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August 13th, 2012, 01:11 PM   #1
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Express the height of the crate in terms of distance

A 20m ropes passes over a pulley 4m above the ground and a crate on the ground is attached at one end. The other end of the rope is held at a level of 1m above the ground and is drawn away from the pulley. Express the height of the crate over the ground in terms of the distance the person is from directly below the crate. Sketch the graph of distance and height. NEGLECT THE CRATE HEIGHT

I am having a really tough time with this. I have assumed that I need to use the tangent of the angle created to solve this but I am really not finding any solutions that work.
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August 14th, 2012, 04:58 AM   #2
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What does "the distance the person is from directly below the crate" mean? Also, I don't understand how it is possible to sketch the graph of distance and height whilst ignoring the crate height. What height is to be graphed if not the crate height?
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August 14th, 2012, 08:57 AM   #3
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Re: Express the height of the crate in terms of distance

Well, if Skip is confused, then I'm twice as confused!
Does it all work something like:
Code:
          P                        P


       


          B                B       A

          A
On left is initial setup: P is pulley, B is the "puller" 1m above floor at one end of rope,
A is the crate on floor, so we assume A is other end of rope: so AB = 1m and BP = 9.5m:
means AP = 10.5, AP + BP being the full rope.

ON right: let's assume B moved distance p to left, such that A moved "up" 1m:
then PAB is a right triangle: AP = 9.5 and BP = 10.5 (rope totals 9.5 + 10.5 = 20).
So p = SQRT(10.5^2 - 9.5^2) = 2SQRT(5) ; is that the way it works?

If so, you need to come up with a "formula" that accomodates ALL created triangles,
not only the right triangle that I lazily picked...RIGHT?

If you answer yes to all the above, are you OK now?
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