 I am Swati Gadge, Assistant Professor, Department of Civil Engineering from Walton Institute of Technology, Solarpur. Topic for today's session, Ladder against wall friction. At the end of this session, learner will be able to solve problem of ladder against wall friction. In the figure one, the person is standing on the ladder and ladder is on horizontal floor and rest against vertical wall. We will draw the free body diagram of that ladder. This is a line showing ladder. The top point is A and the base point is B. At point A, it is supported with the help of vertical wall and at point B, it is supported with the help of floor. Self-heat of the ladder will act at the center of ladder and in vertically downward direction. The person is standing near to the upper end of the ladder and the self-heat of that person will act in vertically downward direction. For the free body diagram, we have to remove the support and instead of that support, we have to show the support reaction. So at point A, after replacement of vertical wall, the horizontal reaction will be act and at point B, after replacement of horizontal floor, the vertical reaction will be act. The point A will have the chances to slip down and the point B will have the chances to slip in right direction. So to avoid that motion, the frictional force will act in the opposite direction of motion. So frictional force is also shown in this figure. So free body diagram of this ladder is complete now. Now we will solve one numerical on ladder. Question is, a ladder of 5 meter long rests on horizontal ground and leans against a smooth vertical wall at an angle of 70 degree with the horizontal. Weight of the ladder is given 300 Newton and the man weight is 750 Newton and it stands on it 1.5 meter high. Question is, calculate coefficient of friction mu B between ladder and the floor to prevent the slipping of the ladder. So all the given data are highlighted here. So with the help of that, we will try to draw the free body diagram of the ladder. Ladder of length 5 meter and the upper end we will take as A and lower end of ladder we will take as B. Support at A vertical wall is supporting the ladder and at B the horizontal floor is supporting the ladder. Self height of the ladder that is 300 Newton given, it is shown at the center of the ladder in vertically downward direction. Angle is given 70 degree. So shown here in the figure. And the person is standing on the ladder at a height 1.5 meter high. So after replacement of the support, we have to show the support reaction. So support reaction at A is horizontal and support reaction at B will be vertical. And here only the frictional force in horizontal direction will act at B point because we have to determine mu B. And it is given that the vertical wall is a smooth wall. So friction offered by the vertical wall will be 0. So this is the complete free body diagram of this ladder. Total how many forces are there acting on the ladder? One is the support reaction at A. 300 Newton, self height of the ladder 750 Newton is the person weight standing on the ladder. RB and FB are the support reaction at point B. So here ladder length is given 5 meter. This is a non-concurrent force system. So we have to use a moment equation. The moment means force into distance. So we have to determine the distance of all the forces with one reference point. So for here we will take B as a reference point. Let us determine first the distance of RA from point B. So this is the force RA and here the point B is. So why is the distance of RA from point B? So if we form a triangle by considering ladder as a hypotenuse. So y is the opposite side of 70 degree. So for opposite side we always use a sign. So y is hypotenuse into sign 70 according to the trigonometry. So we will get y 4.698 meter. Similarly we will determine the x1. X1 is the distance of 750 Newton force from the support beam. So consider this small triangle. We have that opposite side of 70 degree is 1.5 and angle is 70 degree. So we will use here tan 70. Tan 70 is equal to opposite upon adjacent. Adjacent is x1. So x1 is equal to 1.5 divided by tan 70. So 0.546 is the x1. Now we will determine x2. X2 it is a distance of 300 Newton force from support beam. So for that we will consider that small triangle here. Ladder part is 2.5 meter and x2 is the adjacent side of 70 degree. So we will go for the cost formula. So x2 is equal to 2.5 into cost 70. So x2 is equal to 0.855. Now we will use equilibrium equation that is summation of vertical forces is equal to 0. So RB is equal to 300 plus 750. Only three forces are there in y direction RB 300 and 750. So RB must be equal to 300 plus 750. So it is 1050 Newton. Now we'll take a moment about point B. For the reference, we select point B for taking moment. And we have already determined all the distances from point B. So first take a moment of RA about point B. So RA into y is the moment. And to determine whether it is clockwise or anti-clockwise, just rotate your hand in the direction of four. So it is a clockwise moment. So plus RA into y. So similarly, determine the moment of 300 Newton. So 300 Newton into x2 and rotate your hand in the direction of four. So it is anti-clockwise moment. So minus 300 into x2. Similarly, 750 also giving you the anti-clockwise moment. So 750 into x1, it is anti-clockwise moment. So it is minus 750 into x1. For that, we have used the sign convention. For clockwise moment, we use a positive sign. And for anti-clockwise moment, we use a negative sign. All the calculated values of x1, x2, and y put in the equation and find out RA. So RA will come 141.762 Newton. Now we will use this third equilibrium equation, that is summation of fx is equal to zero. RA and RB are in the x direction. So these two forces must be same. So RA is equal to RB is equal to 141.762 Newton. Now here we have FB and RB. So from the coefficient of friction formula, the coefficient of friction is equal to FB upon RB. So put the values of FB and RB in this formula and find out the coefficient of friction at B. It will come 0.135 Newton. Now we'll solve now second numerical. A four meter ladder is weighing 200 Newton. It's placed against a vertical wall. A man weighing 800 Newton reaches a point 2.7 meter from A. The ladder is about to slip. Assuming the mu between ladder and the vertical wall is 0.2. Determine the coefficient of friction mu between the ladder and the floor. So you pause video here. Similar to the first problem, you try to solve and find out the value of mu. Let us see the solution of that problem. So free body diagram is shown here. So similar way we will determine all the distances y, x1, x2, and x3 here because here vertical wall is also offering some friction. It is in this numerical it is not given that vertical wall is smooth. So FB will be there. So we have to determine the distance of that FB with respect to A. That is a reference point, okay? So y, x1, x2, and x3 determine like this by using a trigonometrical formula. The coefficient of friction at vertical wall and the ladder is given that is 0.2. So the FB is equal to 0.2 into RB. Now we take a moment about point A. Here we cannot use directly summation fy is equal to 0 or summation fx is equal to 0 because we are carrying two unknown in y direction and two unknown in x direction. So first we will take a help of moment equation. So RB moment is RB into y. It is clockwise moment, so plus. FB moment is FB into x3, it is also clockwise moment. 200 force moment is 200 into x2, it is anticlockwise moment. 800 force moment is 800 into x1, it is also anticlockwise moment. So after putting all the values in the equation, we will get RB. RB is equal to 331.263. RB and FA, only two forces are there in x direction, so they must be same. So directly right here FA will be also 331.263 Newton. And we have already determined FB is equal to 0.2 into RB. So after putting this value, that determine FB that is 66.252 Newton. Now last equation we will use that is summation of vertical forces. So RA is equal to 800 plus 200 minus FB. So RA will come 933.748 Newton. So here we have RA and FA. So mu at A that is, it is a ratio of frictional force divided by support reaction that is RA. So mu is equal to FA upon RA, so it will come 0.355. So correct option is 0.354 that is near to our answer. Thank you very much.