 Hello everyone. I am Swati Ghadge, Assistant Professor, Department of Civil Engineering from Vulture Institute of Technology, Solapur. Topic for today's session, Static Fiction on Incline Plane. Learning outcome at the end of this session, learner will be able to solve problem of static friction on an incline plane. In the figure, incline plane is shown with inclination theta and the block is on the incline plane. As the inclination increases, the block will start moving down and the maximum inclination of the plane on which body free from the external forces can repose is called as angle of repose. Now we will see how the forces will act on the body which is on incline plane. As the body is on the incline plane, the self-fit is always act vertically downward. We have to resolve that self-fit in the direction of plane and in the perpendicular direction of plane. So like this, we have to resolve that. As the plane is making theta with horizontal, so its perpendicular will make theta with vertical. So w cos theta, it is perpendicular to the plane and w is vertical. So w and w cos theta will be an angle theta in between them. So the fourth component which is perpendicular to the plane is w cos theta and the fourth component which is parallel to the plane which is w sin theta. Surface will offer a normal reaction always in perpendicular direction that is n. As the block is moving down to the plane, so frictional force will act in the opposite direction of motion. So it is acting up to the plane shown in the figure. Here n and w cos theta are along the perpendicular y axis and along the plane it is suppose x axis. So w cos theta and n is acting along y axis. So as it is case of static friction, we can say it is an equilibrium. So summation f y is equal to 0 if we consider n must be equal to w cos theta. So it is written here n is equal to w cos theta and in the x direction there are two forces w sin theta and f. To maintain equilibrium that two forces must be same, so f is equal to w sin theta. We have seen definition of coefficient of friction in the previous session that f and n bears a constant ratio that is called as mu. So f by n is equal to w sin theta divided by w cos theta. w get cancelled. So sin theta divided by cos theta is equal to tan theta. So mu is equal to tan theta is equal to f upon n. This is the coefficient of friction. Now you pause video and solve this question. Question is a block is rest on a rough inclined plane and is connected to an object with the same mass as shown in figure. The rope may be considered massless and the pulley may be considered frictionless. The coefficient of static friction of the block and the plane is mu s. So what is the magnitude of static friction force on the block? So we have to find out the mu s. These are your options. You solve it and select correct option. Let us see the solution of the question. The block of weight mg is given which is on the inclined plane and the rope is pulling that block with same force that is mg. The self weight mg resolve in x and y direction, mg sin theta in x direction and mg cos theta in y direction. Normal offer by the support and the frictional force which is acting opposite to the motion shown in the figure. Two forces are there in a perpendicular direction that is y direction. So mg cos theta and n. So n is equal to mg cos theta. F is equal to mg minus mg sin theta. So this is the addition of all the horizontal forces. F is equal to mg minus mg sin theta. So get mg common. So f is equal to mg into 1 minus sin theta. So frictional force is mg in bracket 1 minus sin theta. So this is the correct option. Now we will solve one numerical on inclined plane. A 3000 Newton block is placed on an inclined plane as shown in figure. Find the maximum value of w for equilibrium. If tipping does not occur, assume coefficient of friction as 0.2. So figure is given and the inclination of the plane is given 30 degree. The weight of the block on inclined plane is given 3000 Newton and it is connected to the rope and to the another end of the rope there is a load w. So we will solve that. First we will draw the free body diagram. So this is the block given to us. Self-heat of the block always vertical that is shown in the figure 3000 Newton acting vertically downward. W is acting horizontally. N and F, N is the support reaction offered by the support and F is the frictional force always act in the opposite direction of motion. Resolution of forces. See how we resolve the forces 3000. So vertical is 3000 that is a component along y direction is 3000 cos theta. Component along x direction is 3000 sin theta. Similarly we resolve w. Component along x direction is w cos 30 and component along y direction is w sin 13. So in y direction there are total 3 forces 3000 cos 30 N and w sin 13. So N is equal to 3000 cos 30 minus w sin 30. So we will apply here the definition of mu F is equal to mu into N. So mu is given 0.2. So 0.2 into N and is 3000 cos 30 minus w sin 30. So this is 0.2 into N and is 3000 cos 30. So this will be the value of F. Summation of Fx is equal to 0 that second equation we will use now. So in the x direction there are forces 3000 sin 30, w cos 30 and F. So w cos 30 minus 3000 sin 30 minus F is equal to 0 that is the equation we get in x direction. So w cos 30 minus 3000 sin 30 minus minus F that is 0.2 into 3000 cos 30 minus w sin 30 is equal to 0. So in this equation only w is unknown. So find out w. So solve for w. So we will get w 2636.57 Newton. These are my references for this video. Thank you very much for listening.