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I usually don't want to do this, but please go to this link, the solution is too big to post it here

http://engineering.union.edu/~curreyj/MER-201_files/Exam2_2_26_09_Solution.pdf

And go to page 5 of the pdf.

Briefly, the problem say

Determine the velocity of the 60-lb block A if the two blocks are released from rest and the 40-lb block B moves 2 ft up the incline. The coefficient of kinetic friction between both blocks and the inclined planes is $\mu_k$ = 0.10.

Things I am confused with the solution

1.First of all, I seriously thought lb was mass not force. after some googling, it turns out they are used interchangeably...

2.Where did they even get $2s_a + s_b = 0$ from? Why did they determine the change in distance this way? My first assumption was that if block B moved up 2ft, then block A should move down 2ft (the rope must "move" 2ft too right?). Then I wasn't sure, so I did a few triangles and found that the angle made a difference

3.Where did $2v_A = -v_B$ come from?

4 The FBD for block A is confusing, why is the friction force $F_A$ in the direction of the ropes? I thought it was block B that is going down? Am I the only one who had trouble deducting that the pulley and block A are the same object?

5.Look at the final answer, how could $v_b$ be negative? The problem says block B goes UP.

If you are wondering, this is not homework. I am just interested in this problem and it is out of curiosity and very confused with the concepts.

I've had at most introductory physics experience, but I think I should have been able to solve this still.

Thank you for reading

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"2.Where did they even get 2sa+sb=0 from?" From the assumption that the length of the rope does not change. "Why did they determine the change in distance this way? My first assumption was that if block B moved up 2ft, then block A should move down 2ft (the rope must "move" 2ft too right?)." No, there is a difference between a movable pulley and a fixed pulley, so A moves down 1 ft.

"3.Where did 2vA=−vB come from?" From the same assumption as above. However, the direction of vA is shown incorrectly (or, alternatively, the signs in the formula are wrong).

"4 The FBD for block A is confusing, why is the friction force FA in the direction of the ropes? I thought it was block B that is going down?" It is writen in the statement of the problem that block B went up. " Am I the only one who had trouble deducting that the pulley and block A are the same object?" I don't quite understand what this phrase means exactly and how it is relevant.

"5.Look at the final answer, how could vb be negative? The problem says block B goes UP."

See the answer to your item 3.

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  • $\begingroup$ @Akhmetli, for 3) (and 5), you mean in the key right? That's what I am confused too. Forget 4), block A is going down, so it make sense for the friction to be in the direction of the ropes $\endgroup$ – Hawk Mar 11 '12 at 4:18
  • $\begingroup$ @jak: I mean in FBD for block A - I guess this is the only place where they show the direction of vA. $\endgroup$ – akhmeteli Mar 11 '12 at 4:34
  • $\begingroup$ Just one last thing, so the numerical answers is right, but just the signs are wrong? Sorry for being repetitive $\endgroup$ – Hawk Mar 11 '12 at 20:53
  • $\begingroup$ @jak: Sorry, don't have time to check the solution. $\endgroup$ – akhmeteli Mar 11 '12 at 21:32
  • $\begingroup$ Oh it's fine. Is it possible to solve this without knowing what is "mechanical advantage" $\endgroup$ – Hawk Mar 13 '12 at 1:29
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2 - the block 'A' is on a pulley, when it moves 1 ft the rope moves 2ft. It has a 2:1 mechanical advantage, normally you would use this to move block 'A' half the distance you pull the rope and with twice the force - the point of a pulley is to use a smaller force for a longer distance.

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  • $\begingroup$ sorry, I'll look up this "mechanical advantage" stuff on my own. Don't know why I expected you to explain everything to me. $\endgroup$ – Hawk Mar 11 '12 at 4:27
  • $\begingroup$ Do you mind if I bug you once more? Without knowing "mechanical advantage" could I have solved this? $\endgroup$ – Hawk Mar 13 '12 at 1:28
  • $\begingroup$ @jak - you don't have to know the word, but you do have to knowhow pulleyswork $\endgroup$ – Martin Beckett Mar 13 '12 at 2:58
  • $\begingroup$ I never learn the concept, but is it just something you could see "oh B moves 2ft up, so A must go down 2ft!". I have difficulties wrapping my heads around this $\endgroup$ – Hawk Mar 14 '12 at 18:52
  • $\begingroup$ @jak if B goes down 2ft then the rope must move 2ft. If the rope on A's side gets 2ft shorter than the block can only move 1ft - because the rope is doubled by the pulley. see en.wikipedia.org/wiki/… $\endgroup$ – Martin Beckett Mar 14 '12 at 18:55

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