welcome to electron align here's our first example of how to work with that equation that deals with belt friction what we have here is we have a rope that's wrapped around the post it's wrapped around twice the Rope begins to make contact right here with the post goes all the way around once goes all run again the second time and then of course after this it no longer makes contact with the post so it's exactly twice around the post and there's a force of 1000 Newton's pulling on this side of the rope the question is how much force is required over here to hold a rope from slipping so that the Rope is stationary we're going to calculate it three times with three different coefficient of static friction point two point four and point eight you notice that each time we double the coefficient of friction from before to see what effect it will have on the amount of tension required they're forced required to keep the rope from slipping so what we can say here is that the equation we found on the previous video t2 divided by t1 is equal to e raised to the coefficient static friction times the angle of contacts in this case the angle of contact would be to complete rotations that would be twice 2pi or a total angle of 4pi radians of contact now we're looking for t1 so this equation can now be solved for t1 say that t1 is equal to t2 divided by e to the MU sub s times beta and we're going to calculate that for three these coefficient of frictions so we're gonna use the first one where mu sub s is equal to zero point two and that means that T 1 will be equal to 1,000 Newton's divided by e to the zero point two times four pi two complete revolutions there's 2 pi radians and yes we do have to use the radians for the angle and that with the calculator we get the following 0.2 times four times pi raise that 2 as the exponent and times one equals and that gives us exactly 81 Newton so it requires t1 equals 81 Newton's the force required to keep this rope from slipping which is kind of nice you see here that when you apply a force of a thousand and here you only need 81 Newton's to keep that rope from slipping and that is if their coefficient of friction is 0.2 what happens if the coefficient of friction is 0.4 how does that change things and initially you might think well if it's double the coefficient of friction you expect maybe half the tension but that's not the case this is kind of a unique situation it's not a linear function and so let's see what that happens again mu sub s now equal to 0.4 so we get t1 is equal to 1000 Newton's divided by e to the zero point four times four pi and let's see what happens now so we have 0.4 times 4 times pi e to the x inverse times a thousand equals so it's only six point five six newtons t1 is equal to six point five six newtons notice by doubling the coefficient of friction we now require less than one tenth attention to keep the rope from slipping now let's go one more let's say the coefficient of friction is 0.8 that's not a usual case but just to see how this works let's try mu sub s equals 0.8 that gives us t1 is equal to 1,000 Newton's divided by e to the 0.8 times 4 pi and let's see what that now becomes so we get 0.8 times 4 times pi as an exponent take inverse of that times 1,000 and now we takes just a small fraction of a Newton 0.043 newtons that's almost hard to believe and that's indeed the case it turns out that when you wrap strings or ropes or cables or anything around the post a number times the amount of tension you need on the other side to keep the rope from slipping is actually quite small and becomes very small as the coefficient of static friction increases so now in the next example we're going to do something similar but now we're going to change the number of times we wrap the rope around the post to see how that affects how much things you need to keep your rope from slipping that's how it's done