Morin - Problem 2.3

Adar Kahiri
May 2020

Problem: A frictionless tube lies in the vertical plane and is in the shape of a function that has its endpoints at the same height but is otherwise arbitrary. A chain with uniform mass per unit length lies in the tube from end to end. Show, by considering the net force of gravity along the curve, that the chain doesn’t move.


Solution: This is a pretty interesting application of the mean value theorem. Consider an infinitesimal piece of the chain of length \(dl\) with weight \(dm\) (given by \(\rho dl\) where \(\rho\) is mass of the chain per unit length). The force of gravity at that point would be given by \(gdm = \rho g dl\). The component of this force with which we are concerned is the component that is parallel to the tube's surface at this point, as the perpendicular component would be cancelled out by the normal force.

Assuming the inside of the tube's surface is given by a differentiable function \(f\), the slope of the tangent to any point in the tube is given by its derivative, \(\frac{df}{dx}\). Now, in order to capture the steepness of the slope at this point as an angle (which will let us calculate the proportion of \(dmg\) which acts parallel to the tube's surface), we can think of the derivative as the height of a right-triangle with a base of one.

The hypotenuse of such a triangle would be \(\sqrt{1+(df/dx)^2}\), and since the component of gravity which acts parallel to an incline is given by \(mg\sin\theta\) where \(\theta\) is the angle the incline makes with the horizontal, the component of gravity which acts parallel to the tube at any point is given by \[\rho g dl \sin\theta = \rho g dl \frac{df/dx}{\sqrt{1+(df/dx)^2}}\] Our goal now is to prove that the sum of tangential components of gravity at every point add up to zero. Now, notice that the infinitesimal length \(dl\) is actually equal to the hypotenuse of the triangle above multiplied by \(dx\). After all, \(dl\) is the resultant displacement when one takes a step \(dx\) in the horizontal direction and a step \(\frac{df}{dx}dx\) in the vertical direction, and factoring out \(dx\) leaves us with the triangle we had before. Thus, we have \[\rho g dl \frac{df/dx}{\sqrt{1+(df/dx)^2}} = \rho g \sqrt{1+(df/dx)^2} dx \frac{df/dx}{\sqrt{1+(df/dx)^2}} = \rho g \frac{df}{dx} dx\] Remember that this gives us the tangential force of gravity at any point along the chain, meaning that integrating this function along the entire chain must give us the net tangential force of gravity: \[\rho g \int_a^b \frac{df}{dx} dx = \rho g [f(b) - f(a)]\] where \(a\) and \(b\) are the locations of the endpoints along the \(x\) axis. Remembering that the two ends of tube are the same height, \[f(b)-f(a) = 0,\] proving that all the forces on the chain cancel out, and that the chain doesn't move!