and the left end will slide upward.
If you are not certain which movement to use in your solution in finding the required value of alpha, simply solve the two impending cases (to the left and to the right) and compare the results of the two. Which value of alpha is larger, that would be your answer.
ah, i got it now..
there is no easy way to assume friction on inclined object. you need to solve different possibilities if your not sure.
thank you very much sir.it helped me a lot.
Add new comment