## JEE MAIN & ADVANCE 11th PCM Physics -Friction-2 Demo Videos

Welcome back students so students we were talking about one beautiful example which we left in between let’s continue that example well an example was that there was this system and we were supposed to find acceleration  well be received with three still this state wherein we determined the FBD we saw what were the directions of friction and also applying constraint equations constraint method using principle of virtual work we found that if acceleration of capital m is a right words then acceleration  of small m will be to 2a down and since captain small I’m are connected therefore the acceleration of small M will be also small a right so we reach  still this point now it’s all about writing or using laws of motion correctly so let’s talk about laws of motion first let’s analyze small M. now along horizontal direction students n1 is responsible for acceleration so long horizontal what can we see n1 will be small ma and long vertical direction  we can say  acceleration  is down then we can say mg minus f1 minus T will be M into 2a but since it is a case of kinetic  friction because there will be an elective slipping hence f1 will be u1 *n1 you see interrelation now this is small M case let’s talk about or let’s first Analyzed let’s first write down net equation if you use n1 from ma and put in the second equation what we’ll get we’ll get mg minus u1 ma plus T is going to m into 2a so let’s name this equation to be our equation number 2 now if we talk about capital insurance this was the FB of capital now a long Vertical direction Direction first we can see that net force long vertical will zero  because there is no acceleration and what can we say n2 write this is the last upwards-  mg  minus f1 minus T will be 0 that is n 2 will be equal to T plus have f 1hich was u 1n1 plus capital mg along horizontal what can we see we can see 2t tension right we’ll try to pull the M block right n1 friction will oppose it So 2t- n1 then – f 2that is u2 n 2 will be capital m a now we put into into this equation what do we get we’ll get this particular expression so on solving what will gate on solving will get 2t minus whole bracket small M a plus u2t plus mu* mu2small ma plus mutual capital mg is going to M1me on further solving from all the questions we get the result acceleration  to be two times smaller2m –u2 M plus small m into G upon capital M plus moment of 5 plus 2 mu 1 minus mu two students of course I i was very fast from Equation number 4 onwards and the solving of it you can always do it but the main character our main heart of this problem is to write laws of motion equations correctly to write constrained relations correctly to mention all the forces correctly and to make sure the direction of friction has been taken correctly so that’s all for on this model certain please what previous module this module again the again, again because this is a very beautiful case which will help you grow in friction chapter very nicely well that’s all till then thank you students

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2017-04-18T04:55:27+00:00 Categories: IIT – JEE Main & Advanced XI - Physics||0 Comments
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