 Problem on Equilibrium for Non-Corcoran Force System. Myself, Bipin Patil, Assistant Professor, Department of Civil Engineering, Vulture Institute of Technology, Solapur, Learning Outcomes. At the end of this particular session, students will be able to learn two points that is explain the concept of equilibrium, second one, analyze and calculate the forces present in a system. First, we will discuss what do we mean by equilibrium. My dear friend, this is very important in civil engineering, in mechanical engineering. Whenever we are connected with the system of forces, first point that is very important to keep the body it is in a rest condition. If the body it is in a rest condition means your whole system it is an equilibrium. So, you observe your any system of forces that keeps the body at a rest is said to be in equilibrium. In other language you must consider this particular statement, the state of the body is not affected by the action of the force system. The state of the body means if P, Q, R, S forces are present all over your body and one force you consider this particular one force it keeps the body it is in a rest condition there is no effect on the body by considering this particular remaining forces. So, this particular action we call as equilibrium, the state of the body is not affected by the action of the force system we call as equilibrium. Second parameter that is equilibrium is applicable to those forces those system of forces whose resultant action is 0. So, let us consider this particular picture it gives you the idea number of forces system of forces for example you consider P and Q this whole particular system of forces keeps the body it is an equilibrium by one force that is nothing but your resultant and what do you mean by equilibrium force that brings equilibrium is nothing but such types of force that brings the whole system of forces in equilibrium is known as equilibrium and this particular equilibrium is always equal opposite and collinear with resultant of the force system. So, E is equal to R next these are the some steps how to identify the unknown forces total four steps we are going to consider here first step select such a joint where one known force should be there one known force should be there means if I consider any one joint so at that particular joint if one force is known remaining all unknowns are there easily we can identify the remaining unknown forces. Second step draw the free body diagram by considering any one joint. So, you observe it your problem setmate how many joints are there with respect to that we just try to identify and then we apply the equations that is three equation one is for horizontal forces second one is for vertical forces third one that is for movement. So, here also you assume the directions for vertical force positive sign horizontal force right hand side positive clockwise rotary effect positive and vice versa negative and then find out the unknown forces let us we will discuss one problem a bracket with forces shown in a figure three forces 400 150 200 and one couple is there find out the magnitude direction and line of action of the resultant node A node B is present three forces one couple so first you mentioned the component of this particular inclined forces 200 as its 10 meter clockwise as it is 150 Newton you mentioned the two components moves away from the node so one is left hand side one is downward it moves away from the node one moves in a poor direction second one in a left hand side now you apply the equilibrium equations first summation f of x is equal to 0 means all horizontal forces so here you do not equate it to 0 you just identify the horizontal forces left hand side forces negative sign right hand side forces positive sign so only two forces are present one is 400 cos 45 that is negative second one 150 cos 30 so f of x is equal to minus 412.7 Newton narrow indicates indicates negative sign means left hand side forces so f of x that is 412.7 Newton and the arrow is this one means negative sign second one consider equation of equilibrium that is f of y is equal to all vertical forces 200 upward positive 400 sin 45 positive 150 sin 30 you consider negative calculate it and mention f of y is equal to 407.8 Newton upward indicates positive now find out the resultant this is the formula resultant and the direction so r is equal to under root horizontal that is f of x square plus f of y square or you consider h square plus v square and two direction you consider time inverse summation v upon summation h or vertical forces divided by horizontal forces so put this particular value in the formula so resultant is 580.2 Newton 412.7 square that is horizontal 407.8 square that is vertical the output is 580.2 Newton similarly here also vertical 407.8 divided by 412.7 so the direction alpha is is 44.76 degrees there so mention it is in a quadrant system that is f of y upward f of x that is negative value so your resultant lies in this particular quadrant so the value of r is 580.2 and with respect to horizontal it comes 44.76 degrees there so this is one type of problem is that for more clarification we will discuss movement means we require the direction distance from support a so I am considering at point a it is 0 so movement at point a first one you consider this minus 400 sin 45 it is upward and it creates anticlockwise root defects and the distance is 3 meter minus 400 cos 45 into this particular distance 0.6 plus you consider 150 now this one 150 cos 30 this distance is 1 meter because we are considering at support a movement second one 150 sin 30 so it is present here and the whole distance is 6 meter and it creates clockwise root defect so positive sin plus 50 that is the couple so movement at a is equal to 438.3 Newton meter so you identify the distance how to identify the distance from support a movement at point a divided by summation f of y vertical so 438.3 divided by 407.8 so the distance from support is 1.074 meter so from support a it acts 1.074 and the direction is present according to this way let us we discuss one more problem determine the resultant of pole system as shown in a figure same point two point loads one inclined load so for inclined you consider these two components similarly summation f of x is equal to 0 all horizontal force is only one force minus 20 cos 60 so f of x is equal to minus 10 so direction you consider 10 kilo dot f of y you consider negative positive minus 20 minus 30 minus 20 sin 60 minus 30 okay so these particular forces so f of y is equal to minus 67.3 so either upward force is positive or lower force is negative so according to that you identify here next one the same formula put the values in this particular formula so time square 67.32 square r is equal to 68.06 and alpha how to calculate it this is v so you consider 67.32 this is divided by 10 so alpha comes 81.55 degree so you consider horizontal force right hand side positive second one f y negative so it moves in a downward direction so your resultant lies here likewise easily you can calculate the resultant force in their direction now the moment is there so consider moment at point a similarly 20 into 1.5 30 into 3 all it creates in clockwise rotary effect so positive sign 20 sin 60 into total 6 so m a is 223.92 similarly distance m a divided by f y so put the value and find out the answer so x that is 3.3 to 6 meter from support a so easily you can identify the distances now let us you pause the video try to give the answer of this particular question this is the answer to prepare this particular session I refer this particular books references thank you.