Carrying a Ladder around a Corner

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greatwhiteshark

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A ladder of length L is carried horizontally around a corner from a
hall 3 feet wide into a hall 4 feet wide. What is the length of the
ladder?
 
greatwhiteshark said:
A ladder of length L is carried horizontally around a corner from a hall 3 feet wide into a hall 4 feet wide. What is the length of the ladder?
Click to expand...
That's ALMOST a problem statement.

First, can we assume the ladder is infinitely thin? Width = 0?

Second, can we assume that ladder BARELY FITS around the corner? Maybe MAXIMUM LENGTH of the ladder?

The present problem statement mentions neither issue. I'm going with L = 2 feet. That will work.

Maybe I'm overthinking it. How about "L"? I like it.
 
Hello, greatwhiteshark!

TK is right: the problem is poorly worded.
Lucky we recognize it as a classic problem.

A ladder of length L is carried horizontally around a corner from a hall 3 ft wide into a hall 4 ft wide.
What is the maximum length of the ladder?
Click to expand...
Code: 
           B 
   +-------*---
   |      /:            The ladder is segment AB = L.
   |     / :4
   |    /t :           It touches the corner at P.
   |   /*------
   |  / |P          Note the similar right triangles
   | /  |                and the angles t (theta).
   |/t  |
  A*----|
   | 3  |
In the lower right triangle: . AP .= .3 secθ
In the upper right triangle: . PB .= .4 cscθ

Hence: . L .= .3 secθ + 4 cscθ

Differentiate: . L' . = . 3 secθ tanθ - 4 cscθ cotθ

. . . . . . . 3 . . . sinθ . . .4 . . . cosθ
Then: . ------ . ------ - ------ . ------ . = . 0
. . . . . . cosθ . .cosθ . .sinθ . .sinθ

Multiply through by sin<sup>2</sup>θ cos<sup>2</sup>θ: . 3sin<sup>3</sup>θ - 4cos<sup>3</sup>θ . = . 0

. . . . . . . . . . . . . . . . . . . . . . . . . . sin<sup>3</sup>θ . . . .4
Then: . 3sin<sup>3</sup>θ .= .4cos<sup>3</sup>θ . ---> . -------- .= .--
. . . . . . . . . . . . . . . . . . . . . . . . . . cos<sup>3</sup>θ . . . 3

And: . tan<sup>3</sup>θ .= .4/3 . ---> , tanθ .= .(4/3)<sup>1/3</sup> .= .1.100642416

Hence: . θ .≈ .47.743<sup>o</sup>


Therefore: . L . = . 3 sec(47.743<sup>o</sup>) + 4 csc(47.743<sup>o</sup>) .≈ .9.87 feet
 
It's thickness is still zero. It can't be a very strong ladder.
 
tkhunny said:
It's thickness is still zero. It can't be a very strong ladder.
Click to expand...
Who would dare climb it?
How would you carry it anyway?
(Storage is not a problem, however.)

I know . . . it's used in Flatland.
 
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