Recent content by marksyncm

##M## is the underlying incline, while ##m## is the sliding block. Is the net force on the underlying incline to the left?

Just a quick thread bump, hoping someone can still help. Thank you.

Actually, I have another related problem that's similar enough I thought I'd add it here rather than start a new thread - hope that's OK: Homework Statement A particle's movement along a curve is described in polar coordinates by ##r(t) = bt## and ##\varphi (t) = \frac{c}{t}## (##b## and...

Nevermind; it was a silly question :) (It was necessary to differentiate the unit vector ##\vec{e_p}##.)

Homework Statement [/B] A particle is moving along a curve described by ##p(t) = Re^{\omega t}## and ##\varphi (t) = \omega t##. What is the particles transverse acceleration? Homework Equations [/B] None The Attempt at a Solution [/B] The position vector is ##Re^{\omega t} \vec{e_p}##...

To make sure, in this part above, ##a_n## refers to any (and all) term of the sequence, and ##a_{n+1}## refers to the term that follows it?

Sorry for being dense, but doesn't this mean that we need to find out IF ##0 < a_n < 2##? Because ##0 < a_n < a_{n+1} < 2## seems to follow from ##0 < a_n < 2## (isn't that what implication means? IF ##A## THEN ##B##, meaning we want to check IF ##A## because if it is, then ##B##?), then don't...

Thank you for the response. I am not sure I follow. Could you show what you mean here? I can't seem to find a way to show that ##0 < a_n < 2## without first showing that ##a_{n+1} < 2## (because the ##a_n## term is "hidden" under the square root term that describes ##a_{n+1}##.

Decided to ask a follow-up question here rather than start a new thread, hope that's OK. I'm reading up about finding limits of recursive functions. For example, ##a_1=\sqrt{2}, a_{n+1} = \sqrt{2a_n}##. As far as I understand, the procedure is as follows: 1) ##\lim_{n \to \infty} a_{n+1} =...

Thank you.

I understand that when a sequence is described recursively, for example: ##a_1=2, a_{n+1} = \sqrt{3a_n}## then we mean that the first term is 2, the second term is ##\sqrt{3*2} = \sqrt{6}##, the third term is ##\sqrt{3*\sqrt{6}}##, and so on. What I do not understand is how to interpret the...

To add, for now I'm assuming above that block ##m## is stationary with respect to block ##M##.

I decided to approach the problem from a non-inertial frame of reference as it seems easier (to me) for this problem: For starters, I would like to see if I finally managed to get the normal force right. Assuming the incline ##M## is accelerating to the right, the sliding object ##m## is...

Thanks to you both, I must rest now but will give this another serious shot tomorrow based on your suggestions.

I think I see the issue. I assumed that ##N- mgcos(\alpha)=0##. Now that the incline ##M## is moving, it is exerting a force ##ma## onto block ##m##, which has a component of ##masin(\alpha)## perpendicular to the incline surface. This makes ##N = mgcos(\alpha) + masin(\alpha)##. Is this correct?