Classical Mechanics ...looks easy

Classical Mechanics Level 4

A ladder of length 1 1 initially stands vertically against a wall. Its bottom end is given a sideways kick, causing the ladder to slide down. Assume that the bottom end is constrained to keep contact with the ground, and the top end is constrained to keep contact with the wall. Describe the envelope of the various positions of the ladder.

x 3 2 + y 3 2 = 1 x^{\frac{3}{2} } + y^{\frac{3}{2} } = 1 x 2 3 + y 2 3 = 1 x^{\frac{2}{3} } + y^{\frac{2}{3} } = 1 x + y = 1 x + y = 1 x y = 1 xy = 1 ( x + 1 ) ( y + 1 ) = 2 (x+1)(y+1) = 2 x 1 2 + y 1 2 = 1 x^{\frac{1}{2} } + y^{\frac{1}{2} } = 1 x 2 + y 2 = 1 x^2 + y^2 = 1 ( x − 1 ) ( y − 1 ) = − 1 (x-1)(y-1)=-1

Rahul Singh by rahul singh , 22, India

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1 solution

Rahul Singh
Mar 24, 2018

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Why does this process give the envelope of the curve?

Calvin Lin Staff - 3 years, 1 month ago

But taking laders lentgh of 1 to be constant....can't we say x sq plus y sq equal 1?

Vilakshan Gupta - 3 years ago

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Note that we're talking about the envelope of the curve.

I'm not sure how you get "length of ladder =1 implies x+y = 1". How is "x+y = length of ladder"? The point (x,y) is just a point on the (envelope of the) ladder.

Calvin Lin Staff - 3 years ago

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