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A rope 28 feet long is attached to a block on level ground and runs over a pulley 12 feet above the ground. The rope is stretched taut and the free end is drawn directly away from the block and pulley at the rate of 13 ft. per sec. How fast will the block be moving when it is 5 feet away from the point directly below the pulley?

My solution:

I start with relating the three sides of the triangle, holding one side constant at 12 feet. My aim is to find the rate of movement of the rope, that is, the rate of change of the hypotenuse, h, which determines, and is therefore identical to, the rate of change of the position of the box.

$$h^2 = x^2 + y^2$$

$$h = \sqrt{x^2 + 144},$$

and since the rate of movement of the rope dictates the rate of movement of the box,

$$\frac{dh}{dt} = \frac{x}{\sqrt{x^2 + 144}} \cdot \frac{dx}{dt}$$

Given that $h = 28 - 5 = 23$ feet when the box is 5 feet from the pulley, and that $$ x = \sqrt{h^2 - y^2} = \sqrt{23^2 - 12^2} = \sqrt{385},$$

it follows that

$$\frac{dh}{dt} = \frac{\sqrt{385}}{\sqrt{385 + 144}} \cdot 13 \approx 11.1 \textrm{ ft/s}.$$

The book's answer is $8 \textrm{ ft^2/s}.$

Did I do something wrong? My answer is so close to the book's.

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  • $\begingroup$ You say 'after some manipulation'. Can you show what that manipulation is ? I think there is something wrong there. Also you have $\sqrt {383}$ instead of $\sqrt {385}$. $\endgroup$ – Kavi Rama Murthy Aug 26 '19 at 6:32
  • $\begingroup$ For me, the details of the geometry are a bit unclear. Of course the sides form a triangle, but I don't really see why the length 28 ft is important in the beginning. Surely, the length of the hypotenuse ($h$) is longer than 28 ft, but why would it be equal to $h+y$ ? There's nothing that states so in the text. Also, is the size (physical dimensions) of the block non-negligeable? $\endgroup$ – Matti P. Aug 26 '19 at 6:33
  • $\begingroup$ Hey, Kavi. I changed it. I didn't find an error. Would you have a look? thanks for your response. $\endgroup$ – Rafael Vergnaud Aug 26 '19 at 6:39
  • $\begingroup$ That its 28ft allows you to determine the length of the hypotenuse when the box is 5 feet from the pulley. Since the rope is 28 ft, and when the box has reached the pulley, the entirety of the rope has been pulled and the hypotenuse is 28 ft long. I can draw my picture and attach it to my post. Do you recommend that? Thanks for your response, too, Matti :) $\endgroup$ – Rafael Vergnaud Aug 26 '19 at 6:40
  • $\begingroup$ The size of the box isn't significant (at least how I tried to solve the problem). I draw a right triangle where the right angle is on the right side of the triangle. The right side remains constant and is 12 feet long (from floor to pulley), the hypotenuse is the rope (which is being pulled in the direction away from the box), and the last side (the bottom) is the movement of the end of the rope that is being pulled (that is not attached to the box). I hope this makes sense. Sorry if I'm unclear! $\endgroup$ – Rafael Vergnaud Aug 26 '19 at 6:43
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Given that $x$ and $y$ are functions of time, the following relationship should always hold true: $x^2+12^2=(16+y)^2$. Thus:

$$ 2x\frac{dx}{dt}+0=2(16+y)\frac{dy}{dt}\implies\\ \frac{dy}{dt}=\frac{x}{16+y}\cdot\frac{dx}{dt}. $$

When the block is $5\ \text{ft}$ away from the point directly below the pulley, $y=7\ \text{ft}$. $\frac{dx}{dt}$ is known. It's $13\ \text{ft/sec}$. What is $x$ at the time when the block is $5\ \text{ft}$ away from the point directly below the pulley? It should be the positive solution to this equation: $$ x^2+12^2=(16+7)^2\implies\\ x=\sqrt{385}\ \text{ft}. $$

Therefore: $$ \frac{dy}{dt}=\frac{\sqrt{385}}{16+7}\cdot 13\approx 11.1\ \text{ft/sec}. $$

I don't know, but that's the answer I get. I don't see anything wrong with my logic.

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$h^2 = x^2 + y^2$

$y = 12$ as identified.

When $x = 5, h = 13$

$h^2 = 5^2 + 12^2 = 13^2$

That the length of the rope is 28' seems to be a red herring, except insofar as it establishes an upper bound for $x.$ We really only care about the length of rope from the pully to the mass. The rope from the pully to the free end is irrelevant.

When we differentiate:

$2h\frac {dh}{dt} = 2x \frac {dx}{dt}$

$\frac {dh}{dt} = 13$

$2(13)(13) = 2(5)\frac{dx}{dt}\\ x = \frac{13^3}{5}$

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  • $\begingroup$ Hey, Doug. Thanks for your response. Why did you get x = 5? Is it not the case that h + y must be equal to 28, since h + y is the length of the rope? The number 5 is associated with the side that describes the movement of the box. In my problem, x is the side that describes the movement of the individual who is pulling the rope away from the box. $\endgroup$ – Rafael Vergnaud Aug 26 '19 at 7:08
  • $\begingroup$ Should not dh/dt = 8, as the book says it should? (assuming dh/dt is how fast the block is moving) $\endgroup$ – Rafael Vergnaud Aug 26 '19 at 7:08
  • $\begingroup$ Why does x = 5? because we want to know how fast the block is moving when it is 5 feet away from the point directly below the pulley. Why does $\frac {dh}{dt} = 13$ because that is the speed the free end is moving. $\endgroup$ – Doug M Aug 26 '19 at 9:46

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