PANTOGRAPH CHALLENGE FROM PATHFINDER CAN YOU SOLVE THIS IN ONE STEP?

[@Fighter_Believer_Achiever]

Thank you very much sir🥳🥳🥳🥳

[@alphaiitd1]

Beautiful explanation sir

[@GallingEssay]

Sir Tension is coming to be 3F/2 by constraints and net torque=0 method. It may be because the work done on the string, half is kinetic energy and half is potential energy. Can you please confirm?

[@Chole67767bhature]

Thank u sir🙏🙏👌👌

[@KapidhwajSonithB]

Sir, the string is inextensible. And if the points move dx and 3.dx/2 ... String will be streched🤔

[@PHYSICSSIRJEE]

dx is infinitesimal . For finite force , if extension is small then k of string is tending to infinity... That's what inextensible means ... Extension is not finite and visible

[@90shiphop-gl4zo]

thank you sir

[@srinivaskoneru4136]

Excellent

[@twistedargument4308]

Sir I had a doubt ,sir the smaller rhombus should not extend due to string so why not the both sides of rhombus have same velocity?

[@TrishaBafna-en8vq]

Sir can you pls tell why work due to hinge reactions was ignored i read your reply to @Vineet Sir's comment below But in actual scenario, even tension is internal force and wouldnt do work on system. But still for infinitesimal displacement, we considered its work too Then why not consider the same for hinge rxn (for infinitesimal displacement)

[@ponnapu]

sir is it possible to solve this problem by force analysis

[@endlesseducation2845]

#21 For my revision

[@steelcubes723]

Sir, could you please do a solution video on pathfinder problem- electrostatics MCQ -21

[@dasaradhramgottapu]

Super sir 🙏

[@sakshampaliwal8170]

And the rods also have tension, so why their work is not considered??

[@PHYSICSSIRJEE]

Please read my reply to "Vineet" sir's comment below 👇

[@sakshampaliwal8170]

@@PHYSICSSIRJEE ohkk I got it because we are not considering the virtual work by rods, we are seeing only string virtual work...

[@jobbyjoseph3641]

@@PHYSICSSIRJEE Sir can you please tell if my Interpretation is correct....The virtual work done by the rods on the first rhombus is equal to but opposite of the work done by the rods on the second rhombus..So their work adds to zero...also for entire system Leftmost hinge is not moving So it's virtual work is zero. So we consider only tensions and force's work...

[@jobbyjoseph3641]

@@PHYSICSSIRJEE Sir I have one more doubt regarding Virtual work method....In pulley System's when we use the sum of the virtual works =0,, the relation of Tension.accleration is applied only on blocks....But in one Infinite pulley problem we use sigma T.a=0 for the blocks as well as for one of the acclerating pulleys.....So finally what is right?? Can we use in one acclerating pulleys and blocks also ???

[@vivek31290]

Sir can you pls explain how displacement of the two hinges are different. DX and 3/2 DX . Since connected by same string shouldn't the infinitesimal displacement be same

[@PHYSICSSIRJEE]

The idea of virtual work is to assume virtual displacement. So front and back ends won't have same virtual displacements

[@sanjayreddy2916]

Actually diagram is little wrong second rod length is 2a not 3a

[@WatchManofDoom123]

I am bit late to reply lol. But 3a is 2a of left rhombus+ a of right rhombus

[@tridibeshsamantroy9837]

Sir but in virtual work method we try to put the submission of tension and velocity as 0

[@jeeboi347]

It's the same thing as this. Just differentiate both sides with time and we'll get what you are saying

[@nimitjain8163]

revised 16/12/21

[@mamtamittal3051]

Sir looks like you have done a calculation error the tension must be 2F not 3F

[@PHYSICSSIRJEE]

No error, please check again. Kindly Go through the solution with my voice again slowly

[@vineetguleria]

why are we ignoring the hinge reactions or the forces from remaining part of blue rods

[@PHYSICSSIRJEE]

That was a good query sir, the virtual work equation was written for entire pantograph with external forces F and two T 's only doing work. ( even though diagram shown was of right rhombus in Solution) The hinge forces of the system are internal non-deformative , so W by them is zero

[@vineetguleria]

Would it not really take a little more elaborate argument or maybe a set of equations of equilibrium to be able to prove that the rods are non deformative especially if we replace the string by a hinged rod.

[@vineetguleria]

But the tension would not change even if we take a rod instead of a string .

[@invincible9240]

@@PHYSICSSIRJEE sir even I have the same query ,can we prove this argument .also how can we approach this q without virtual work (I am getting stuck at fbd pls give some hint)

[@invincible9240]

@@vineetguleria sir u can check out some standard text on virtual work it will be cleared

[@praneethreddykoyagura4020]

how to do it by normal method??

[@PHYSICSSIRJEE]

Write equations of force and torque balance for every rod by assuming appropriate hinge reactions and solve the 6-7 equations with equal number of unknowns