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shubhamsrg
 3 years ago
what is the maximum length of rod that can pass through this Lshaped corridor :
shubhamsrg
 3 years ago
what is the maximum length of rod that can pass through this Lshaped corridor :

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shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1347549016758:dw

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1its the same question which @siddhantsharan asked here : http://openstudy.com/study#/updates/504e2cd7e4b064acfea5f1ae there arent much people there,,and i liked this ques,,so i bring this here..

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1@mukushla @eliassaab

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3dw:1347549611645:dw By SIMILARITY of Triangles [ three similar , rightangle triangle] we obtain (X+a)/(y+b) = x/b = a/y So ab = xy and bx + ba = yx + bx

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3y = ab/ x Substitrute int Length^2 = (y+b)^2 + (x+a)^2 AND Make derivation , then equate to 0 and solve for x  here is your SOLUTION !!

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3(ab/x + b)^2 + (x+a)^2 = a^2b^2/x^2 + b^2 + 2ab^2/x + x^2 + 2xa + a^2

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1well in the position you mentioned,,the rod cant move,so wont exit out.... dw:1347549761719:dw

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3(a^b^2/x^3) * (2)  2ab^2/x^2 + 2x +2a = Derivative

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Now Deivative = o means the following should be solved

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1you're concept it wrong buddy,,see my last comment again,,and correct me if am wrong..

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3dw:1347549859821:dw Well  sorry to be so direct. But you ARE wrong. Rotation is NO PROBLEM

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3To be precise Rotatio + Simultaneous sliding of the ends

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3You see  the interval (lets call it "stick") that is the solution is THE SHORTEST POSSIBLE OF ALL ANGLES. Thus, a little motion provides it with MORE SPACE , NOT LESS !

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Imagine a stick that fits into this space at "the most restrictive" angle. Well, it will obviously fit into positions at LES restrictive angles, wouldn't it ?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1347550109862:dw for up ward movement,,downward movement should also be there,,which is not possible..

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1sorry didnt get this figure..

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Firstly the LEFT END RISES, and the right end SLIDES into positions 3 and 3

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1it cant slide like that.. dw:1347550476551:dw

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Look forget  drawing for a second. If Some position is the SHORTEST interval  all its small motions will create SPACE AT BOTH ENDS or AT ONE END at least !

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Come on  after rotation the OLD SHORTEST STICK IS NOT FILLING THE NEW  LONGER SPACE

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3More SPACE  NO PROBLEM FITTING SHORTEST STICK !

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3You need a stick that will move inside the SHORTEST place. Sooo just take the SHORTEST LENGTH  some infintisimal "epsilon"

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1you mean.. dw:1347551005335:dw ?