Suppose we have some inclined plane, and there is some chain of balls of length $l$ and mass $m$ lying on it. No friction at all in the system.

1) What is $x_0$ (the vertical hanging part of the chain) needed for the chain to be in equilibrium?

Answer: $l/3$

2) If we push the chain a little bit to the right, what is the acceleration of the chain, at instant when the vertical part of the chain is $x$ long ($x > x_0$)?

Answer: $a(x)=\frac{g}{l}(\frac{3}{2}x-\frac{l}{2})$

3) What is the velocity of the chain at instant when all of it is vertical (e.g. all the left part on the inclined plane "became vertical")? (no answer)

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1) I do not understand, why the component of the weight of the left part should be equal to the weight of the right hanging part. Their vectors are not along the same line, so why should this condition be necessary for equilibrium? Also, the left part is being pulled by the right, so there should be another force along the incline.

2) Based on the first question, this one should be:

$(1-\frac{x_0}{l})mg \sin 30^{\circ}-\frac{x_0}{l}mg=ma$

However, this leads to a wrong $a$, according to the answer.

3) Based on the book's answer of 2), is this right?

From Torricelli's equation:

$v^2=v_0^2+2a \Delta x$

$v(x)=\sqrt{v_0^2+2 \int_{l/3}^{x} a(x) dx}$



  • $\begingroup$ If you have access to the Feynman lectures (volume 1), there's a very nice physical argument explaining aspects of this problem. $\endgroup$ – Siva May 13 '13 at 3:26
  • $\begingroup$ @Siva - Feynman talks about Stevin's machine, with a chain loop of balls, as a concept of perpetuum mobile. It is not the same. $\endgroup$ – grjj3 May 13 '13 at 12:38
  • $\begingroup$ @grjj3 -- it is not the same, but "Stevin's loop" does provide a direct answer to your first question. Sometimes you have to morph the problem a bit to find an easy route to the solution... $\endgroup$ – Johannes May 13 '13 at 17:30

Stevin's loop (or the "epitaph of Stevinus") gives us the answer to the first part. That is, the loop will not move if its two sides span the same vertical distance down the triangle.

So its center of mass lies at the $y$-axis point one-half of the way down the vertical chain length (as well as one-half of the way down the slanted chain length -- the numbers are the same.) So from the top corner of the triangle, the center of mass is $l/6$ meters down while the chain is in the starting position.

After it slides so that the chain is totally vertical, the center of mass is $l/2$ meters down from the top corner of the triangle.

We have assumed the triangle to be a frictionless surface so that no energy is lost. Let us define the zero of potential energy to be the top corner of the triangle. Then conservation of energy tells us that $$PE_i+KE_i=PE_f+KE_f$$ $$(0-mgl/6)+0=(0-mgl/2)+mv^2/2$$ $$\sqrt{2gl/3} =v.$$

  • 1
    $\begingroup$ +1 for Stevinus alone. I first saw that argument in Feynman's books, and Never has so simple and short a bit of text and a scribbled picture brought me so much joy. $\endgroup$ – David H May 14 '13 at 2:14

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