 I am Swati Ghargay, Assistant Professor, Department of Civil Engineering from Walton Institute of Technology, Solapur. Topic for today's session, Static Friction on Horizontal Plane. Learning outcome of this session, at the end of this session, learner will be able to solve problem of static friction on horizontal plane. The person is trying to move that block with a force F on horizontal plane. The block is getting a velocity in the direction of force. So in the opposite direction of force, the frictional force will act, that is, FR. The self-fit of the block is mg, acting vertically downward. And that surface is offering a support reaction and it is vertically upward, that is n. So all the forces are acting on this block, shown in this figure. And this is showing the friction on horizontal plane. Now we'll see what is mean by friction. When the body moves or tends to move over another body, a force opposing a motion developed at the contact surface is called as friction. There is a limit beyond which the magnitude of the friction force cannot increase. This limiting value is known as limiting friction. The same figure, the person is trying to pull the block with a force F. But it is not moving. It is at a rest position. So this is a case of static friction. In another figure, the boy is trying to pull the block with the force F and it is moving in the direction of force. So in that case, the friction is called as dynamic friction. The types of friction. Static frictions and the dynamic friction are the types of friction. Dynamic friction again classified into two types, that is rolling friction and the sliding friction. In this situation, we will see the static friction. First, we will see the coefficient of friction, mu. So consider a body having a self weight w. The w is acting vertically downward. And the P force is the pulling force, which is trying to pull the block in the direction of force. And the support will offer a normal reaction in the vertical direction. And the frictional force will act in the opposite direction of motion. So in the opposite direction of P, the frictional force will act. Here, frictional force is indicated by F. And F and N, we are a constant ratio. And that constant ratio is called as the coefficient of friction. So definition of friction is a ratio of F by N, frictional force to the normal. Or F is equal to mu into N. We'll solve a numerical friction on the horizontal floor. A block shown in figure of weight 2000 Newton. Here the block is of weight 2000 Newton, rest on the horizontal floor. A chord attached to A passes over a frictionless pulley. So at point A, this chord is attached and it is passing over a frictionless pulley. And support of weight equal to 800 Newton. So at the another end of the loop, there is a weight 800 Newton. The value of coefficient of friction between the horizontal plane and A is 0.35. So at this point, at this contact surface, mu is given 0.35. Solve for the horizontal P if the motion is impending towards left. So we have to determine the force P, the motion is impending towards the left side. So we will first draw the free body diagram of block A. It is given that the block of weight 2000 Newton, so it is acting vertically downward. The pulling force is there towards the left direction. So we have to show that in a free body diagram. And there's a chord which is inclined. And thus another end of chord, there's a force 800 Newton. So total 800 Newton force running in that chord. So this 800 Newton force shown in the figure. It is rest on the horizontal surface. That horizontal surface provide a normal reaction which is in the upward direction. And it is given that the motion is towards the left direction. So against a motion, the friction force will act. So it is towards the right direction. All the forces 2000 Newton, P, F, and N are in the standard direction that is x and the y direction. But only 800 Newton force which is not in x and y direction. So we have to resolve that in x and y direction. For that 800 Newton force is making an angle 30 degree with horizontal. So we will resolve that force in x and y direction with the help of the angle 30 degree. So it is given with horizontal. So horizontal component will be 800 into cos 30. That is 692.82 Newton. And the vertical component is 800 into sin 30. That is 400 Newton. So like this, now all the forces are in x and y direction. So first we will consider the summation of all the vertical forces is equal to 0. As it is a case of static friction. So we can apply the equilibrium condition here. Now here in the y direction, there are three forces. 2000 Newton, N, and the 800 sin 30. That is 400 Newton. The three forces are there in a y direction. In that only N is unknown. So we will determine N by using the equation summation of 5 is equal to 0. So N is equal to 2000 minus 400. N will come 1600 Newton. According to the definition of coefficient of friction, we will determine F. F is equal to mu into N. So F is equal to 0.35 into 1600. So F is equal to 560 Newton. Now there is only one force P which is unknown in x direction. Now we will use summation Fx is equal to 0 and we will determine P. So P is equal to 800 into cos 30. That is 692.82 plus 560. That is F is equal to 1252.82. So this is the value of P. Now you read the question carefully and answer the question. Question is refer same problem. Solve for the horizontal P if motion is impending towards right. You pause video and solve for the given question and be ready with correct answer. Only change is that in the first problem motion is towards the left side. Now it is given that motion is towards the right side. So what will be the value of P? Solution is same first we have to draw the free body diagram. So self weight 2000 Newton acting vertically downward. P is acting toward left side and 800 Newton inclined with 30 degree horizontal. Everything is same and horizontal surface will offer a normal reaction in vertically upward direction. But the frictional force is acting against the direction of motion. And direction of motion is given it is towards the right side. So direction of motion is towards the right side. So frictional force will act towards the left side opposite of motion. So this is only a change in this numerical. So same we will resolve the force. The x component is 692.82 and y component is 400. Similar to the first numerical. We cannot go directly for the summation fx equation because we are carrying two unknown in x direction that is p and f. And in y direction there is only one unknown n. So obviously we have to use summation fy first. So n is equal to 2000 minus 400 is 1600 Newton. As per the definition of coefficient of friction f will come 560 Newton. Then only one unknown is there in x direction that is p. Now we will use summation fx is equal to 0. So p is equal to 692.82 that force 692.82 minus f that is 560. So it is 132.82 Newton. So the answer of question is p is 132.82 Newton when block is impending towards right. For this video these are my references. Thank you very much for the listening.