 Hello everyone, myself Sandeep Javeri, Assistant Professor, Department of Civil Engineering from Valtryan Institute of Technology, Sholapur. In today's session, we are going to discuss problem on DLMR principle for lift which is in motion. This is a part of dynamics and the subject is engineering mechanics. The learning outcome at the end of this session students will be able to solve problem on DLMR principle for lift which is in motion. Now before starting the problem on DLMR principle, we first know or we should understood what is the DLMR principle. So the statement is the DLMR states that the impressed forces which are acting on the body are in dynamic equilibrium with inertia forces of the body under motion. So to understand this, let us consider this body having mass m and which is moving in the right foot direction and it is subjected to external forces f. If there is a friction exist or if you consider the DLMR principle then it says that during the motion of this body there exist inertia force which try to tend it to the opposite direction. So this inertia force is acting in the opposite direction. So the mass is replaced by the weight that is mass into access into the gravity mg. So or we can say this mass is represented by W by G. This mass is represented by W by G. So DLMR principle states that the resultant of external forces and the inertia force is always 0. This is the condition of dynamic equilibrium. So if you write down the expression for this, you are getting f minus W by G into a equal to 0. So we are considering summation of force in x direction equal to 0. So we are getting f that is a force which is acting on a body having mass m is equal to W by G into a. So here this DLMR principle says whenever the body is in motion there exist an inertia force which is opposite to the reaction of motion and there exist a dynamic equilibrium. So now let us consider the problem, a lift carries a weight of 100 Newton and is moving with a uniform acceleration of 2.45 meter per second square. Determine the tension in the cables supporting the lift when lift is moving downwards and lift is moving upwards. Take accession due to gravity as 9.80 meter per second square. So here there are two cases given. One is lift is moving in downward direction and the lift is moving in upward direction and we are supposed to find out the tension in the cables in both the cases. Let us consider the solution for this. We require the free board diagram for the lift which is in motion that is in downward motion. So as we know let T be the tension in the string this is because of the cable which is attached to the lift. This is a lift this is say rectangular shape. Now the motion of the lift is given as downward direction. So this is the accession direction is given that is nothing but the direction of motion. Now according to DLM principle there exists an inertia force that is W by G into A that is W is the weight of the lift that is 100 Newton. Now the weight of lift is acting in a downward direction that is shown here. So what are the forces which are acting on the body that is shown by the free board diagram. So there are different forces which are acting on the lift one is T there is a tension in the cable then there is a weight of the lift itself that is 100 Newton and the inertia force according to DLM principle that is acting in opposite to the direction of motion that is W by G into A. Now from this expression we are getting one equation that is if you consider all the forces in an equilibrium and applying the DLM principle we are getting T plus W by G into A equal to W we are considering upward forces are positive and downward forces are negative. So these are the upward forces and this is the downward force. So T plus W by G into A is equal to W if you put the value of W as 100 and G as 9.81 acceleration as 2.45 then we are getting the tension in the lift or tension in the rope as 24.97 Newton. Now consider the case 2 that lift is moving in the upward direction for that again we have to consider the free board diagram. So in this diagram the lift is moving in upward direction so this is the direction of motion and T is a tension in the string and W is a weight which is acting what you can downward through the CG of this body. Now inertia force is acting opposite direction of motion that is in downward so it is given by W by G into A. Now rewriting the equation now we are supposed to write the equation by using the DLM principle. So according to the DLM principle considering all the vertical forces in the equilibrium so T minus W by G into A minus W equal to 0 so T that is the tension in the string is equal to W into 1 plus A by G equal to 0 put the value of A as 2.45 G as 9.81. So we are taking out the W that is a weight common that is 100 so we are getting the tension in the cable as 2249.949 say 250 Newton. So this tension is almost 10 times than that of the previous case so in previous case we are getting the value of tension as 24.97 it is nearly about 25 Newton when the lift is moving in a normal direction and when lift is moving in upward direction we are getting the tension in the cable is almost 10 times that what we obtained in the previous case. So it is obviously more because when lift is moving we are entering in the lift so we feel that there is a heavy weight when we are inside the lift because our direction of motion is opposite to the acceleration due to gravity and when lift is moving downward that is in the direction of acceleration due to gravity so the tension which is created in the lift is in the cable is less now you are supposed to pause the video and answer this question so here in this problem a lift has an upward acceleration of 1 meter per second square and the question is what reaction offered by the man having weight 600 Newton on the floor of the lift. So for this we had to first draw a freeboard diagram let us consider the freeboard diagram a man is entering the lift having weight is 600 Newton and this is a floor which is in contact with that man and there is a resistance is offered from this floor that is R so let us consider A be the acceleration of the system that is 1 meter per second square given W is the weight of the man that is 600 Newton and R is the reaction offered by the man on the floor and mass of the man in kg that is written as W by G so that is 600 by 9.81 now using the freeboard diagram of the lift which is moving upward direction we are getting this freeboard diagram this is the direction of motion direction of inertia force is opposite to that direction of motion now using the DLM principle at dynamic equilibrium considering the vertical forces so consider upward forces are positive now what as negative so R minus W minus W by G into A equal to 0 so W is 600 put the value of W as 600 9.81 as G we are getting R is equal to 661.16 Newton so when lift is moving upward we are feeling heavy weight or we are offering the more resistance reaction offered by the man during upward motion of the lift is equal to 661.16 Newton now you are supposed to pause the video and answer this question these are the answers as we know DLM principle direction of inertia force is opposite to direction of motion and when lift is moving upward direction of the reaction offered by the man having weight W is on the lift is given by R is equal to W plus W by G into A so A is the right answer these are the references which we are considering for the creation of this video thank you.