 Hello everyone, myself, Sandeep Javeri, Assistant Professor, Department of Civil Engineering from Valchin Institute of Technology, Sholapur. In today's session, we are going to discuss the problem on DLMB principle for the emotion of connected bodies on inclined plane. The learning outcome of this session. At the end of this session, the students will be able to solve problem on DLMB principle for the emotion of connected bodies on inclined plane. So this is a problem. The statement is two rough planes which are inclined at 30 degree and 60 degree to the horizontal are placed back to back as shown in the figure 1 below. The blocks of weights 15 Newton and 100 Newton are placed on the faces and are connected by a string which is running parallel to planes and passing over a frictionless pulley. If coefficient of friction mu is equals to one-third that is 0.66. So we are supposed to find the resulting acceleration and tension in the string. So here, now while looking to the figure, we can say that this block is trying to move in this direction. So which is connected to the another block having magnitude 15 Newton block. So this 15 Newton block is trying to move in upward direction while the 100 Newton block is trying to move in downward direction as this inclination of the surface is more and the weight is also more. So first of all consider freeboard diagram for a body which is moving down the plane. So this 100 Newton block is trying to move in a downward direction. So according to DLMR principle, there exists an inertia force which is opposite to the direction of motion. Now the direction of motion is in the down to the plane and the inertia force is opposite to that that is shown here. It is given by mass into acceleration. This mass is replaced by 100 by G. This 100 is nothing but the weight of this block that is 100 Newton. Similarly, this block is having self-weight 100 Newton. This weight can be resolved in two components. One is normal to this inclined surface and another is parallel to this inclined surface. The component which is normal to this inclined surface is 100 cos 60 degree which is acting towards the CG of this weight of the block and another which is acting along the direction of the plane that is in a downward direction that is 100 sin 60 degree. So one of the component which is acting parallel to this plane that is 100 sin 60 degree, the direction is shown as in the figure. The second is a vertical component of this force or this is a normal to the plane, inclined plane which is having magnitude 100 cos 60 degree, the direction is shown in the figure. Now as this block is connected to another block by means of a string. So if you consider the free board diagram, we can show the tension in the string in this way. Similarly, there is a coefficient of friction that is friction exists between this block and the surface. So this direction of frictional force is opposite to the direction of motion. And by law of friction, there exists normal reaction as it is in contact with the surface. So the direction of frictional force is opposite to the direction of motion. Direction of inertia force is also opposite to the direction of motion and this is normal reaction which is due to the point of contact of this block and the surface. So here all these forces we are shown here. Now let us consider free board diagram for the body which is moving up the plane. Now this is a 15 electron block. This is moving up to the plane. Now as it is moving up the plane, the inertia force exists according to DiM principle and the direction is opposite to that of direction of motion. So this is moving in a downward direction. Now this inertia force having magnitude is mass into acceleration and this mass is given by w by g into a that is w means the weight of the block which is given as 50 newton. So 50 by g into acceleration, now acceleration of the system is constant that is a. So for both the free board diagram this acceleration is same and for both the free board diagram the coefficient of friction value is given same. Now this 15 electron block is going to resolve into two components. One is normal to this inclined surface and another is parallel to this inclined surface. The normal component is 50 cos 30 degree and this parallel component is 57 30 degree. The directions are also shown in the figure. Now there exists friction force. So this value is F2 which is having direction opposite to the direction of motion and also there is a point of contact of this 15 electron block and this inclined surface. So there exists a normal reaction that is N2. So this is the direction of motion of this block of 50 newton and the direction of block of 100 newton is also given. Now this free board diagram is very important for writing the equation and from that equations we will obtain the value of unknowns that is acceleration of the system and the tension in the string. Let us look to another slide. Now as we know the T be the tension in the string, A be the acceleration of the system, value that is the coefficient of friction which is same for both the inclined surface which are in contact with the 100 newton and 15 newton block. Now by law of friction we have coefficient of friction is equals to the F by N. F is a frictional force and N is a normal reaction. Now let us consider Fbd of 100 newton body which is moving down the plane. So here from this free board diagram we can obtain the equation. If you consider summation of forces normal to the inclined plane, so N1 minus 100 cos 60 degree equals to 0, N1 will work out to be 15 newton. Similarly by using law of friction and knowing the value of mu as one third we are getting F1 is equal to mu into N that is equals to 16.67 newton. Now we can apply the real-life principle at dynamic equilibrium summation of forces parallel to inclined plane equal to 0. So we have T plus 100 by g into A that is inertia force minus 100 sin 60 degree plus F1 equal to 0. Now putting all the values of g as 9.81 and F1 as 16.67 we are getting one equation which is in the form of tension in the string and acceleration that is T plus 10.193 into acceleration of the system that is equals to 69.932 that is expression number 2. If you consider the free board diagram of 50 newton body which is moving up to the plane, so consider summation of forces which is normal to the plane equal to 0, so we are getting the expression as so we are getting N2 minus 50 cos 30 degree equal to 0. So N2 will be 43.301 newton since we know F2 is equal to mu into N2, so mu is one third and N2 is 43.301. So we are getting the frictional force for the body of 50 newton block is 14.433 newton. So this is from law of friction that is mu is equal to frictional force divided by the normal reaction. Now applying DLM principle at dynamic equilibrium summation of forces parallel to inclined plane equal to 0. So we have this T minus 57.30 degree minus 50 by g into A minus F2 will be equal to 0. So here T minus 50 by g into A minus 57.30 degree minus F2 equal to 0, so by putting the value of g 9.81 and the value of frictional force as 14.433 we are getting the equation in terms of tension in the string and acceleration of the system as T minus 5.096 into A which is equal to 39.433. So by solving equation number 2 and by solving equation number 3 we are getting the two variables that is two unknowns, we are getting the two unknowns that is tension in the string and acceleration of the system as 49.598 newton and acceleration is equal to 1.994 that is 2 meter per second square. So we are supposed to pause the video and answer this question, so this is a free board diagram. So as the body is moving towards rightward direction the tension in the string is T and the inertia force is opposite to the direction of motion that is 200 by g into A, the weight of the body is 200 ton which is acting vertical downward, the frictional force is acting towards leftward direction that is opposite to the direction of motion. Similarly for 800 ton block the direction of frictional force is opposite to the direction of motion and the direction of inertia force is also opposite to the direction of motion and the external force is shown as 400 ton, the tension in the string is shown here towards left direction for this n 1 block that is for 200 ton block it is showing to rightward direction. So after drawing this free board diagram we are able to solve the equations and from that equation we are getting T and A, these are the references which are used for the creation of this video. Thank you.