welcome to our lecture line our next example on friction has to do with a bracket on a pole we have the bracket here that has two rings there attached to the bracket and we're trying to put a load on the bracket now the closer we bring the load to the pipe rear the less torque we're going to have and the less friction we create between the two rings and the pipe the far the way the load is the greater the torque degraded the friction because the greater the torque the greater the normal force is you're going to have on the bottom and the top ring and of course the greater the normal force to grave the friction force because by definition the friction force is equal to the normal force times the coefficient of friction in this case we're going to assume that nothing is sliding so we'll take the static coefficient of friction so what we're trying to figure out is how close can we bring the load here before the whole bracket begins to slide what is the minimum distance X at which we can still apply a load of 100 Newtons to that bracket we're going to start by figuring out the normal forces here at the two brackets and we do that by saying that the friction force and a is defined as the normal force at a times the coefficient of friction we can also say that the friction force at B is going to be equal to the normal force at B and times the coefficient of static friction now the only forces acting in the X direction are these two forces right here so we can say that the sum of the forces in the X direction must add up to zero which is equal to the positive force and sub a minus the negative force and so B and so from this we can conclude that the normal force at a equals the normal force and B we can also sum up all the forces in the Y direction and see what we get sum up all the forces in the Y direction and those are made up of the two friction forces right here plus the load force right there so we know they should add up to zero we have the positive friction forces were which are acting upward starts friction force at a plus friction force and B minus the load force and so we could say that the load force F sub load is simply equal to the sum of the two F sub A plus F sub E and we can replace the friction force oops I should actually have friction force I'll write it like that that's better friction force we got at B and of course this can be written as the normal force at a time's the coefficient of friction plus the normal force at B times the coefficient of friction of course realizing that these must be equal to one another you can also write this as the normal force at a time's the coefficient of friction plus the normal force at a time's the coefficient of friction or twice the normal force at a times the coefficient of friction and we know that the load force is equal to 100 Newtons from this we can conclude that the normal force at a is equal to 100 Newton's divided by 2 times the coefficient of friction the coefficient of friction to find is 0.2 so this is 100 Newton's divided by 2 times 0.2 and so that would be 0.4 or 250 Newtons and if the normal force a is 250 Newtons then we know that is also equal to the normal force at B so from that we've concluded the two normal forces at a and B which means we can now calculate the moment about one of these points so we probably want to take the moment about point eight let's do that I don't think it really matters we could do it about B with it doesn't really matter which one we pick so let's take the sum of all the moments about point a and we know this also must equal to zero if nothing is sliding if things remain static and this is equal to now of course these two forces are canceled out because they go right through the point here about which we take the moment we do have the force here the load for is 100 new acting over distance of well would be X minus half the diameter of the pipe so that would be a clockwise moment which is a minus 100 Newton's times X minus and that would be 2 and 1/2 centimeters converted to meters 0.025 meters I guess I'm running a little bit out of room here and what else do we have we have the normal force of B which is acting in a counterclockwise direction so that would be plus the normal force at B which is 250 Newtons and that is acting over a distance of 10 centimeters that would be 0.1 meter and then we have the force at B which will give us a clockwise torques that's minus the force at B now the force at B would be the normal force at B times mu sub s so it would be equal to 250 Newtons multiplied times 0.2 that would be equal to 50 Newtons so that would be the force at B at 50 Newtons and we have to multiply that times the perpendicular distance from the line of action of the force to the pivot point which would be 5 centimeters so 0.05 meters that gives us a clockwise moment or clockwise torque with therefore that's where the negative comes from from this we should be able to calculate X because X in this case will be the only unknown so let's go ahead and work everything out so we have 0 is equal to 100 mmm let's see I'm going to move this to the other side so that makes it easier so multiply this times this move to the left side becomes positive so positive 100 Newtons times X and then we leave everything else on the other side is equal to a negative times the negative makes that a positive 100 times point zero two five which would be 2 point 5 Newton meters C is that correct 100 times that there would be yep 2.5 Newton meters and let's see here then we have this that would be plus 25 Newton meters and then here we have minus 50 times this that would be again 2 point 5 Newton meters notice that the positive to point 5 the negative 2 point 5 Newton meters cancel out and that all we have left to do is divide both sides by 100 Newtons so x equals 25 newton meters divided by 100 Newton's which is equal to 0.25 meters or 25 centimeters so that's the closest you can apply the 100 Newton load force to the of course X is relative to the center of the pipe here but the closer you can get is 25 centimeters you get 1 in 25 centimeters the bracket will begin to slide because you're not applying enough torque to create enough friction force anything greater than 25 centimeters the bracket will hold the friction force will hold up the load force and that's how you do that