can someone help me with this?

Question

Two pipes are bound tightly together with a very thin wire as in the diagram. The radii of the pieces are 2in and 6in. Neglecting any practical allowance for security (meaning only wrapped once), determine the minimal length of the wire.

i tried my best to do the diagram and upload it here, its rough but i think you can get the general idea. could someone please help me with it?

http://www.imagehosting.com/show.php/1174338_circle2.GIF.html

wow thanks a whole lot for your answer gp4rt. lol, by the time i checked this though i had already got that same answer lol, so at least now i know its correct.

Answer 1

The answer above has the right idea but the wrong diagram and numbers. The wire is tangent to the pipe edge at each pipe, and a perpendicular from the tangent point will go through the center of the circle. See the diagram here:

http://img233.imageshack.us/img233/1013/pipeswireuq3.jpg

The geometry is symmetrical, so only the top half must be determined, then multiplied by 2.

The lengths in the diagram are:

L is the leg of a right triangle with hypotenuse 8 (6+2) and one leg 4 (6-2) giving L = √[8^2 - 4^2] = √48 = 4√3

The angle ø is arccos(4/8) = arccos(.5) = π/3. The wire arc on the large pipe S1 subtends the angle π - π/3, or 2π/3. The arc length is then 2π/3 * 6 = 4π. The wire arc on the smaller pipe S2 subtends the angle π/3, so the length is π/3 * 2, or 2π/3

Add them all together to get 4√3 + 4π + 2π/3 = 4√3 + 14π/3

Multiply this by 2 to get 8√3 + 28π/3 for the total wire length.

Answer 2

break it down into peices, like a race track. the two turns and the two straight pieces. The turns are going to be half the circumference of the circle on that side, so one will be 6 pi, the other will be 2 pi. work the sides like right triangles. here is my diagram. then just add it all together.

http://i17.photobucket.com/albums/b74/amberns1/circle2.gif

Answer 3

well, you need to know the diameter of both the circles, but the wire isn't centered...I wish I could help. Why not go to a tutor and ask for their help?