If you have a body on a table,with a velocity $u_0$, going from a non friction area to a friction area, how do I find the function for velocity and acceleration due to the variable friction force(T). I say it is variable, because at any time the mass on the friction side of the table changes.

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Assuming the body comes to rest before it completely enters the region, the body will trace out a partial sine curve in the space of $(t,x)$. Suppose the body is of length $L$ has uniform mass $M$. We let $x$ correspond to the positive horizontal distance from the interface. Then the force of kinetic friction (with coefficient $\mu_k$) is

\begin{cases} -\mu_k Mg\frac{x}{L} & 0 \le x \le L, \quad \dot{x} > 0 \\ -\mu_k Mg & x > L, \quad \dot{x} > 0 \\ 0 & \text{else} \end{cases}

The equation of motion in the region $0 \le x \le L$ is then

$$ M \ddot{x} = - \mu_k Mg\frac{x}{L} $$

In this region, the solution is

$$ x(t) = A \cos(\omega t + \delta), \quad \omega^2 = \mu_k g/L, $$

with constants $A, \delta$ to be fixed by the initial conditions on $x(t)$ and $\dot{x}(t)$. Note that this result is only valid for solutions that stay within the region $\mathcal{D} \in \mathbb{R}$, where

$$ \mathcal{D} = \left\{(x, \dot{x}) \middle | \ 0 \le x \le L, \quad \dot{x} > 0 \right\} $$

  • $\begingroup$ I found the same result, but this means that the body oscillates(?).It has to stop. $\endgroup$ – cianan1705 _c_ Jul 28 '15 at 20:22
  • $\begingroup$ I think is -μMg else, and no zero $\endgroup$ – cianan1705 _c_ Jul 28 '15 at 20:28
  • $\begingroup$ Sorry, fixed up my post. Let me know if there are any other mistakes. $\endgroup$ – Ultima Jul 28 '15 at 20:31
  • $\begingroup$ I can't understand why the equation does not depend on the mass of the object.Has to do with the costand A? $\endgroup$ – cianan1705 _c_ Jul 28 '15 at 20:38
  • $\begingroup$ To first order, friction is proportional to the perpendicular force a surface exerts on a massive object. In this case, the normal force is equal and opposite to gravity. Since objects fall with the same acceleration in a gravitational field regardless of mass, so too do they experience friction (to first order) with a horizontal surface without regard to mass. $\endgroup$ – Ultima Jul 28 '15 at 20:44

This is a subtle problem. Assuming that the normal force that is involved in friction is strictly dependent on how much of the block is over that surface, it is seen that the friction force increases steadily as more and more of the block moves over that surface. This means that the amount of friction force and the acceleration are dependent on the fraction of the block that are on the frictional surface. Any time that the amount of something is dependent on how much of a related variable is there, I would expect the answer to be an exponential form, not a sinusoidal form. This suggests that the answer for acceleration is of the form a = a-final * (1 - k1 e ^ (-k2 t)). My differential equations are a bit rusty at this point, so a little math help would be appreciated.

My apologies to Ultima. After developing a numeric solution for this problem, it does in fact require a trigonometric solution. The distance that the block travels would be described by a sin function, the velocity would be described by a cos function, and the acceleration would be described by a -sin function. The BIG difference between this answer and the "normal" trigonometric answer occurs because the solution to this problem is NOT periodic. The trig functions mentioned above describe the motion of the block while it is entering the frictional area of the diagram. Once the block is completely in the frictional area of the diagram, the trig functions no longer apply, and the problem becomes a standard "sliding block with friction" problem.

  • $\begingroup$ You mean that you did an approximation of the solution of the differential equation?Thanks for getting me out of troubles, because i was ready to perform the experiment... I had the same induition about the expodential form. Your second paragraph was very helpfull. $\endgroup$ – cianan1705 _c_ Jul 30 '15 at 7:45

Mathematically, the friction is described by a couple of Heaviside functions $\Theta$, asuming your body is of length $d$ and the transition from non-friction to friction occurs at $x_0$ with $x_1 = x_0 - d/2$ and $x_2 = x_0 + d/2$.

$~~~~~~~~~F_\mu (x,v) = -\mu \cdot v \cdot \left[ \Theta(x-x_1) \cdot \Theta(x_2-x) \cdot \cfrac{x-x_1}{x_2-x_1} + \Theta(x-x_2) \right]$

The rest is fairly simple: just integrate $F = \dot p$ (as usual) with your specific integration constants (starting conditions).

  • $\begingroup$ Really intresting approach, i'll try it.Thanks a lot! $\endgroup$ – cianan1705 _c_ Jul 30 '15 at 7:38

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