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# Masses on incline system problem

## Video transcript

so just to show you how powerful this approach is of treating multiple objects as if there were a single mass let's look at this one this would be a hard one we've got a nine kilogram mass hanging from a rope that rope passes over a pulley then it's connected to a fourth kilogram mass sitting on an incline and this incline is at 30 degrees and let's step it up like let's make it hard let's save the coefficient of kinetic friction between the incline and the four kilogram mass is 0.2 and that's the coefficient so there's gonna be friction as well again if you try to solve this the hard way you'd be challenging it's doable but you're going to have multiple equations with multiple unknowns if you try to analyze each box separately using Newton's second law but because these boxes have to accelerate at the same rate well at least at the same magnitude of acceleration then we're just going to be able to find the system's acceleration or at least the magnitude of it the size of it right this fourth kilogram mass is going to have acceleration this way of a certain magnitude and this nine kilogram mass is gonna have acceleration this way and because our ropes are not going to break or stretch these accelerations are going to have to be the same so we get to use this trick where we treat these multiple objects as if there were a single mass and the acceleration of the single mass only depends on the external forces on that mass so we're going to only look at the external forces and we're going to divide by the total mass so what would that be if we wanted to find the acceleration of this four kilogram mass let's say what's the magnitude of the acceleration well look at this nine kilogram mass is much more massive than the four kilogram mass and so this whole system is going to accelerate in that direction so let's just call that Direction positive so that's one weird part about treating multiple objects as if they're a single mass is defining the direction which is positive is a little bit sketchy to some be with them they might find the sketchy we're just saying the direction of motion this way is what we're calling positive and that works just fine so when I plug in up here when I go to solve for what is the acceleration I'm going to plug in forces we go this way is positive and forces which go the other way as negative so let's do this what do i plug in up top what forces make this go well the force of gravity on this nine kilogram mass is driving this system this is the force which makes the whole system move if I were to just let go of these masses that start accelerating this accelerating this way because of this force of gravity right here so that's going to be nine kilograms times 9.8 meters per second squared and that's going to be a positive because it's making this system go there's there's no other forces that make this system go so now I'm only going to subtract forces that resist the acceleration what forces resist the acceleration well the gravity of this four kilogram mass resists acceleration but not all of the gravity the gravity of this four kilogram mass points straight down but it's only this component this way which resists the motion of the system in this direction and what is this component this is mg sine theta so if that doesn't make any sense go back look at the videos on inclines or the article on inclines and you'll see that the component of gravity that points down an incline parallel to the surface is equal to mg sine theta so I'm going to have to subtract 4 kilograms for kilograms times 9.8 because that's G meters per second squared times sine of the angle and my angle is 30 so I'm going to do sine 30 degrees okay we need some more room up here because there's going to be more forces that try to prevent the system from moving there's one more force the force of friction is going to try to prevent this system from moving and that force of friction is going to also point in this direction it's not equal to mg sine theta it's equal to right remember the force of kinetic friction going to be equal to MU K times F n and the MU K is going to be this zero point two you got to be careful though the F n is not just equal to mg and the reason is that on an incline remember the normal force points this way so the normal force doesn't have to counteract all of gravity on an incline it just has to counteract that component of gravity that's directed perpendicular to the incline and that happens to be mg cosine theta for an object on an incline and again if that makes no sense go back look at the video on inclines or look at the article on inclines and you'll see that this component of gravity pointing in to the surface is mg cosine that means the normal force is mg cosine because there's no acceleration in this perpendicular direction and after multiplied by 0.2 because I'm not really plugging in the normal force up here or the force of gravity in this perpendicular direction I'm plugging in the kinetic frictional force this 0.2 turns this perpendicular force into this parallel force this way so I'm plugging in the force of kinetic friction it just so happens that it depends on the normal force that's why I'm plugging that in so I'm going to need a negative 0.2 times 4 kilograms times 9.8 meters per second squared so that's a point 8 and then I need to multiply by cosine of the angle and in this case the angle is 30 degrees alright and now finally I divide by my total mass because I have no other forces trying to propel the system or to make it stop and my total mass is going to be 13 kilograms you might object you might be like wait a minute there's other forces here there's this tension going this way why don't we include that well that's an internal force and the whole benefit the appeal of treating this to mass system as if it were a single mass is that we don't have to worry about these internal forces yeah that force is there that tension but you can also have that tension over here and on this side it's resisting the motion because it's pointing opposite the direction of motion and this side is helping the motion it's an internal force the internal force is cancelled that's why we don't care about them that's what this trick allows us to do by treating this to mass system as a single object we get to in neglect any internal forces because internal forces always cancel on that and so if we just solve this now we just solve this calculate we're going to get four point seven five meters per second squared is the acceleration of the system so this four kilogram mass will accelerate up the incline parallel to it with an acceleration of four point seven five meters per second squared and this nine kilogram mass will accelerate downward with a magnitude of four point seven five meters per second squared remember if you're going to then go try to find out what one of these internal forces are I mean they're there we neglected them because we treated this as a single mass but you could ask the question what is the size of this tension oftentimes that's like a part two because we might want to know what the tension is in this problem well if we do that now we can look at the nine kilogram mass individually so I can say all right just the nine kilogram mass alone what is the tension on it and what are the forces I can find the forces on it simply by saying that the acceleration of the nine kilogram mass is the net force on the nine kilogram mass divided by the mass of the nine kilogram mass now this is just for the nine kilogram mass now I'm done treating it as a system this trick of treating this two mass system as a single object is just a way to quickly get the magnitude the acceleration now that I have that and I want to find an internal force I'm looking at just this nine kilogram box now and I could say that my acceleration is not four point seven five but negative four point seven five at least if we want to treat downward as a negative and upward is positive which is often done then I have to plug this magnitude of acceleration in as a negative acceleration because the nine kilogram mass is accelerating downward that's going to equal all right what forces are there on the nine kilogram mass I called downward negative so that tension upwards got to be positive but then minus the gravity force of gravity on the nine kilogram mass is nine kilograms times 9.8 meters per second squared and then I divided by nine kilograms I don't divide by the whole mass I'm done treating the system as if it were a single mass I'm now looking at an individual mass only we go back to our old normal rule tools for Newton's second law where up is positive down is negative and I only look at forces on this nine kilogram mass I don't worry about any of these now because those aren't directly exerted on the nine kilogram mass and at this point I'm only looking at the nine kilogram mass so if I solve this now I can solve for the tension and we solved for the tension I'm going to get forty five point five Newton's which is less than nine times nine point eight it's got to be less because this object is accelerating down so we know the net force has to point down that means this tension has to be less than the force of gravity on the nine kilogram box so recapping treating a system of masses as if there were a single object is a great way to quickly get the acceleration of the masses in that system once you find that acceleration you can then find any internal force that you want by using Newton's second law for an individual box you're done treating as a system and you just look at the individual box alone like we did here and that allows you to find an internal force like the force of tension