 Hello friends, Myself Sandeep Javeri, Assistant Professor Department of Civil Engineering from Valchance Stop Technology, Sholapur. In today's session, we are going to discuss a problem on Varegnon's Theorem. The learning outcome, at the end of this session, students will be able to solve problem on Coplanar force system using Varegnon's Theorem. Now this is a problem. In this problem, a triangular plane is shown and it is subjected to forces at different vertices. At A, there is an inclined force of 100 Newton is acting and having inclination of 60 degree with the horizontal x axis. There is another force which is acting at B, having magnitude 120 Newton and having inclination with x axis as 30 degree. There is another force acting at point C, one is acting vertical downward, having magnitude 80 Newton and another force which is acting at point C, having magnitude 80 Newton which is acting rightward direction as shown in the figure. Now the problem is to find resultant force and locate its position from point A. So we have to find out the position of resultant force from point A and also we have to find out the magnitude of resultant force. Now these are the steps involved while solving the problem. Step one, first of all we have to resolve the forces into x and y components. Step two, we have to find out the summation of all forces in x direction and summation of all the forces in y direction. For this, we have to refer the sign convention. If you are going to find out summation f of x, the rightward direction force is taken as positive and leftward direction force is considered as negative. While calculating summation of f of y, we have to consider upward direction force as positive and downward direction force as negative. Then step three, we have to find out the resultant force r and its direction. So for this, we have to use the equation r is equal to under root of summation of x square plus summation of f of y square. And after that, we have to locate position of r, for this we have to use very nonce theorem. So the solution for this problem, first of all we have to find out summation f of x using the sign convention. So f of x equal to minus 100 cos 60 degree plus 80 minus 120 cos 30 degree. So you can see here, after resolution of forces, we can see the minus 100 cos 60 degree that is x component of this force 100 Newton and minus 120 cos 30 degree that is a x component of 120 Newton and there is a rightward direction of force of 80 Newton. So it is plus and after solving this equation, we are getting summation f of x is equal to minus 73.92 Newton. Similarly, we can find out summation f of y that is equals to 100 sin 60 degree minus 80 minus 120 sin 30 degree and which is equals to minus 53.40 Newton. So you can see in the figure, this is a y component, it is positive here. The another y component for 120 Newton is in downward direction and it is minus and this is a 80 Newton which is in downward direction, so it is also minus. So after putting all these values, we are getting the summation f of y is equal to minus 53.40 Newton. So magnitude of resultant we can obtain by using the expression under root of summation f of x square plus summation f of y square by putting the values of summation f of x and summation f of y, we are getting the value of resultant force as 91.190 Newton. Similarly we can find out direction of resultant force theta is equal to tan inverse of summation of f of y upon summation f of x. So this is in the mod. So after putting the values of summation f of y and summation f of x, we are getting the value of theta as equals to 35.84 degree. Now to locate position of resultant force, we are going to use a Varyon's theorem. So applying Varyon's principle summation of m at a, for this we are going to use the sign convention as clockwise movement as positive and anti-clockwise movement as minus. So summation of m at a we can work out as 80 multiplied by 50 plus 80 times 100 sine 60 degree plus 120 sine 30 degree multiplied by 100. So here 80 is a force and 50 is a propellant distance from point a. Similarly here 80 is a force and 100 sine 60 degree is a propellant distance and 120 sine 30 degree is a force and 100 is a propellant distance. So you can see here this 80 into that propellant distance that we have worked out. Similarly this 80 into this propellant distance that we have worked out. This is with respect to a and for this 120 sine 30 degree we have worked out the propellant distance AB and is equal to 100 mm. So we are getting the movement at a is equal to 16928.203 Newton mm. So by putting this value in the expression r into d equals to summation of movement at a which is 16928.203 Newton mm. So by putting this value of r we are getting the value of d that is propellant distance as 185.635 mm. Now to find x intercept we are going to use the expression summation of y into x that is equals to summation of m at a. So by putting the value of summation of y 53.40 so we are getting x equals to 317.007 mm. So after finding a resultant force we can show this case for resultant force by showing summation of x and summation of y. So we draw these rectangular components then this force represents resultant force and this angle is theta that is direction of resultant force. So here r lies in third quadrant. Now after finding the x intercept we can also show the location of resultant force. Now this is our x direction this is a vertices of triangle this is point a. So our resultant lies in third quadrant so I can show here the perpendicular distance from a it is d and also the x intercept from point a. So in this way we can show the position of r from point a. Now here we are supposed to pause the video and answer this question now this is the answer if summation of m of all forces is given as 5000 Newton meter and perpendicular distance of resultant force from a movement center is 2 meter find the magnitude of resultant force. So this value work out to be 2500 Newton these are the references that we are using for conducting this session thank you.