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NEET & AIIMS 11 th PCB Physics -Friction 1 Demo Videos

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Welcome back, students.

So, students, we were talking about friction problems let us continue with few more examples.

Now in this given example we have two blocks, A and B. There is a force of 120 Newton which is acting on the left of A, and a force of 100 Newton which is acting on the right of B. The coefficient of friction between A and surface is 0.9 and coefficient of friction between B and surface is 0.3.  Mass of A is 10 kg 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 is this situation is given we have to find tension. Now, students, this particular scenario is a very beautiful situation where it will teach us many things. Now what all possibilities can arise?

Now as a result of these two forces, 120 and 100, the possibilities that can arise are (a) A and B move, (b) A and B don’t move. Now for any possibility what we need to know is what exactly will be the limiting value of frictions because in this case MuS and MuK are not mentioned separately, we will take MuS and MuK to be the same and if the friction is static we need MuN, if the friction is kinetic then again we need MuN. So basically first thing is to know what exactly will be the limiting value or Mu times N. Now for A since mass of A is 10 therefore mg will be 100, and Mu between A and surface is 0.9 therefore fA max will be 90, similarly fB max is 60. So the thing is if both A and B move then friction will be kinetic and values of kinetic friction will be 90 and 60, and if both A and B don’t move then whatever the value of friction will be that value of friction should be less than 90 and 60, pretty simple.

Now let us assume, in first place let us assume that both bodies move. Now if both bodies move then it is pretty clear that 120 minus 100, that is 120 left and 100 right, so 120 minus 100 is the driving force, which is trying to drive A and B towards left. Now when 120 minus 100 that is 20 is driving A and B towards 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, their values are 90 and 60, therefore the opposing force or the resisting force will be MuA into NA plus MuB into NB that is it will be 150 Newton. 120 Newton and 100 Newton were acting opposite therefore we can say 120 minus 100 is the driving force, friction will be opposing force because friction opposes motion, relative motion, kinetic friction opposes relative motion.

So over here, what can we see, we can see that since resisting force is greater than driving force therefore how can motion happen. Therefore this case is not possible. We can conclude that system will remain stationary it will not move. Our assumption was incorrect, right, students. So now when this assumption is incorrect that both A and B move then what is the next possiblity, A and B don’t move. Now when A and B don’t move, in that as well there can be two possibilities, which one. One possibility will be that friction will be static for both A and B, when I say static friction I mean to say friction will be less than limiting friction for both A and B. This will be possibility 1.

Second possibility will be that friction will be static for one and limiting for another. That can be the second possibility. That is first A will attain its limiting friction and then 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 in 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 Newton friction itself is sufficient, tension will not, even if two bodies are not connected with string, still both A and B will not move, right.  So if friction is static then we can say that tension in the string should be zero, right, students. Now if the tension in the string is zero, then to stop A, the friction on A should be 120, and to stop B the friction on B should be 100. But what do we know? We know if friction on A is 120 and friction on B is 100 this value of friction is just not possible because maximum values of friction were 90 and 60 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 limiting value of friction let us see.

Now let us first assume that 10 kg block has limiting value of friction, it reaching its limiting value first, then what can we say, what will be the limiting value – 90. Therefore on A, 120 is the force left, 90 that is the limiting friction becomes right, balance will be taken tension that is tension’s value will be 30. Now if tension is 30, then on B from FBD we can say that T plus f is 100 that is 30 plus f is 100, that is f becomes 70. But again if friction on B is 70, it again is greater than its maximum value which is 60, again this case is not possible. So the only possibility we are left with is B will attain its limiting value first and then balance friction will take care of A.  So if B attains limiting value that is on B friction will be limiting and that will be 60. So on B balance will be taken care by tension and it becomes 40 Newton. So if we take the same tension on A then we can say that f will be 80. And if friction is 80 we can easily see that this 80 Newton 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 40 Newton.

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 is not a very simple case always when we are dealing with multiple bodies connected with threads, right, students. Remember, first friction will try to oppose the driving force, then comes tension. Now just make sure friction never exceeds its limiting value.

Well, that’s all for now in this module, we will get back, till then, thank you, students.

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2017-04-18T04:55:21+00:00 Categories: XI-NEET & AIIMS|Tags: , , , |0 Comments
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