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## NEET & AIIMS 11th PCB 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, that example was that there was this system and we were supposed to find accelerations. Well, we had reached till this state wherein we determined the FBD. We saw what were the direction of friction and also applying constrained equations, constrained methods by using of principle of virtual work, we found that if acceleration of capital M is a right wards, then acceleration of small m will be 2a down, and since capital M and small m are connected therefore acceleration of small m will be also small a right. So we reached till this point, now it is all about writing or using Laws of Motion correctly.

So let’s talk about Laws of Motion, first let’s analyse small m. Now along horizontal direction, students, N1 is responsible for acceleration. So along horizontal what can we say, N1 will be small ma. And in vertical direction we can say if acceleration is down, then we can say mg minus f1 minus t will be m into 2 a but since it is a kinetic friction because there will be relative slipping hence f1 will be Mu1 into N1 you see inter relation.

Now this is small m, let’s talk about or first let’s analyse let’s first write down net equation, if we use N1 from ma and put in the second equation what we will get. We will get mg minus Mu1 ma plus t is equal to 2 into ma. So let’s name this equation to be our equation no.2.

Now we will talk about capital M, students, this was the FBD of capital M. Now along vertical direction first we can see that net force along vertical will be zero because there is no acceleration. Hence what can we say N2, right, this is thus upwards minus capital Mg minus f1 minus T will be zero. That is N2 will be equal to T plus f1 which was Mu1N1 plus Mg. Along horizontal what can we see, we can see 2 T tension right will try to pull the capital M block right. N1 and friction will oppose it. So 2T minus N1 minus f2 that is Mu2N2 will be capital Ma. Now we put N2 into this equation what do we get, we will get this particular expression, so I am solving what we will get. On solving we will get 2T minus whole bracket small ma plus Mu2T plus Mu1Mu2 small ma plus Mu2 capital Mg is equal to capital Ma. On further solving from all the equations we get the result acceleration to be two times small m minus Mu2 into capital M plus small m into g upon capital M plus small m into 5 plus 2 Mu1 minus Mu2.

Students, of course, I was very fast from equation no. 4 onwards and the solving of it, you can always do it. But the main crux or the 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 now in this module, students, please watch previous module and this module, again, again and again because this is a very beautiful case which will help you grow in friction chapter very nicely.

Well, that’s all for now till then thank you, students.

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2017-04-18T04:55:20+00:00 Categories: XI-NEET & AIIMS||0 Comments
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