Pulling a tablecloth from under a glass

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In summary, the conversation discusses a question involving the acceleration of a cloth and a glass on a table. The question is stated and the conversation includes a typo, a clarification of the forces involved, and a discussion of the published solution. The summary concludes that the published solution is incorrect by a factor of √2 due to a missing distance calculation.
  • #1
Leo Liu
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Homework Statement
This is a statement.
Relevant Equations
##f=\mu_{k} N## ##F=ma##
1596128579431.png

The question is stated above.

I tried to solve it on my own and I obtained the following three questions:
$$\begin{cases}
a_{cloth}=\frac{F-\mu mg} M\\
\\
a_{glass}=\frac{\mu mg}m=\mu g
\end{cases}$$
Where small m is the mass of the glass and capital M is the mass of the cloth. However, the first equation does not appear in the solution given by my answer book. So I think I must have wrongly interpreted the question. I would like someone to explain the question to me. Many thanks.
P.S. The official answer is shown below:
1596129361786.png
 
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  • #2
Your second equation should be ##=\mu g##
The detailed acceleration of the cloth doesn't matter...only the time it takes. Presumably the "puller" can supply whatever force is required.
 
  • #3
hutchphd said:
Your second equation should be
It was a typo. Sorry.
hutchphd said:
The detailed acceleration of the cloth doesn't matter...only the time it takes. Presumably the "puller" can supply whatever force is required.
You have got the point. Thanks.

Update: Yet if I cross out the second equation, the time it takes to yank the cloth out is independent of the force. Since there is nothing to be maximized, what's the point of asking for the "longest time"?
 
  • #4
What "force" are you solving for?
 
  • #5
Leo Liu said:
Where small m is the mass of the glass and capital M is the mass of the cloth. However, the second equation does not appear in the solution given by my answer book.

Surely it is the second equation that does appear in the given solution?

As @hutchphd alluded to, the dynamics of the cloth are irrelevant. We essentially have a glass that accelerates uniformly at ##\mu g## and then decelerates, in a symmetrical manner, with a magnitude of ##\mu g##. The symmetry allows you to deduce the times for each stage are equal. The speed time graph is an isosceles triangle!

It is the longest time because the cloth can be pulled out arbitrarily quickly so that negligible impulse is initially imparted to the glass, causing it to come to rest well before the end of the table.
 
  • #6
Can the answer T = 1/4 s be correct? If the cloth accelerates the glass for 1/4 s, where will the glass be located on the table at the instant the cloth is completely out from under the glass?
 
  • #7
etotheipi said:
We essentially have a glass that accelerates uniformly at μg and then decelerates, in a symmetrical manner, with a magnitude of μg. The symmetry allows you to deduce the times for each stage are equal.
Not quite. I think the assumption is that without the cloth the friction is big so it decelerates "instantaneously"...makes the problem easier.too
 
  • #8
TSny said:
Can the answer T = 1/4 s be correct? If the cloth accelerates the glass for 1/4 s, where will the glass be located on the table at the instant the cloth is completely out from under the glass?

Did they forget to add the distance traveled during the acceleration phase?

hutchphd said:
Not quite. I think the assumption is that without the cloth the friction is big so it decelerates "instantaneously"...makes the problem easier.too

I will bite the bullet... doesn't the question say the coefficient of friction without the cloth (i.e. glass on table) is 0.5?
 
  • #9
etotheipi said:
It is the longest time because the cloth can be pulled out arbitrarily quickly so that negligible impulse is imparted to the glass, causing it to come to rest well before the end of the table.
Oh this makes sense. So we only pull the tablecloth during the first half of the journey. After the external force is removed, the friction changes its direction. Thanks James!
etotheipi said:
Surely it is the second equation that does appear in the given solution?
Meh. It's another typo :-p .
 
  • #10
etotheipi said:
Did they forget to add the distance traveled during the acceleration phase?
They let ##s_0## represent the distance traveled during just the deceleration phase. But they took this distance to be 6".
 
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  • #11
etotheipi said:
I will bite the bullet... doesn't the question say the coefficient of friction without the cloth (i.e. glass on table) is 0.5?
Oops. You are indeed correct. Doesn't that make the published answer incorrect by 2?)
 
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  • #12
TSny said:
They let ##s_0## represent the distance traveled during just the deceleration phase. But they took this distance to be 6".
hutchphd said:
Oops. You are indeed correct. Doesn't that make the published answer incorrect by 2?)

Yeah, they should have ##s = \mu g T^2 = 6\text{in}##
 
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  • #13
hutchphd said:
Oops. You are indeed correct. Doesn't that make the published answer incorrect by 2?)
Reading the official solution, it looks like they left out the distance covered while being accelerated by the cloth.
It makes the the answer too much by a factor √2.
 
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1. How is it possible to pull a tablecloth from under a glass without knocking it over?

This phenomenon is possible due to the principle of inertia. Objects at rest tend to stay at rest, and objects in motion tend to stay in motion. When the tablecloth is pulled quickly, the glass on top of it will remain in its original position due to its inertia.

2. Is there a specific technique to successfully pull a tablecloth from under a glass?

Yes, there are a few techniques that can help increase the chances of successfully pulling a tablecloth without knocking over the glass. These include pulling the tablecloth quickly and smoothly in a straight line, ensuring that the tablecloth is completely flat and not hanging over the edge of the table, and using a slippery or lightweight tablecloth material.

3. Can this trick be performed with any type of glass?

Yes, this trick can be performed with any type of glass as long as it is placed on a stable surface and the tablecloth is pulled quickly and smoothly. However, it is recommended to use a lightweight or plastic glass to reduce the risk of breakage in case the trick does not go as planned.

4. Why does the glass sometimes fall when attempting this trick?

If the glass falls when attempting this trick, it is likely due to a lack of practice or using the wrong technique. It is important to pull the tablecloth quickly and smoothly in a straight line to minimize any disruption to the glass's inertia. It may also be helpful to use a lightweight or slippery tablecloth material.

5. Are there any safety precautions to keep in mind when attempting this trick?

Yes, it is important to take necessary precautions to prevent any potential accidents when attempting this trick. Make sure the glass is placed on a stable surface and there are no sharp objects or obstacles in the way. It is also recommended to wear protective gear, such as gloves, in case the glass does fall and break. Additionally, it is important to practice this trick in a safe and controlled environment before attempting it in front of an audience.

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