Solving the Winch Problem at 12m High

[dontdisturbmycircles]

Homework Statement

A winch at the top of 12 meter high building pulls a pipe of the same length to a vertical position. The winch pulls in rope at a rate of -0.2 meters per second. Find the rate of the vertical change and the rate of the horizontal change at the end of the pipe when y=6.

Homework Equations

http://www.imagehosting.com/out.php/i249658_ahhh.JPG

The Attempt at a Solution

I simply don't have enough information..

I set up an equation relating the three variables..

$$s^{2}=x^{2}+(12-y)^{2}$$

and differentiate with respect to time and end up with

$$s\frac{ds}{dt}=x\frac{dx}{dt}-(12-y)\frac{dy}{dt}$$

But I simply don't have all of the variables needed. I can solve for x (with pythagorean theorem), but not dx/dt...

Any help is appreciated. :yuck:

 

[arildno]

Hint:
First find the ANGULAR rate of change by a smart law of cosines.
THEN, find the rates of change of (x,y) by using a right-angled triangle.

 

[dontdisturbmycircles]

Thankyou arildno

I was able to get the correct numerical answer but something is bothering me...

Take the angle made at the top of the wall to be $$\theta$$...

We know that $$\frac{d\theta}{dt}$$ must be positive since the angle is tending towards 90deg right?

I got $$\frac{d\theta}{dt}=\frac{/sqrt{3}}{80}$$... but this leads to something that is incorrect. Because when I set up the following equation to relate x and s, I get this..

$$sin\theta=\frac{x}{s} = x(s)^{-1}$$
$$cos\theta\frac{d\theta}{dt}=x(-1)s^{-2}\frac{ds}{dt}+s^{-1}\frac{dx}{dt}$$

which eventually leads to

$$\frac{dx}{dt}=s(cos\theta\frac{d\theta}{dt}+x*s^{-2}\frac{ds}{dt})$$

Now, solving for x using the right triangle at the base, $$x=\sqrt{12^{2}-6^{2}} = 6*\sqrt{3}$$

and s=12 by similar reasoning

Then $$tan\theta=\frac{6\sqrt{3}}{6}$$

$$\theta=\frac{\pi}{3}$$

So solving $$\frac{dx}{dt}=s(cos\theta\frac{d\theta}{dt}+x*s^{-2}\frac{ds}{dt})$$

I get $$\frac{dx}{dt}=\frac{\sqrt{3}}{15}$$

Which is the correct magnitude, but the incorrect sign... :/

 

[arildno]

Well, I used the lowest angle, with the variable length s(t) as the opposite side. I didn't calculate it through, but my signs looked good.

 

[dontdisturbmycircles]

Ok, let me try that and see if I can get it to work.

 

[dontdisturbmycircles]

Wow it worked very smoothly using the lower angle... Correct sign and all. I'll have to think about this problem for a bit. Thanks again.