superprismatic Posted July 13, 2009 Report Share Posted July 13, 2009 This problem actually came up in real life for me. This is definitely not a trick question. So, if there's something you don't understand about the problem statement, I'll be happy to clear it up. I was trying to store a ladder in a shed.....well, you'll see: I would like to store a ladder inside a shed. I want to store it flat against the wall using hooks. The shed wall is 16 feet long (192 inches). The ladder is 16 feet and 1 inch long (193 inches) and it is 18 inches wide. What is the minimum height that the shed wall must be to allow me to mount the ladder flat against this wall without poking through the floor, ceiling, or side walls? My shoot-from-the-hip initial guess was way off. So, I was surprised at the height needed after I worked it out. I would ask the solver to give 2 answers: (1) An initial guess and (2) The actual value (to the nearest inch or so). I would be particularly interested in an "AHA!" solution. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 13, 2009 Report Share Posted July 13, 2009 about 47 or 48 nches Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 13, 2009 Report Share Posted July 13, 2009 (edited) I did the math and got exactly 37.528152621644 inches is the minimum height Sorry, I suck at guess work Edited July 13, 2009 by Mistwalker Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 13, 2009 Report Share Posted July 13, 2009 My immediate guess is that you will have to add at least 2 feet to the width to make it fit. Just a guess, 3 1/2 feet. But that's probably wrong. It must be more than half the height of the ladder, at a secondary guess, 8 feet 10 inches. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 13, 2009 Report Share Posted July 13, 2009 If you raised one side up 18 inches that would give you the diagonal of the ladder, then another 18 and you would get the same length as if it were flat on the ground, and then another 18 and you would probably lose that inch, but then you need to account for the width of the ladder so add another 18 or so. So my guess is 18*4 = 72 inches which should be on the high end. 2) I got 62.028 inches Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 13, 2009 Report Share Posted July 13, 2009 I did the math and got exactly 37.528152621644 inches is the minimum height Sorry, I suck at guess work will you show the math? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 13, 2009 Report Share Posted July 13, 2009 85 inches Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 13, 2009 Report Share Posted July 13, 2009 61.67247666", or 62" to the nearest inch. This is astonishingly similar to that of Bobifishi, yet different. Perhaps a rounding error on your part? (I used AutoCAD, and therefore did no rounding.) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 13, 2009 Report Share Posted July 13, 2009 I did the math and got exactly 37.528152621644 inches is the minimum height Sorry, I suck at guess work Math is correct!! got the same figure. My answer is 38" (37.5281... + s*** happens) Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 14, 2009 Author Report Share Posted July 14, 2009 Wow! Three essentially different calculated answers! Did anyone try to make a scale model of the problem out of paper to get a ballpark on this? I'll say one thing: The off-the-cuff guesses some of you made were all closer to the truth than my original guess. Thanks for your interest in this problem. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 Well, I did the math (I thought) and got somewhere around 37 inches. However, using inverse sines, etc. the angle of the ladder to the floor comes out to be around 5.8 degrees, which seems a little low. However, the wall is only one inch longer than the ladder, so perhaps... Anyway, my math (what little there is of it) is: Pythagorean theorem to figure distance from floor to bottom of latter at its angle: sqrt(193^2 - 192^2) = about 19.6 inches. So that's the distance from the floor to the bottom of the tip of the ladder resting on the wall when set in its place. Adding 18 inches for the width of the ladder, you get 37.6 inches, but you don't need all of that, so it's a little less. Using inverse sines, you can find that the angle of the ladder to the floor is 5.8 degrees, but I didn't feel like putting that anywhere to find out a real answer. So there it as, for what little it's worth. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 16 feet 1 inch. The ladder actually gets longer when you turn it. It's shortest length is when it's straight. If your ceiling is 16'1", you can store it straight up. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 Using the centerline of the ladder requires adding 18 inches to the overall length and then using 211 and 192 inches respectively the height works out to 87.5" That is for the ladder to rest the bottom leg end on the ground and the top edge of the ladder at the top of the wall.ras Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 I agree with postmaster. 61.67247 inches is correct. x = ((((193^2 + 18^2))^.5)^2 - 192^2)^.5 + (2 * (18 * sin(invsin(193 / (193^2 + 18^2)^.5 ) - invsin((((193^2 + 18^2)^.5)^2 - 192^2)^.5 / ((93^2 + 18^2)^.5)))) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 The leangth of the ladder is the hypotenuse of a right triangle, and the leangth of the wall is the base of the triangle. We are looking for the height, therefore we use Pythagorean's Therom (192 in)2+X2= (193 in)2 36864 in2+X2=37249 in2 37249-36864=X2 385=X2 19.62141687=X We could take this farther and use trigonometry to determine the angle of the ladder's storage to calculate the effect the width would have, but I am feeling lazy so I am just going to add it on... 19.62141687+18 in (?)= 37.62141687 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 53.378 Inches. The ladder makes an angle of 10.656 degrees with horizontal Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 14, 2009 Author Report Share Posted July 14, 2009 The length of the ladder is the hypotenuse (193 inches), but the base of the triangle is a bit smaller than 192 inches because it ends where it touches the bottom edge of the ladder, at the vertical line segment on the right side of your graphic. So the base is 192-y inches for some y. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 The length of the ladder is the hypotenuse (193 inches), but the base of the triangle is a bit smaller than 192 inches because it ends where it touches the bottom edge of the ladder, at the vertical line segment on the right side of your graphic. So the base is 192-y inches for some y. Oh duh, I was wondering why that just didn't look right... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 16' 1'5' 9.3" rounded off Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 16' 1'5' 9.3" rounded off sorry had the wrong angle went back and put it in AutoCAD it is 62" Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 14, 2009 Report Share Posted July 14, 2009 The length of the ladder is the hypotenuse (193 inches), but the base of the triangle is a bit smaller than 192 inches because it ends where it touches the bottom edge of the ladder, at the vertical line segment on the right side of your graphic. So the base is 192-y inches for some y. Okay I think I got it now: Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 14, 2009 Author Report Share Posted July 14, 2009 Okay I think I got it now: A small mistake leads to a big error in the final answer, as you know. So far, we have essentially 5 different answers to this problem in this forum. Lots of people are making lots of different mistakes! I also got around 62" as the answer. You were very quick to recover from your error. By the way, I couldnt fit the ladder in the shed because the wall was only 48" high before it hit the A-frame of the shed roof. Quote Link to comment Share on other sites More sharing options...
Question
superprismatic
This problem actually came up in real life for me.
This is definitely not a trick question. So, if
there's something you don't understand about the
problem statement, I'll be happy to clear it up.
I was trying to store a ladder in a shed.....well,
you'll see:
I would like to store a ladder inside a shed.
I want to store it flat against the wall using hooks.
The shed wall is 16 feet long (192 inches). The
ladder is 16 feet and 1 inch long (193 inches)
and it is 18 inches wide. What is the minimum
height that the shed wall must be to allow me to
mount the ladder flat against this wall without
poking through the floor, ceiling, or side walls?
My shoot-from-the-hip initial guess was way off.
So, I was surprised at the height needed after
I worked it out. I would ask the solver to give
2 answers: (1) An initial guess and (2) The
actual value (to the nearest inch or so).
I would be particularly interested in an "AHA!"
solution.
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