Recent content by phantomvommand

I currently see ##T \frac {d\phi} {ds}## as normal force = ##Td\phi##, and this is spread over ds. Is this how it should be seen, or is there a way to see it as ##\frac {d\phi} {ds}## multiplied by T? Also, I think the it balances ##p\sigma + \sigma gcos\phi##

Thanks. I've used the tangential equation to obtain the text's solution, with the approximation ##s \sim r\phi##. However, I am curious about how you obtained the normal equation. I get that the quantity ##p + \sigma gcos\phi## must be balanced, but I do not see how it is balanced by ##T \frac...

@Chestermiller yep, I meant px as a product Your equation is correct, though of course under a different definition of theta.

Yes, p is uniform. but I do not see how the equations can be reduced to the text’s solution.

This has to do with your definition of ##\theta##. I defined ##\theta## to be the angle swept out from the bottom of the cylinder, so under my definition, in the diagram in #1, ##\theta## is obtuse.

##\theta## is the angle swept out from the base of the cylinder to the fabric at height x. Under my definition of ##\theta##, ##\theta## will be an obtuse angle in the provided diagram.

A cylindrical bag is made from a freely deformable fabric, impermeable for air, which has surface mass density σ; its perimeter L is much less than its length ##l##. If this bag is filled with air, it resembles a sausage. The bag is laid on a horizontal smooth floor (coefficient of friction µ =...

But how does ##\sqrt {2Rr} >> r## mean that pressures at heights above r will be smaller? Sorry, I wasnt clear about my final question. It should be why ##strain \sim \frac {\delta} {r}##. But I suppose if the claim that pressures above r are substantially smaller, then r will be the effective...

A solid ball of radius R, density ρ, and Young’s modulus Y rests on a hard table. Because of its weight, it deforms slightly, so that the area in contact with the table is a circle of radius r. Estimate r, assuming that it is much smaller than R. I have no issues understanding the...

I see, then the ratio of tensile to shear stress in the spaghetti rod would far exceed 2:1 (generally), since ##l >> d##. However, could you further explain why "thin rods usually break by snapping, not by shearing or pulling apart". Shouldn't a larger tensile stress cause it to pull apart...

Are both tensile and shear strength not fixed values for a particular material? My view is that while horizontal stress is indeed larger in this case, it depends on the maximum horizontal and vertical stress that the spaghetti can withstand too.

TL;DR Summary: A thin piece of spaghetti of diameter d is balanced horizontally from its middle. It can have a length ℓ ≫ d before it snaps under its own weight. How does ℓ scale with d? Let the spaghetti rod have density ρ, and consider ONLY its right half. (1) The spaghetti rod experiences...

Meaning to say, the x and y axes in the solution rotate, right? That clears it up. Thanks!

I am able to understand the textbook solution, except for its very first assumption: We use the coordinate system shown in the figure, and find the shape ofthe spring (assumed to have already attained its stable configuration) in this frame. Why is it fair to assume that the slinky will ever...

Oh right, forgot about that! Thanks!