A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire that passes over a frictionless pulley (Fig. E10.16). The pulley has the shape of a uniform solid disk of mass 2.00 kg and diameter 0.500 m. After the system is released, find

(b) the acceleration of the box, and

Video transcript

Hi everyone today, we are going to determine the acceleration of the box or off the system of this particular practice example. So, first I am probably just going to start with creating a list of given. So we know that M one here is going to be weighed at then newton. So I'm just gonna write down as double U. One which is going to be equals to M one multiplied by G. Which is equals to 10 newton like. So next we are given the weight of the second box here, which is going to be six Newton. So W two is going to be M two multiplied by G. Which is going to be six newton like. So these two boxes are going to be connected with a lightweight court. So the weight of the court is going to be negligible. Attached to these two boxes to a cylindrical pulley with a mass of one kg. So the mass of the pulley is not negligible. Which is going to make this problem a little more interesting. And the pool itself will have a diameter. I'm just gonna write down as DP of 18 cm, which in S. I. units is going to be 0.0, sorry, I mean 0.18 m and R P or the radius is going to be 0.9 m like. So, Okay, so next we have to identify the other forces that is going to be acting upon our system here, we know that the cylindrical pulley itself is not going to be negligible. So therefore there will be two different tension here acting upon our system one it's going to be T one and one is going to be T two. T one is going to be pulling em one and T two is going to be pulling em too with the cylindrical pulling itself with the diameter or the radius of R P is going to actually be pulled by T one and T two. Okay, so looking at the system, we can know that the system is actually not at equilibrium because this box right here is hanging is not at equilibrium and therefore it is actually going to be moving. So pulling itself is going to be moving clockwise especially because this box right here is going to be pulled out, pulling down the system and this weight right here is not actually going to be acting upon the system because it is perpendicular to Discord right here. Okay, so now that we know that the system itself is moving, we will know that we will have a certain acceleration on our system. So I recall the acceleration is going to be represented with an A and however we also need an angular acceleration because the pulley is going to rotate on its own angular is exploration is going to be represented by this alpha right here and this too is actually going to be different but related with this particular formula. So exploration is going to equals two R P multiplied by alpha and alpha is actually going to be a over R. P. Like so so R P remember is going to be the radius of the bully. Okay, so now that we actually have identified all of this, we want to start solving this problem right? So to do so we need equations. And the equation that we're going to use on this particular system is actually going to be the Newton's second law. Why? Because the system itself is not at equilibrium and we know that it is moving with a certain acceleration. So recall that Newton's second law is going to be uh sigma F equals to M multiplied by A. So the M is actually going to be just the mess of an isolated system and the acceleration itself is also going to be the acceleration of the isolated system. The force is going to be all the forces acting on that system. Okay, in this system right here, you want to actually identify multiple different smaller systems inside. So the first system that we have is going to be system of box one, which is actually going to be having uh we should look at it on a horizontal access. So recall the Newton's second law again, which is just going to be C. F by M one times A. Because the A of the overall system is going to be the same while we're only looking at M one. That's why this is only the mass considered. Okay, the only force that is going to be acting upon this M on the access where the system is moving is only going to be T1. So the 10 Newton here is not going to be considered because it is not acting upon the access that the system is moving. Okay? So this is going to be T one equals to M one multiplied by A. And that will be our first equation like. So okay now we have our second system which is box two and we have to also recall the Newton's second law and two multiplied by A. Which is the same age as this. And then this is going to be equals to the weight minus P two because we know that M two is going to be moving downwards. So we want to start with the force acting along the direction that the system is moving minus the one opposite to it. So we know I'm just gonna represent this with W two minus T. Two equals M two multiplied by A. And then W two is M two multiplied by g minus T. Two equals two, multiplied by A. I'm just gonna rearrange this so that it is pretty similar to this first equation that we have. So this is just going to be T two equals two. Um M two G minus M two A. Like so and then I can take all the M two out which is just going to be T two equals M two deployed by g minus A. Like. So. Okay, so now that we have this two equation, we still need a third equation to actually determine our T. S. And R. A. So we need a common equation where we can combine all of this into one. Okay, so the next equation that we need is to actually look, we can obtain it by looking