1
$\begingroup$

Consider a mass $m$ at position $x(t)$ on a rough horizontal table attached to the origin by a spring with constant $k$ (restoring force $−kx$) and with a dry friction force $f$

\begin{equation} f= \begin{cases} F\quad \quad \quad \quad \quad \quad \,\,\,\, x<0\\ -F \leq f \leq F \quad \quad x = 0\\ -F \quad \quad \quad \quad \quad \quad x>0 \end{cases} \end{equation}

What is the range of $x$ where the mass can rest? Show that if the mass moves, the maximum excursion decreases by $2F/k$ per half cycle. Discuss the motion.

Up until now i have only dealt with SHM with no friction so I am a bit lost. Any help is appreciated.

$\endgroup$
  • $\begingroup$ Your friction term is not written correctly, you have two cases for $x>0$. $\endgroup$ – Ian Nov 28 '15 at 15:51
  • $\begingroup$ See this. $\endgroup$ – A.S. Dec 7 '15 at 12:40
0
$\begingroup$

Some Hints

The governing differential equation according to Newton's second law of motion will be

$$\begin{cases} - kx - F = m\ddot x & \dot x \gt 0 \\ - kx + F = m\ddot x & \dot x \lt 0 \end{cases}$$

If $\dot x=0$ then we should investigate that whether the particle moves or not due to the laws of static and kinetic friction. Rearranging the terms will give

$$\begin{cases} \ddot{x} + \omega^2 x = - \frac{F}{k} & \dot x \gt 0 \\ \ddot{x} + \omega^2 x = \frac{F}{k} & \dot x \lt 0 \end{cases}$$

where $\omega = \sqrt{\frac{k}{m}}$ is the natural frequency. The general solution to this ODE is

$$\begin{cases} x(t) = A \cos (\omega t) + B \sin (\omega t) - \frac{F}{k} & \dot x \gt 0 \\ x(t) = A \cos (\omega t) + B \sin (\omega t) + \frac{F}{k} & \dot x \lt 0 \end{cases}$$

Try to discuss the problem based on initial conditions.


Reference

Here is a nice article that you can refer to it. It explains everything from A to Z. You can take a look at part IV. There is a little mistake in equations 22a, 22b and 23a, 23b. The $\dot{x}$ should be written instead of $\ddot{x}$ for the signs. The text mentions this issue by words right above equations 22a, 22b so it seems to be a simple typo.

$\endgroup$
  • $\begingroup$ Could you provide a complete solution? $\endgroup$ – KingJ Dec 7 '15 at 0:21
  • $\begingroup$ @RSparkes: I added more hints. I want you to solve it by your self. :) $\endgroup$ – H. R. Dec 7 '15 at 9:18
  • $\begingroup$ You solution is incorrect as it's discontinuous. Also $sgn'=2\delta\ne 0$. $\endgroup$ – A.S. Dec 7 '15 at 10:04
  • $\begingroup$ What does F and f signify, also what is a dry friction force? Also what is sgn? $\endgroup$ – KingJ Dec 7 '15 at 10:42
  • $\begingroup$ @A.S.: Maybe you are right but I should think more. :) However, we are taking the derivative with respect to $t$ and not $\dot x$! Physically, we have a force that its direction depends on the sign of velocity. Please feel free to edit my answer if you are sure about something or post a new answer that we can learn from it. :) $\endgroup$ – H. R. Dec 7 '15 at 11:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.