Recent content by physicsman2

But why is it cos and not sin for the perpendicular Fg? I don't understand why in this case the angle is sin(pi/2 - theta) and not just sin theta like in the perpendicular Fw. Edit: I think I see why now, I thought theta would be the angle between the ladder and mg but I think theta would be...

I found the diagram online: https://egp.rutgers.edu/gifs/CJ/9-7.gif I originally did make the point where the ladder touches the ground the pivot point, but for some reason I don't understand why the force has a component when it's just the same force exerted. Wait, is it because you need to...

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Homework Statement A uniform ladder of mass m and length l leans at an angle theta against a frictionless wall. If the coefficient of static friction between the ladder and the ground is mu, determine a formula for the minimum angle at which the ladder will not slip. Homework Equations...

The problem is that π((14-2x)/π)2 = n[(14-2x)2/n2] and not n[(142 + (-2x)2)/n2] the way you solved it. Edit: Got beat to it.

when you apply the limit, you should get 1/infinity + 1/infinity, which is 0 + 0 = 0. A small number over a much larger number goes to 0. Edit: Instead of using L'Hopital's Rule, you could just say that the power of y in the denominator is larger than the power of y in the numerator, so since...

The first one is infinity because you're essentially dividing by a very small positive number since you're approaching 3 from the right. When applying the limit, you will get something like 4/(1 x 0^+), meaning your dividing 4 by a very very small number therefore the limit is infinity. The...

Pretty much this. Sorry for the confusion earlier, I wasn't thinking all that well until I was corrected. Remember that |t| also equals sqrt(t^2). |t| can also equal either t or -t the way Dickfore showed. Since you want to take out a t^2 and cancel to apply the limit, you change |t| to...

You're right, I feel stupid now for not realizing my mistake from the beginning. They should have left the t^2 in there and multiplied the expression by |t|/|t| to take the t^2 away in the numerator and leave a -t to be multiplied through in the denominator and apply the limit to get the answer.

That too. So the above is a better answer. Multiply by |t|/|t| and look at the situation, ie. where the limit is going off to, if you have a problem like the one you have with a square root, etc. in order to see if you need a +t or -t.

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If a = 0, 0! = 1

I know, just showing you another way of going about it, with thrill3rnit3 giving you a huge hint on how to go about it.

I would try: sec2x(sec x)(tan2x)(tan x)

You can do it whatever way you want, but I would use the quotient rule. It seems less confusing to me.