I found this link with a collection of very interesting physics problems from Duke University's physics department. Maybe some of the people in this forum will find it useful either as educational reference, or simply entertainment.
There are some puzzling problems there. I'm curious if anyone knows what the answer to #3 is? (You can either post it hidden as a spoiler or PM me.) I don't see how such a heavy sandbag could possibly be blown that distance with just a drinking straw.
I would simply blow pulses on the sand bag in resonance with its swing (a constant) until it knocks the bottle over. Might take awhile. John Edit: @cmartinez Darn, you type fast. I didn't have time to even read the question twice. JPA
Ah, that gave it away. I don't think I would have ever thought of that. I guess I have a lot to learn.
For beginners in electronics, I found problem #34 very interesting. EDIT: Problem #52 is also worthy of careful consideration
Here's my answer to #34. Couldn't find a symbol for an incandescent bulb, but the principle is the same. The diode across the bulb shorts it, so it doesn't light with that part of the cycle. Damn Eagle took forever to draw. John
I had a hard enough time finding the SPST switches, and I use Eagle all the time. I did think about a resistor, but then I would have wanted to make a new device with little arrows to indicate the light. We make our own misery. Bottom line, I am not real fast at the computer. John
This is a bit OT. I lasted a little more than a month after retirement before I was begging for something to do. Volunteered teaching for a couple of years, then bought a hobby farm that needed lots of work. My involvement in electronics is proportional to how cold it is. John
I find Challenge #18 "Volume of a Holey Cube" intriguing. Consider a solid silver cube whose side has length L=4 cm. If three holes of diameter D=3 cm are drilled completely through, and perpendicular to, the centers of all the faces of the cube, what is the volume V of the remaining metal in the cube? I haven't found a solution yet.
I hadn't thought of that... down here's 16°C (61°F) and by Thursday it's going to be a rather nice 22°C (72°F) Did I tell you that I lived in Ohio for a year back in 1981? I stayed in LaGrange, that's about 20 miles south of Cleveland and Elyria ... and I DO remember how cold it gets up there... good Lord... I still get the shivers from those memories... Come to think of it... if I had stayed there I'd probably be an electronics expert by now if your rule applied to me too!
Neither have I... the question is: what's the volume of three equal cylinders that intersect orthogonally?
Lot of this, the cube, stuff is covered in a copy of Machinery's Handbook. Or at least the formulas for solving it are there. An older copy is a good thing for anyone interested in "engineering" to have. The new ones are expensive, but the common everyday stuff never changes, so an old one will serve for most things.
As a start, I would reduce it so a simpler problem. Imagine an inner cube that is the same size on each side as the diameter of the holes and this is centered in the main cube. Now imagine that the holes are only bored down to the face of the inner cube. The amount of material removed at this point is trivial to determine. Now we can work with just the inner cube. If bore our three holes in it, we are left with just the material in the eight corners and finding the limits of integration for one of them is quite straightforward since we can locate the cube relative to a coordinate system to make the one we are interested in convenient. I'm guessing that cylindrical coordinates will be the easiest one to carry out the integration in. Note that there is symmetry out along the axis from the center of the cube through a corner that might be exploitable, too.
The volume of the holey cube is equal to the volume of the solid silver 4 cm cube minus the quantity of three times the volume of a cylinder 3 cm in diameter and 4 cm long, minus three times the volume of a 3 cm sphere, plus the volume of 1, 3 cm sphere. If you take the volume of one cylinder and subtract the volume of the middle sphere.....that gives you the two end cap volumes. Repeat for remaining cylinders....add one sphere volume for the center, subtract all that from the starting volume..... and presto.......you have the volume of the remaining holey cube. Right?
I was thinking something similar, but I don't think it's that simple. If it were, then you could make a sphere by taking a cube and turning it in a lathe to remove one cylinder, then turn it 90 degrees and repeat removing another cylinder, and then do that one more time for the remaining axis. I could be wrong, but I don't think the result will be a sphere. Even if it were, I think that there is some double counting going on.
It seems simple to me. That's why I subtracted 1 sphere from each cylinder. And then only added 1 back in. Because there is only one. I read some of the links' examples.....I believe a lot were trick questions.
So you are claiming that if I take a cube in turn it on a lathe along three orthogonal axes that the result will be a sphere?