# Tablecloth Trick

1. Apr 2, 2014

### jbunniii

1. The problem statement, all variables and given/known data
This is problem 3.6 from K&K, 2nd edition. I think I solved it correctly, but I want to check my reasoning.

If you have the courage and a tight grip, you can yank a tablecloth out from under the dishes on a table. What is the longest time in which the cloth can be pulled out so that a glass 6 inches from the edge comes to rest before falling off the table? Assume that the coefficient of friction of the glass sliding on the tablecloth or on the tabletop is 0.5.

2. Relevant equations
$F = ma$, and when there is relative motion between two surfaces with friction, the friction force is $\mu N$.

3. The attempt at a solution
I assume that the glass is not moving initially, and that its initial position is $x = 0$.

The tablecloth pulls the glass in the positive $x$ direction for $0 \leq t \leq T_0$. I assume that the motion is such that there is relative motion between the tablecloth and the glass, i.e. the tablecloth moves faster than the glass. Thus the friction force is $\mu N$. The direction of this force is in the positive $x$ direction during this time interval.

Key assumption 1 I am neglecting the fact that it's impossible to have a step function in velocity, so there's no way for the tablecloth to instantly start moving fast enough to achieve the maximum friction force of $\mu N$. [Edit: actually, maybe that's not an issue. As long as the tablecloth is pulled with a force which exceeds the friction force against the glass, then it should move relative to the glass. This is a bit tricky because the tablecloth is acted upon by friction against both the glass and the table. In any case, I am sweeping this complication under the rug and am assuming that the tablecloth moves relative to the glass starting immediately at $t=0$.]

Key assumption 2 Also, I believe the assumption is that the friction force is $\mu N$ as long as there is relative motion, regardless of how fast the motion is. In particular, it doesn't matter whether the velocity of the tablecloth is constant, and we don't need to know how it varies vs. time, as long as it is fast enough.

The tablecloth is free of the glass at time $T_0$. After this, the glass skids to a halt at time $T_1$. During this time, the glass is acted upon by friction against the tabletop, again with force $\mu N$ (same coefficient of friction). This time the force is in the negative $x$ direction.

As soon as the glass stops moving, the friction force jumps immediately from $-\mu N$ to zero. I think. This is because when there is no motion, the friction force counteracts any other horizontal forces. But there are no other horizontal forces in this case.

Therefore:

For $0 \leq t \leq T_0$, we have $m \ddot x = \mu m g$, or $\ddot x = \mu g$.

For $T_0 \leq t \leq T_1$, we have $\ddot x = - \mu g$.

We want to find $T_0$ and $T_1$ that will ensure that the glass is stopped at time $T_1$ (i.e. $\dot x(T_1) = 0$), and that at that time, it has reached the edge of the table ($x(T_1) = d$). In this problem, $d$ is 6 inches, but we'll do it symbolically for now.

Assuming this is correct, I integrate the acceleration equations to get velocity and position. This gives me
$$\dot x = \begin{cases} \mu g t & 0 \leq t \leq T_0\\ \mu g T_0 - \mu g(t - T_0) & T_0 \leq t \leq T_1\\ \end{cases}$$
and
$$x = \begin{cases} \frac{1}{2} \mu g t^2 & 0 \leq t \leq T_0 \\ \mu g T_0 t - \frac{1}{2} \mu g (t - T_0)^2 - \frac{1}{2} \mu g T_0^2 & T_0 \leq t \leq T_1\\ \end{cases}$$
where the constants $\mu g T_0$ in the second case for $\dot x$ and $-\frac{1}{2}\mu g T_0^2$ in the second case for $x$ were chosen to ensure continuity at $t = T_0$.

From here it's straightforward. We want $\dot x(T_1) = 0$ and $x(T_1) = d$. Solving two equations with two unknowns $T_0$ and $T_1$ gives us $T_1 = 2T_0$ (expected because the coefficient of friction is the same for both surfaces, so the velocity vs. time plot is an isosceles triangle), and
$$T_1 = 2\sqrt{\frac{d}{\mu g}}$$
The units make sense, and plugging in the numbers $d = 0.1524$ meters, $\mu = 0.5$, and $g = 9.8$ meters/sec^2 we get $T_1 \approx 0.35$ sec, which seems plausible. The tablecloth must be pulled in half that time: $T_0 = T_1/2 \approx 0.18$ sec.

Last edited: Apr 2, 2014
2. Apr 2, 2014

### TSny

:thumbs: Looks very good to me

3. Apr 2, 2014

### jbunniii

Great, thanks for checking it out.

4. Jul 17, 2015

### rivendell

hi, I did this problem too and got sqrt((2d)/(ug)) instead of what the first post said, 2sqrt(d/(ug)). The way I did it was just F=ma=friction, so umg=ma and a = ug. So using d = 1/2 (at^2), plugged a=ug in to get d = 1/2 (ug)t^2. finally, solved for t to get sqrt((2d)/(ug)). The first post's way looks beautiful but I don't totally understand how the integration was done for the T0 < t < T1 part or the last step. Is the result I got with my way right, or what am I missing here? Thank you!

5. Jul 17, 2015

### TSny

Hi, rivendell. In your solution you appear to be assuming that the acceleration of the glass is constant for the entire distance d. But the glass has two different accelerations during the time it travels the distance d. There is the acceleration while the tablecloth is still underneath the glass and then there is the deceleration as the glass slides against the bare table surface (after the cloth has been pulled from under the glass). The deceleration brings the glass to rest right at the edge of the table.