 Hello everyone, I am Sachin Rathur working as an assistant professor in mechanical engineering department from college, wall chain stop technology, Sholap. So today we will see the three cases of the break. So last time we have seen how the break will work. So in that three cases are there. So our learning outcome is at the end of this session student will able to understand the working of single block or shoe break. So this diagram already we have seen in last session that is a drum rotated in the clockwise directions and this is the lever on which the block is rigidly mounted and if you are applying the force and the drum will get stopped. So here the free board diagram of this and here we have seen the breaking torque Tb is equal to mu Rn into the radius of the drum. So here we have to calculate the value of Rn for getting the breaking torque. So in that three cases are there. So the first case, the first case when the line of the action of tangential breaking force passes through the fulcrum O of the lever and the break will rotate in clockwise direction. So we will see the first case, this one is a rotating drum, this is the block or shoe and this is the lever, here we are applying the force P. Now the line of the action of the tangential breaking force passes through the fulcrum O. So this is a line of the action of the breaking force so it is rotated in the clockwise direction. So the breaking force or the frictional force on the block will be acting like this and the value is mu Rn, this is a normal reaction Rn and it passes through this fulcrum O, this is the O point, this is the length of the lever and this is the distance between fulcrum O to the center of the block. So right now in this current condition you have to find out how much amount of the breaking torque is there. So for finding the breaking torque we have to calculate the value of Rn. So we have to take the moment about fulcrum O, so take the moment of about the O point. So we are getting P into L, so minus Rn into A and this force it is acting from the moment about O. So we have to neglect this is equal to 0, therefore Rn is equal to Pl upon A, this is the value of normal reaction Rn. So we are knowing that breaking torque Tb is equal to mu Rn into radius of the drop R is equal to mu into Pl upon A into the radius R. So like this we can easily find out the value of breaking torque Tb is equal to mu Pl upon A into O. So this is a breaking torque. So we will see the second case that is when the line of the action of the breaking force passes through the distance B below the fulcrum O of the lever, brake wheel rotates in the clockwise direction. So this is a rotating drum, this is a block, this is a lever. Now here the breaking force below the distance B below the fulcrum O. So this is a fulcrum point O and here the breaking force that is a B distance. Here the force P is applied as the drum rotates in the clockwise direction the frictional force F is equal to mu Rn is acting like this and this is a normal reaction Rn. So for finding the value of Tb when the drum rotates in the clockwise direction and the breaking force that is the frictional force passes through the distance B below the fulcrum O. So we have to take the moments about the fulcrum O P into this is the length of the lever L and this is the A distance. So P into L minus Rn into the distance A and minus mu Rn into distance B is equal to 0. Therefore we are getting P into L is equal to Rn into A plus mu Rn into B. Therefore Rn is equal to Pl upon A plus so this is the value of normal reaction Rn. Therefore the breaking torque Tb is equal to mu Rn into radius of the drum is equal to mu into Pl upon A plus mu B into the radius of the drum R. So like this we can easily calculate the breaking torques under this condition. Now it is rotated in the clockwise direction if the drum is rotated in the anti-clockwise direction. So the diagram will appear like this force P. So same condition is here that is the now the drum rotates in the anti-clockwise direction so the frictional force will appear like this that is the mu Rn and the fulcrum position is O is at that point this is the distance B distance A this is the length of the lever here the force P is going to apply. This is a normal reaction Rn. So under these circumstances we are getting take the moments about the fulcrum O. So the same condition is there when the line of the action of the breaking force passes through the distance B below the fulcrum O and the brake wheel rotates instead of the clockwise it rotates in the anti-clockwise direction. So due to the anti-clockwise direction the frictional force on the block will appear like this that is the mu Rn. So we have to take the moments about the O point therefore we are getting P into L that is acting like this minus Rn into perpendicular distance A that is plus mu Rn into B is equal to 0 therefore P into L is equal to Rn into A minus mu Rn into B. Therefore P into L is equal to Rn into A minus mu B therefore Rn is equal to P L upon A minus mu B. Therefore breaking torque T B is equal to mu Rn into R is equal to mu into P L upon A minus mu B into radius R. So this is the breaking torque under these circumstances. So in that breaking torque we will observe that the force P is equal to Rn into A minus mu B divided by P L sorry L. So this is the amount of the force required to stop the rotating wheel. So here we are getting this equation. So from this equation we will get the concept of self energized brake and the self locking brake that we will see in the next session. So these are my references. Thank you.