Tablecloth Trick

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.

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.

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 equations of motion. 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## we get ##T_1 \approx 0.35## sec, which seems plausible.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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