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what is the maximum length of rod that can pass through this Lshaped corridor :
 one year ago
 one year ago
what is the maximum length of rod that can pass through this Lshaped corridor :
 one year ago
 one year ago

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shubhamsrgBest ResponseYou've already chosen the best response.1
dw:1347549016758:dw
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
its 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..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
@mukushla @eliassaab
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
dw: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
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
y = 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 !!
 one year ago

MikaelBest 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
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
well in the position you mentioned,,the rod cant move,so wont exit out.... dw:1347549761719:dw
 one year ago

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

MikaelBest ResponseYou've already chosen the best response.3
Now Deivative = o means the following should be solved
 one year ago

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

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

MikaelBest ResponseYou've already chosen the best response.3
To be precise Rotatio + Simultaneous sliding of the ends
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
You 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 !
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
Imagine 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 ?
 one year ago

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

shubhamsrgBest ResponseYou've already chosen the best response.1
sorry didnt get this figure..
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
Firstly the LEFT END RISES, and the right end SLIDES into positions 3 and 3
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
it cant slide like that.. dw:1347550476551:dw
 one year ago

MikaelBest ResponseYou've already chosen the best response.3
Look 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 !
 one year ago

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

MikaelBest ResponseYou've already chosen the best response.3
More SPACE  NO PROBLEM FITTING SHORTEST STICK !
 one year ago

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

shubhamsrgBest ResponseYou've already chosen the best response.1
you mean.. dw:1347551005335:dw ?
 one year ago