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JEE MAIN & ADVANCE 11th PCM Physics -Friction-1 Demo Videos

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Welcome back students so students you’re talking about friction problems let’s continue with few more examples now in this give an example we have two blocks A and B there’s a force of 120 Newtons  which is acting on the left of a and a force of 100  Newtons ’s which is acting on the right of B the coefficient of friction in the A end surfaces 0.9 and coefficient of friction between B end surfaces 0.3 mass of a is 10kg and mass of B is 20 kg now it is given that we have to find the tension in the string in this particular scenario that does this ,this situation is given we have to find tension now students this particular Scenario  that is a very beautiful situation where it will be teach us many Things now what all possibilities can arise now as a result of these two forces when 120 and 100 the possibilities that can arise are, a and B move and  a and b don’t move now for any possibility what we need to know is what exactly will be d limiting value of frictions because in this case Newest  meu k are not mention separately will take meu s and meu K to be seen and if the friction and static we need meu n and if the friction is kinetic  then again we need meu n in so basically first thing is to know what exactly will be the limiting value our meu times n now for a since mass of a is 10 therefore mg will be hundred and meu between a surfaces 0.9 therefore a max will be 90 similarly FB max is 60 so the thing is if both a and B move in friction will be kinetic  and the values of kinetic friction will be nineteen sixty and it worked in we don’t move then whatever the value of friction will be that value of friction should be less than nineteen sixty pretty simple now let us assume in first place assume  that both bodies move well both bodies move then it is pretty clear that 120 minus hundred that is 120 left and hundred right so 120 minus hundred is the driving force which is trying to drive a and b the words left now when 120 minus 100 that is 20 is driving in with words left friction on both a and B will act right and it will oppose the motion since a and  b move friction will be kinetic  the values are nineteen sixty therefore the opposing force are the resisting force will be meuA into na plus meuB*nb   that is little be 150 newtons 120 newtons and were acting opposite I think are present there for the inside 120 minus hundred is the driving force friction will be opposing force with friction  oppose this motion relative motion kinetic  friction opposes of relative motion so over here what we can see you can see that since resisting  force is greater than driving force their for how can motion happen therefore this case is not possible we can conclude  that’s a system  remain stationary did it not move our assumption was incorrect right students so now when this assumption is incorrect that both a and be moved what is the next possibility A and B don’t move when the A and B don’t move in bad as well there can be two possibilities which one possibility will be that friction will be static for both a and B when I says  that direction I mean to say friction will be less than limiting friction for both a and B this will be possibility one second possibility will be the frictional will be  static for one and limiting for another

that can be a second possibility that is first a in its limiting friction and then Oh on b the friction will not be limiting  but some lower value or vice Versa so which of the cases are true let’s see let’s assume that both the friction is static that is both a and b the value of friction is less than the limiting value of friction now actually this is a very logical thing if friction in both a and B is static right it means to oppose 120 and 100 Newtons  friction itself is sufficient attention will not even if do what is that not connected with string still both a and B will not move right so if friction is that take then we can say tension in the string should be 0 right students now if the tension in the string is 0 then to stop a the friction on a should be 120 and to block  b efficient on be should be hundred but what do we know we know if friction on a is  120 and friction on b is a hundred this value of friction is just not possible because maximum values of friction were nineteen sixty respectively so it means this case is also not possible that is friction on both a and B cannot be static one of them has to have a limiting value of friction now which one of them has a limiting value of friction let us see now let us first assume  10kg block as limiting value of Friction it is reaching its limiting value first then what can we say what will be limiting value 90 their for on a 120 is the force left 90 that is limiting friction becomes right balance will be taken care by attention that is tensions value will be 30 now if the tension is 30 then on b from FBD we can say that T plus F is hundred that is  30 + F is hundred that is f become 70 but again in friction on b 70 it again is greater than its maximum value which is 60 again in this case is not possible so the only possibility we are left with is the b eliminating the  value for first and then balance friction will take care of a so it’s b tense limiting value that is on be friction will be liberating and that will be 60 so on b balanced Will be  taken  can get  tension and it becomes 40 Newtons  so if we take the same tension on a then we can say that F will be 80 and the friction is 80 we can easily see that this 80 newtons is actually less than the limiting friction which was 90 on a hence  what do we conclude we conclude that tension between the two blocks is 14 newtons  students this was a very beautiful problem you need to watch this module any number of times because this will give you a lot of clarity it’s not a very simple case always when we are dealing with multiple bodies connected with treads right  students  remember first friction will try to oppose the driving force then comes tension not just make sure friction never exceeds its limiting value well thats all from on this model will get back to till then  Thank You students

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