Recent content by AzimD

Thank you all! @erobz @kuruman @haruspex

I also attempted to solve part C. From the FBD's of each block I got: ##T_B-m_2g=m_2a## and ##m_1g-T_A=m_1a## using the relationships provided, I was able to obtain the answer: ##a=g\frac{m_1-m_2e^{\mu_k\pi}} {m_1+m_2e^{\mu_k\pi}}##

@haruspex @kuruman I realized I had a couple of mathematical errors as well as an error in conceptualizing the problem. Here's my reattempt at it: ##T_1=T_A=m_1g## and ##T_2=T_B=m_2g## and ##T_B=T_Ae^{-\mu_s\theta}## so ##T_A=T_Be^{\mu_s\theta}## Thus: ##m_1g=m_2ge^{\mu_s\pi}##...

Okay, I understand that now. So to attempt this again I know that: ##T_1=T_A## and ##T_2=T_B##. I also know that ##T_B=T_Ae^{-\mu_s\theta}## ##m_1## at the point where it begins to slide would mean that ##T_1## and ##T_2## are equal correct? So would that mean: ##m_1g=m_2ge^{-\mu_s\pi}## and...

On question B, I've attempted a solution that I have posted. However, I don't think it's correct. Am I allowed to treat this rope that wraps around the rod as a "negative" force and simply attach it to the other side of the equation such that ## m_1g=m_2g+\mu_sN##?

What is the Free Body Diagram on the last block and what is the only Force acting on it? What must be the value of the Force acting on it such that it moves at 10 ## m/s^2##?

I accidentally deleted my initial post so here's the repost: Ask these questions to yourself: 1) How are the force ##F## and the tension ##T_1## related? (Hint: Consider the relation between ##F## and ##T_2## and then ## T_2## and ##T_1##) 2) Of the two masses involved with ##T_1## which of them...

##T(0)## for ##m_1## would be at length ##l## no? Of course, the force caused by ##m_2## isn't added in because I come back to add that in later.

For T(0) I have ##T(0)=-m_1\omega^2l## Should this be a positive value?

So going from here then: ##T(r)=m_1\omega^2(\frac{r^2}{2l} -l)## Then considering the Tension caused by the ball ##T_b=-m_2\omega^2l## We can combine the two tensions and get the actual T(r) to be ##T(r)=m_1\omega^2(\frac{r^2}{2l} -l) - m_2\omega^2l## Am I conceptualizing this right? Or...

Thanks for checking my answer! Yep, I recently found the LaTeX Guide! I'll be attempting to use it from here on out.

For this portion of the problem, I managed to get T(r)-T(0) =(m_1* omega^2 * r^2)/2l

Got it. This is where I learned the constraint condition method. Perhaps you can use it to answer your question! https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/pages/week-4-drag-forces-constraints-and-continuous-systems/12-2-pulley-problem-part-ii-constraint-condition/

I believe I've gotten it right this time.

I'm not sure if I fully understand your question... was the question rhetorical lol