Consider a uniform rod initially kept vertical over a horizontal surface which has some coefficient of friction. The rod is now slightly disturbed from its position. This experiment is conducted for different values of coefficient of friction. When the surface is frictionless the bottom of the rod slips backward. For what condition of the coefficient, the rod slips forward?

**Note:** Mass and length of the rod are given.

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TopNewestBy the way, how does it slip forward...? – Kishore S Shenoy · 4 months, 3 weeks ago

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@Tapas Mazumdar Here is the link to the experiment for better understanding, however no explation is given. http://aapt.scitation.org/doi/abs/10.1119/1.4934950?journalCode=ajp – Aaron Jerry Ninan · 4 months, 3 weeks ago

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Assuming that your question translates to:

Then your answer will be

\[\mu_s = \dfrac{3 \sin \theta \cos \theta}{1 + 3 \cos^2 \theta} \] – Tapas Mazumdar · 4 months, 3 weeks ago

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I don't think such a situation is possible. Assuming that by 'disturbance' you mean the slight amount of angular displacement provided to the rod from the vertical (keeping the bottom end fixed), then the first situation is possible. But no matter how much you increase the coefficient of friction, it will not slip in the direction of the angular displacement. Friction (if sufficient)can only stop the rod from slipping backward in the second case and the rod will fall in a manner such that it's center of mass describes a quadrant. – Tapas Mazumdar · 4 months, 3 weeks ago

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– Kishore S Shenoy · 4 months, 3 weeks ago

Ya. It won't slip forward.Log in to reply

I think one of my questions has a similar concept, possible it will help, if it's a doubt. Cheers! – Kishore S Shenoy · 4 months, 3 weeks ago

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@David Mattingly @Josh Silverman @Harsh Shrivastava @shubham dhull @Kishore S Shenoy@Tapas Mazumdar @Aditya Kumar @Steven Chase @Spandan Senapati @Kushal Patankar

Please join in this discussion!! – Aaron Jerry Ninan · 4 months, 3 weeks ago

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