 In today's session, we are going to discuss a small problem on equilibrium for two cylindrical sphere. Myself Bipin Patil, assistant professor, department of civil engineering, Vulture Institute of Technology, so Labur. So, my dear friends, these are the learning outcomes for today's session. So, at the end of this particular sessions, you may learn concept of equilibrium, then calculate the forces present in a member. So, you consider the sphere and according to that, you identify the forces present in the particular segment of members. Before that, we focus on this particular figure, body here showing resultant and equilibrium. Earlier, we discussed all these particular sessions, what do you mean by resultant, equilibrium. So, you consider this particular system of forces, P and Q, these particular two forces are present to keep this particular body, it is in a rest condition, that is nothing but your equilibrium. So, two forces are present, keep this particular forces. In equilibrium, we required one more force, the additional force that is your resultant. Now, the whole system, you just try to keep as it is in a rest position. So, for that, we required one more force that is your equivalent. So, resultant and equilibrium, it is exactly having same magnitude, opposite direction and collinear with each other. Equilibrium is applicable to those system of forces whose resultant action is 0. So, there is a resultant action is 0 here, that is why we called as A is equal to R. Next, analytical conditions of equilibrium, now same figure you referred here, two conditions are present, one is force law, second one is movement law. So, what it indicates, force law and movement law. It is very simple, in force law we are considering horizontal as well as vertical forces. The algebraic sum of forces, it may be horizontal, it may be vertical, acting on this particular body. For example, I am considering this particular body. So, the algebraic sum of forces acting on this particular body is 0, this particular law is called as force law. Similarly, we are considering third one that is movement is equal to 0, that is movement law. Over the same system of forces, the algebraic sum of movement of forces about any point may be here, may be here, may be anywhere. The algebraic sum of movement of forces about any point is as 0. So, it is called as the movement law of equilibrium. So, we are considering summation of movement as equal to 0, any point here summation h, summation v force law in movement law summation of m is equal to 0. Here also we are using the same sign conventions and some steps to identify the forces. Step a, identify the contact surfaces. First you identify the contact surfaces. Step b is there, draw the free body diagram by removing the contact surfaces. Remove the contact surface and you mention the free body diagram. Step c, apply the equations. So, two equations are present. One is Lamis theorem. Second one is static equilibrium equation. So, Lamis theorem, three forces and proportional that particular sine of angle you consider. Static equilibrium equation earlier we discussed movement law and force law and find out the unknown forces. So, for that we required some sine conventions, all vertical upward forces, positive sine, right hand side forces, positive sine, clockwise rotary effect, positive sine, anticlockwise rotary effect, negative sine. Then left hand side forces, negative sine, all downward forces, negative sine. Let us discuss one problem. Focus on this particular smooth identical sphere. Two identical cylinder, this particular two, one and two. Each weighing 500 Newton means the weight they are mentioning a 500 Newton. Our place in a through means you consider here as shown in figure means the inclination they are mentioned here with respect to horizontal 30 degrees there. And the reaction developed at a contact point A, B, C and D. So, E is present here, B is present here, C is present here, D is here. As if all points of contact are smooths means we are considering frictionless. So, first according to our steps you observe here contact surfaces we identify now free body diagram remaining the contact surfaces. Then we are going to apply the equation and then unknown forces. So, this is your problem set mean and free body diagram of a smooth sphere. So, I am removing this particular two contact surfaces, one, two, similarly from base. So, mention here the reactions. It is always present towards the surfaces. So, if I consider sphere number one. So, free body diagram for sphere number one. This is free body diagram for sphere number two. So, you observe A that is reactions at A or A and this particular angle. So, you consider here x and y axis with respect to this particular base. So, this particular angle is 30 degree because the inclination is 30 degree if I rotate this and you identify the geometry. So, whenever you are preparing a schematic diagram. So, this angle with respect to y axis means with respect to x and y axis it is here 30 degree. So, with respect to y axis this particular angle is 30 degree and this angle is 60 degree. So, for this is same for sphere two also. So, RA this inclination is 30 degree. Sphere way they have mentioned 500 Newton. So, it is in a downward direction 500. Next one contact surface B is present. So, mention this one. So, here also this particular angle with respect to x axis 30 degree. Similarly, for sphere two contact surfaces in one rate 500 Newton downward RC this particular with respect to this particular 30 degree and one more it is exactly perpendicular to your vertical plane. So, this is RD. Let us we apply the equations. First one consider sphere one. So, in your sphere one one moves in a downward direction. So, I am considering I am changing the geometry, but I am not changing the direction. So, I am considering 500 Newton present here. This is RB present here towards the joint. This is either you consider all forces towards the joint or away from the joint. So, sphere one RA RB and next one that is your 500 Newton. So, you calculate this particular angle. So, this particular is 30 degree and remaining one this is your 60 degrees there. So, total 90 degree. Similarly, I am considering 500 up over. So, 30 plus 90 and here also 90 plus 60 is there. So, with respect to that I have calculated 90 plus 60 150 90 plus 30 120 because I require this particular angle for Lame's theorem. So, you apply the Lame's theorem first one 500 Newton force and sign up the angle between remaining two forces. So, 90 sin 90. Similarly, RA and this particular sin 120 RB sin 150. So, in this particular statement one force is known. So, equate 500 sin 90 to RA 500 sin 90 to RB and you calculate this particular to value. So, RB sin 150 equated to 500 sin 90 find out the value of RB it comes 250 Newton. Similarly, consider 500 this particular force divided by sin 90 equated to RA sin 120 find out the value of RA that is nothing but your R2 or RA is there. So, this particular RA is equal to you consider 433 Newton is there. So, according to sphere one by applying Lame's theorem easily can calculate RA and RB. Next one you consider free body diagram for sphere two means this particular sphere. So, all are towards the joint. So, we require this angle 90 plus 30 then we require this particular angle. So, let us see how to solve the problem. So, here we are not applying Lame's theorem we are applying the static equilibrium equation. Already we have calculated the value of this 500 is known RB already we have calculated okay RC is present here and RD is unknown. So, you apply the static equilibrium equation summation h is equal to 0. So, RB cos 30 is there RB sin 30 here also you consider RC cos 60 RC sin 60. So, summation h is equal to 0 all right hand side force is positive sin RD minus RB cos 30 minus RC cos 60 is equal to 0 equation 1. Here two unknowns are there. So, we will move towards equation number 2 summation v is equal to 0. So, all vertical downward forces negative minus 500 plus RC sin 60 minus RB sin 30 is equal to 0. Now, put the value of RB here in calculate the value of RC. So, RC is equal to 721.7 Newton okay we are applying static equilibrium equation. So, from equation number 1 put the value of RC in equation number 1 calculate the value of RD. So, RD minus RB cos 30 minus RC cos 60 is equal to 0 is there. So, RD minus 250 cos 30 minus 721.7 cos 60. So, the value of RD is 577.4 Newton. Likewise easily you can calculate the unknown forces by considering equilibrium equation or laminar. So, my dear friends you just try to pause the video practically give the answer of this particular question. This is the answer of this particular question. To prepare this particular session I refer these are the references. Thank you.