Recent content by Arghzoo

No problem! Good luck with everything else.

Yeah if you switch the bounds you get +5L^4/36

If you switch the bounds, you'd get the same answer, just positive. That should be right... integral(3*x^3 dx) from -L/3 to 2L/3 ?

What are your bounds?

Also, the M in the center of mass is not M given, but mass obtained when integrating the linear density across the rod. Working through the calculations, we find the center of mass at 2L/3, which verifies your guess.

The length could still be L, but you're right on with the -L/3 to 2L/3 if the mass of the rod is 1kg.

Something like that. You know how to perform a center of mass integral, right? integrate 3*x^2 dx from 0 to L and divide by M, giving L^3/M. So that's the center of mass. It seems like it should be from -2L/3 to L/3, but it's from -L^3/M to (L-L^3/M) because the actual calculated center of...

When making the substitution from the original form of the inertia integral you substitute M*R^2/L dr. But M/L is simply the linear density (for any uniform object). So when substituting in dm=ρ*r^2 dr you can use the 3x even though you have a negative lower bound because definite integrals...

Because the line density is ρ=3x, wouldn't the center of mass not be in the center? And so when evaluating the integral of inertia wouldn't the bound not be from -L/2 to L/2? You'd first have to find the location of the actual center of mass, and then evaluate from the endpoints when the center...

No problem! You're exactly right; the rope doesn't shorten, excluding it from the count. The mechanical advantage is still 3.

Yep you got it! It sounds complex but all pulleys do is decrease the effective load by redirecting force. It seems like you're making it easier by adding more direction changes and pulleys, but all that free work manifests in the longer distance you have to pull the rope through.

If you've gotten to the point where you know the rules for resistors, they should tell you how to find the equivalent resistance of two resistors in series and in parallel. 1 and 4 are series, and so are 2 and 5. They're directly connected to each other. In this process, you can reduce the...

By "collapsed" I'm talking about how the resistances of two or more resistors connected in a particular way can be made into one resistor with a resistance that represents the resistance of all of them

CWatters brought up the most important part: if the voltage at the ends are equal, no current can flow, and there might as well be empty space there. So, then the two top resistors and the two bottom resistors... Can be effectively "collapsed" into one top and one bottom.

Instead of drawing it as a triangle, think of the top 5 resistors as connected in two horizontal lines, with the vertical resistor bridging the middle. The bottom resistor directly connects with both terminals of the battery. So if you can collapse the top 5 resistors and find its equivalent...