 Now, let us introduce the universal law of gravitation. I hope there is no doubts, write down universal law of gravitation. The universal law of gravitation basically tells you that there will be a gravitational force, gravitational force of attraction between two point masses. The law is about two point masses, it is not about two big objects. So if two masses are there, if two point masses are there, the dimensions are very small, then you can use universal law of gravitation to find the gravitational force of attraction between the two objects, understood? So what Newton thought or Newton has found out that that gravitational force of attraction is proportional to multiplication between the two masses and same force is also proportional to 1 by r square where r is a distance between the two masses, r is a distance between the two masses. So Newton has used the concept that the planets are moving in a circular orbit. So v square by r should be the acceleration, so using that he has come out with this kind of expression. But then we are not getting into all that. So we are directly assuming that this is known that force is proportional to multiplication between the two masses and force is also proportional to 1 by distance square between the two masses. And of course when you talk about heavenly objects, they are not the point masses, they are spherical, huge spherical objects. As in earth you can consider as huge sphere but soon you will realize that huge sphere can be treated like a point mass that we will discuss little bit later. Right now let us just discuss about the universal law of gravitation. So this is what is observed and force is a vector. So direction of force will be along the line joining the two masses. So it is an attractive force and it will be along the line joining the two masses. So suppose you have these two masses, let us say this is m1 and this is m2, line joining these two masses is this. So this is the gravitational force by m2 on m1. Can you tell me which direction the gravitational force will be on m2 due to m1? This force is on 1 due to 2. You need to find out the direction of force due to 1 on 2 where it will be in opposite direction again along the line joining. And according to the Newton's third law if m1 is pulling m2 with this force, m2 will pull m1 in opposite direction with equal force. So according to Newton's third law these two forces are pairs, equal and opposite pairs. So let us say the distance between them is r. The gravitational force is given as a constant which is universal gravitational constant g into m1, m2 divided by r square and this is attractive in nature. Why I have written attractive in nature is that you do not need to know in vector notation about the direction of the force. Every time you have to represent the gravitational force just assume the mass and see where are the other masses. So the attraction forces will be towards the other masses. Other masses will pull this mass and that is how the direction of force will be determined. So individually case to case basis you can see how the directions can be taken care of. But if you want to write in a vector notation, for example in your school exam if they ask you to write down the universal law of gravitation in vector form, vector notation let us say they ask you to find the vector notation. Then let us say you have two masses given in the space this is m1 and this is m2 then their locations will also be given. So in a way you can say that their position vectors are given. So let us say this is r1 vector and this is r2 vector like this. So can you write down the gravitational force between m1 and m2 in vector form? Can you try that quickly to that? See the magnitude of force has to be equal to g times m1 m2 divided by distance between the masses square. Now how will you get the distance between these two points? Those two points are r1 and r2 position vectors. So distance how will you get? This distance how will you get? You can find out this vector. You can find out this vector let us call it as r12. So you can see that there is a triangle. If there is a triangle you can also write r1 plus r12 some of these two vectors should be equal to r2. So r12 you can write it as r2 minus r1. So magnitude of r12 vector is the distance between the two masses. So magnitude of r2 minus r1. So the force magnitude is g times m1 m2 divided by magnitude of r12 vector the whole square. Now it is a vector we have just written in the magnitude of it. But if you want to write the vector f then you need to multiply a unit vector in the direction of the force. So let us say I am trying to find out force on m1. Force on m1 it is a vector quantity so that multiplied by the unit vector along the direction of r12 like this. Now it is a vector force. Unit vector can also be written as the vector divided by its magnitude. By the way whatever we are discussing the vector notation will not be used in problem solving. So it is just for your school derivation purposes I am doing it. It is hardly used in problem solving. This is unit vector. So when you simplify it further you can write it as g m1 m2 divided by r12 whole cube multiplied by r12. So this is the vector notation of the gravitational force between the two point masses. Any doubts? Anyone? Is this thing clear? No doubts right? Fine. So now let us solve few numericals on something we have learned just now. So here is an equilibrated triangle. You have let us say three masses. This is three equal masses m, m and m. This is let us say what is A, this is B and this is C. You need to find the force net force on mass kept at A. It is an equilibrated triangle. The side length is A, small a. All of you try this? Fine. So of course this mass will get pulled by these two other masses. So one pull is in this direction, other pull is in that direction. So if this is force F that is also F only. So magnitude of force will be same both sides. So magnitude of the force F will be equal to g times m1 m2 will become m square divided by A square. Fine. This is the magnitude of force. Right? So in problem solving I do not write in vector form. I just find out the magnitude and direction is anyway obvious. So that is how I will deal with it. Same force is over here also. So now the scenario is like this on A there are two forces like this F and F. The angle is 60 degrees. Okay, Devdas quickly do this. So it is like vector addition of two forces which are at an angle of 60 degrees. So you can do that it will be A square plus B square plus 2 AB cos of angle between the two vectors. So I am using that formula cosine law. Right? Now theta is 60 degrees. If theta is 60 degrees I can write it as F square plus F square cos of 60 is half. This will also become F square only. So this is root 3 times F magnitude of the resultant. Right? Yes. So resultant will be equal to pi 2 kym aabey no Devdas no resultant will be root 3 times F which will be root 3g m square divided by A square. Okay? And you can easily make out the direction. Right? You do not need to use any law as such. You can apply a symmetry logic. The line of the resultant force has to be perpendicular sorry the angle bisector in this direction because the forces in that direction will get cancelled off. The component horizontally will get cancelled off the only vertical component will remain. So you can solve it like that only you can say F cos F cos 30 plus F cos 30 is the total force. Okay? So like that also you can solve the same thing. You can say that this angle is 30 degrees right? So component along the vertical direction is F cos 30 okay? And component of this force along vertical direction is again F cos 30 only so 2 F cos 30 is the resultant force because horizontally F sin 30 and F sin 30 will get cancelled off. So only vertical component will remain. Look at the same answer. Clear to all of you? Now suppose a mass is kept, a mass of capital M is kept at the centroid okay? This is the centroid of an equilibrated triangle, side length is A. What is the force capital M will feel at the centroid of this triangle? Symmetry? Symmetry is Sukrit logic that is correct. So tell me all these three forces will be equal or not? This capital M will get pulled by this small m, by this small m, by that. The equal amount of force right? This is F, this will be F and that will be F right? So horizontally in this direction you will see that the forces will get cancelled off which is 90 degree to the vertical force right? So the component of this force will cancel out component of that force horizontally okay? Vertical direction if you see this is the vertical direction okay? This angle is 60 degrees okay? So vertical F cos 60 plus F cos 60 because of those two forces it will be F only, F by 2 plus F by 2, F down and there is one force F up okay? So vertical direction also all the forces will get cancelled away okay? So net net everything get cancelled off horizontally and vertically right? And hence the total force on capital M will be 0 because this is your NCRT question alright? Okay now let us talk about few cases. These cases will help you in problem solving they are related to like if you encounter a case in a question then you should not start from scratch. So these scenarios are very common so that is why we are doing it as case study okay? Whatever we have done till now is it clear to all of you any doubts quickly ask then I will proceed.