welcome to I lecture online in this example well looks exactly the same as the previous example with one difference is that we have a forty Newton force pushing against the Box horizontally instead of only a 20 Newton force that will result the whole situation to become a static problem rather than a dynamic problem we're still going to try to find the normal force the force up the incline the force down the incline the maximum friction force you can have the friction force required to keep things static and then of course the various angles involved so let's go ahead and start with trying to find the normal force the normal force is going to be the reaction or force pushing back in this direction so there's the normal force and that is going to be equal to the sum of the reaction Air Force to the perpendicular component of the weight and the perpendicular component of the force so this is going to be the weight times the cosine of 30 degrees plus the force applied on the side here times the sine of 30 degrees because the perpendicular component of this one here and let me use some different colors there's the perpendicular component notice that this here would be the angle of 30 degrees and it's the opposite side to the angle their probe use the sine over here and now we can calculate what that is equal to that would be 100 Newton's times the cosine of 30 degrees plus 40 Newton's times the sine of 30 degrees like so and so this would be equal to 86 point 6 Newtons plus that's 1/2 times 40 or 20 Newtons which is equal to a hundred and 6.6 Newtons so that's the normal force pushing back now that will allow us to have a maximum friction force the maximum friction force in a static situation knowing that the static coefficient of friction is 0.2 so the friction force max is going to be equal to the normal force times the coefficient of friction of course this is coefficient of friction that would be 100 and 6.6 Newton's times 0.2 that's double that that would be 21 just to make sure them mess up here 106 point six times point two is twenty one point three two Newton's so that's the maximum friction force you can get from the situation that doesn't mean that will be the actual friction force but it's the maximum friction force that can be provided here now let's find the force up in kind the force pushing the block up the incline based on the applied force so the force up the incline is going to be equal to that would be F times D that would be the adjacent side that would be the cosine of 30 degrees which is 40 Newtons times zero point eight six six so forty times point eight six six equals and that would be thirty four point six four new tents we have a thirty four point six four Newton force pushing the block of the incline now what about the force down incline the parallel component of the weight of the block what is that equal to the force down the incline is equal to not F but the weight so that would be the weight times the sine of 30 degrees which is the parallel component of the weight which is equal to 100 Newton's times well the sine of 30 which is 0.5 which is 50 Newtons that means that we have a 50 Newton force pushing the block down incline of thirty four point six for Newton force pushing the block of the incline so if the force can be at least the difference of the two then the block will not move so in other words the maximum force F max must be greater than these two forces must be greater than 50 Newton's minus thirty four point six four Newtons and so that means that the maximum of force friction max I forgot the friction their force friction max must be greater than that would be fifteen point three six mutants so that's the friction force required to keep the block from moving the friction force that the maximum friction force you can have here because the normal force being what it is 100 6.6 Newton's is 21 point 3 2 Newton's which is larger than this so therefore we can say we're going to have a static situation the block will not move now let's find out some of these angles so we know that the friction force will be up the incline right there there will be a friction force and then we have what we called the resultant force the resultant reaction force which will be the sum of the normal force and the friction force we call that the resultant force and we want to know the angle right here we'll call that angle fee that's so we have the friction force let's put up here the friction force we have the normal force and we know that those two are perpendicular to each other and then we have the resultant force like so there's my resultant force there's my friction force there's my normal force normal force friction there's my resultant and this angle that's called his angle feet we know that this is going to be a right angle which means that the fee is going to be equal to the arctangent of the opposite side which is the friction force divided by the adjacent side was the normal force so it's equal to the arctangent of the friction force is the normal force times mu divided by the normal force which means the normal force cancels out that means this is equal to the arctangent of the coefficient of friction in this case is going to be the static coefficient of friction so therefore it's the arctangent of zero point two and point two take the arctangent of that which is eleven point three one degrees and then finally we want to find the angle between the resultant or the reactionary force and the vertical which is this angle right here we'll call this angle alpha and alpha is going to be equal to the thirty degree angle minus fee which is equal to the thirty degree angle minus eleven point three one degrees and so this will be equal to eighteen point seven degrees leaving off this second decimal place there so that's how we find the angle between the vertical and the reactionary force which is simply the difference in this case of the angle or the incline angle and the angle we found between the normal and the reactionary force there's one more triangle that we can draw we can draw the summation of these three since the whole thing is static now we can draw this force right here which is the weight the weight which is equal to 100 Newtons we can draw the force right here the forty Newton force and then we can draw the reaction force right here our and because of this now we know what the value of art can be because this is a right triangle and also we can find the angle here and that's interesting what this angle will be equal to let's for now call the angle beta and so we know that beta will be equal to the arc tangent of the ratio of the opposite side which is 40 divided by the adjacent side which is 100 so this is equal to the arc tangent of 0.4 and we get twenty one point eight degrees now the question is why did we not find the same value for beta here as we found for alpha here what's the difference well it turns out that when we found this angle here we use the friction force we use the maximum friction force and of course we didn't want to use the maximum friction force because that's not the actual friction force the actual friction force of fifteen point three six Newtons so when we go back and find this angle right here this is using the maximum friction force so therefore we get a different angle the actual friction force will be less which is found by doing this combination here where we have the actual applied force of 40 Newtons the actual weight of the block and then the response of the reactionary force which actual gives us a twenty one point eight degree angle and so this is the true angle in this situation this is the angle we would get if we were actually utilizing the maximum friction force and that's how you can tell the difference between those two angles and the difference between the two applications of using them the maximum friction force and using the actual friction force so this here is the actual friction force all right and that's the difference and that's how it's done