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I've stumbled across this delightful and difficult collection of problems, by Jaan Kalda. The following problem has stumped me. (It's probem 16 on the sheet, which I have provided as a link)

http://www.ioc.ee/~kalda/ipho/meh_ENG.pdf

A block is situated on a slope with angle $\alpha$, the coefficient of friction between them is $\mu > \tan \alpha$. The slope is rapidly driven back and forth in away that it's velocity vector $\vec{u}$ is parallel to both the slope and the horizontal and has constant modulus $v$; the direction of $\vec{u}$ reverses abruptly after each time interval $\tau$. What will be the average velocity $w$ of the block's motion? Assume $g\tau \ll v.$

diagram

The author also enlists some relevant and helpful ideas, which I'm not listing here but one of them has to do with perturbation method. He also mentions that the last assumption means that the block's velocity doesn't change much in the time interval $\tau$, so the zeroth approximation is that of a block moving in a straight line with constant velocity. He then says that we can calculate the average value of the frictional force based on the motion obtained in the zeroth case.

I don't understand how that can be done, so any hint or idea would be immensely helpful. I've been trying this problem for 2 months now and just to ensure that I am not missing some maths, I've gone and read perturbation theory, but it hasn't been of any help.

How do I proceed now?

The answer given, is $\frac{v}{\sqrt{\mu^2 \cot^2 \alpha -1}}$.

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    $\begingroup$ Hi Aritra and welcome to the Physics SE! Please note that this is not a help site for homework or exercises. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Nov 5 '15 at 17:46
  • $\begingroup$ @JohnRennie I know that, but I don't really want help on this particular problem. Maybe I couldn't make it clear enough, but what I really want is some insight into how perturbation method applies to mechanics problems, apart from the classic examples (like the ones about gravitational orbits). $\endgroup$ – Aritra Das Nov 5 '15 at 17:59
  • $\begingroup$ @Aritra - I'm a bit suspicious of the answer given. It contains the shaking velocity v, the coefficient of friction, and the slope angle BUT it doesn't contain the gravitational acceleration g. Doesn't that strike you as strange? It seems that the answer is saying that the average velocity of the block would be the same if the exact same experiment were performed on either the Earth or the Moon or a planet with hardly any gravity at all. I find that hard to believe. $\endgroup$ – user93237 Nov 5 '15 at 19:58
  • $\begingroup$ @SamuelWeir I think you should visit the link I provided, and have a look at the things mentioned after the problem. There were some comments about disregarding the acceleration due to gravity as $\tau$ is quite small with respect to $\frac{v}{g}$. Maybe that's why $g$ is totally absent from the answer. $\endgroup$ – Aritra Das Nov 5 '15 at 20:16
  • $\begingroup$ @AritraDas - But that doesn't make sense. Gravity after all is the ultimate driving force in determining which way the block travels across the plane, right? So how could the magnitude of g be totally absent? As for the stated assumption that tg<<v, that's not violated in considering smaller and smaller g's. $\endgroup$ – user93237 Nov 5 '15 at 20:50
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The idea is that, roughly speaking, the block moves down the gradient of the slope.

Because the slope changes its direction of motion, it pushes the block left and right in roughly equal measure, and because it has a short period, the block never moves with significant horizontal velocity relative to the fixed axes. Thus, the horizontal speed of the block relative to the plane is always approximately $v$ and the average horizontal velocity is zero.

The velocity in the direction of the gradient may vary a small amount, but let us simply work with the average, $w$. The component of friction pointing up the gradient must be $F_w = m g \sin\alpha$ to prevent the block from accelerating. The friction force points opposite the velocity, so $\frac{F_x}{F_w} = \frac{v}{w}$.

Meanwhile, the friction force is $F = \mu N = \mu m g \cos\alpha$. Using $F^2 = F_x^2 + F_w^2$ we can eliminate the forces algebraically to find the answer you quoted.

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  • $\begingroup$ Why should the velocity go back to zero? And I think we have to ignore the $g \cos \alpha \tau$ change in velocity and that is the perturbed problem. Also, the answer given is $\frac{v}{\sqrt{\mu^2 \cot^2 \alpha -1}}$. $\endgroup$ – Aritra Das Nov 5 '15 at 18:23
  • $\begingroup$ @AritraDas Yes, I see I made a mistake before, and have updated the answer. I still don't see a need for perturbation theory. $\endgroup$ – Mark Eichenlaub Nov 5 '15 at 20:01
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    $\begingroup$ @Samuel, please just do the algebra. There are two expressions containing $F_x$. It can be eliminated. $\endgroup$ – Mark Eichenlaub Nov 5 '15 at 21:31
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    $\begingroup$ Aritra, the answer describes the steady-state solution. There is a transient period during which this analysis does not hold. The block begins sliding because initially, before the shaking begins, the block does not move. Once the slope starts shaking, the block moves horizontally because frictions pushes it. But the friction force is then exactly horizontal, so the gravitational force is unopposed and pulls the block down the plane. It accelerates down the gradient until it gets to $w$. $\endgroup$ – Mark Eichenlaub Nov 5 '15 at 21:33
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    $\begingroup$ @Samuel Thanks for getting me to think about it a bit more! $\endgroup$ – Mark Eichenlaub Nov 5 '15 at 22:57

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