Recent content by Electrowonder

Oh, I see you meant ##d[(R^2 +z^2 -2Rzu)^{-1/2}]##, I was a bit confused with the exponent outside as I haven't done this for 3 years.

I'm not sure what to do with the ##\int v du## term, what is v going to be?

Oh that's right, got it, thank you. That should prove helpful for other problems.

Thank you, I was able to fully solve it! That was a brilliant substitution, and now it seems almost obvious after the fact. But what thought process do I need to adopt to do smart substitutions like that in the future? Can someone advise me how to also do it by partial fractions as the author...

Seems quite obvious but I have never done substitution with a quadratic before. Usually, you substitute with something like ##s = R^2 + z^2 -2Rzcos\theta## but I realized that won't go anywhere nice. Using your way, I did $$d(s^2) = 2sds = 2 \sqrt{R^2 + z^2 - 2RZcos\theta} d(\sqrt{R^2+z^2...

$$\int \frac{z}{(R^2 + z^2 - 2Rzu)^{3/2}} du +\int \frac{-Ru}{(R^2 + z^2 - 2Rzu)^{3/2}} du$$ How would I integrate the second term? I have already done a U substitution, am I allowed to just do another substitution?

Sorry, that should be a +z^2 in the denominator. PeroK The original integral was $$\int \frac{\sin (\theta) (Z-Rcos\theta)}{(R^2+z^2 -2RZcos\theta)^{3/2}} d\theta$$ If I use the substitution s^2, how do I do find ds?

Oh, got it! Thank you!

I was doing this problem from Griffith's electrodynamics book and can't figure out how to do this integral. The author suggested partial fractions but the denominator has a fractional exponent which I have never seen for partial fractions, and so, I am unsure how to proceed. The integral I am...

To Chestermiller, thank you, but the problem is asking to show that the total force along the curve is 0. You did help me out though in writing the sum of the forces along the slope as ehild asked. To ehild: The sum of the forces along the slope is given by" $$ \sum F_{Tangential} = -\rho g...

I'll call the normal force on one piece of the chain attached to the neighboring piece ## N_{ConC}##. Summing the forces in parallel slope direction, one gets ## N_{ConC} - mgsin\theta = 0## Meanwhile, using what you told me: $$sin^\theta + cos^2\theta = 1$$ $$tan^2\theta~ cos^2\theta + cos^2...

##sin\theta = tan \theta cos \theta. ## Can you clarify what you mean by the last sentence? As far as Ia m concerned the force acting on the chain is just the weight of the chain, but I don't know how to incorporate that.

There is ##mg sin \theta## pulling it to the left and the frictional force holding it to right parallel to the slope with a magnitude ##\mu N## So the sum of the forces along the slope would be $$\mu N - mg~sin\theta = 0$$ Upon substituting the ##m=\rho \sqrt{1+f'^2}dx## we get: $$\mu N - \rho...

$$ T My apology, let's start from the beginning. Consider an infinitesimal piece of the chain between x and x+dx. The opposite side is given by f'dx. The hypotenuse of the chain can be found by Pythagorean's theorem. $$ c^2 = dx^2 + f'^2 dx^2 = dx^2 (1+f'^2)$$ Then ##c = dx \sqrt{1+f'^2}##...

Please see attached, c is the component of gravity along the curve (hypotenuse). Have I done anything wrong?