at our pulling itself. Okay, so that will be our third system. So now our next system is go, oops, I'm just gonna go back to black. Our next system is going to be the pulley but remember the pool is rotating not translating. So what we need to do here is to actually use the torque formula for a non equilibrium system. So recall that the torque formula for a system that is moving is going to equals to the inertia, multiplied by the angular acceleration. And we are going to look at it at a certain access which is to see access just because it is on a different rotating access from the way that the system is moving. Okay, so We are going to use this formula right here to finally solve our problem. Okay, so we know that we've mentioned earlier that the two forces I think upon the pool is going to be T2 and T1, recall that the torque formula is going to be our torque is going to be calculated by multiplying uh force multiplied by the distance, which the two has to actually be perpendicular to one another. So okay, uh We know also that the system is going to be moving clockwise from earlier. So we want to do everything, we want to subtract everything going along the direction of what the system is moving into, minus everything that is opposite to it. So the um the first force that we want to do here which is moving clockwise as well is going to be T two multiplied by R. P. Which is going to be subtracted by T one. The other acting force force multiplied by R. P. As well. And the eye, the eye for a cylindrical pulley is always going to be the same, which is going to be half m the mass of the pulley, R. P the radius of the pulley squared, yep, police squared And earlier we have mentioned that the angular acceleration is actually going to be the acceleration over the radius of the pulley itself. Okay, so this is the equation that we have and we can actually simplify this. I'm just gonna write this down as cylindrical pulley because this is something that you guys need to remember which is usually not known. Okay, I can simply but find this just by crossing the square with this, our P down here and at the same time I can rearrange this just so that it is easier for you guys, I'm gonna go down a little bit. Okay. Okay I'm gonna rearrange this by writing this is going to be T. Two uh multiplied by R. P minus T. One, multiplied by R. P equals two. This is just going to be exactly the same as the top by R. P. Supplied by a. Okay, okay. And then we know that this set has all have our peace and there's also there's also an RP there. So I'm just gonna simplify by eliminating the RP or defining all of it by R. P. This is going to be half mp deployed by A and this is going to be our third equation like so. Okay so we have three different equations and if you look at it we also have three different unknowns which is only T. One T two and the exploration which is the one that we wanted here. So now that we have actually um eliminated all the degrees of freedom in this particular problem, we can combine all these three equation and then solve our actual ah so solve our actual problem. Okay so we are going to combine 1, 2 and three. And the way we are going to do that is you know that we have T. One here equals to something and M. One equals to M. One times A. We know that we have T. Two here which is also equal to something. I'm just gonna write this down is the second equation forgot to do that. But yep And then we know T. One. We know T. Two we have T. Two E. F. T. One. We want to actually combine all of it into this equation here and then solve for a. Especially as because the D. A. Is actually what we want. Okay so this is T two minus T. One. I'm just gonna rewrite equation three that we have half mp. Multiplied by a. Okay and then T two is going to be from equation two which is going to be M. Two multiplied by G minus A. That is going to be T two minus T. One which is going to be from equation one and one multiplied by A. And that is going to equal to half M. P times a. Okay so we can actually we only have now one unknown. So the way we want to do this is to actually just block everything in but I'm gonna start redistributing this M. Two just because I think it's going to be easier for you guys to solve it to see the pattern that we have here -M A equals half mp. Okay like so and then I am just gonna scroll down a little bit because we don't really need to use the two equations again. Okay like so and then after this I want to actually pull and put I'm gonna put all the ace into one side and pull it out. So I'm gonna put this to this side. So this side is just going to be M two multiplied by G equals two health mp, multiplied by A plus M two A plus M one A. And all of this age, we can actually go out a of mp plus M two plus M one. So, and then we can simplify and they will equals to M two G over all of this thing, which is half M P plus M two plus M one. Okay. And then we can start actually plugging everything in to here. So we only know what W two is and what W one is, but we know that W will equals to M over G. So essentially M will just be, I'm sorry. Uh, W really goes to M times G. So M will essentially be W over G. So that is what we want into this equation right here. Okay, so, we know M two G. Which is going to be six newton from which is going to be W two and then This side is going to be half MP, which is going to be half of the MP earlier is one kg. And then M two M two is Ashley going to be W two over G which is going to be six newton over 9.81 m per second square plus W M one, which is going to be W one over G, which is going to be 10 newton over 9. m per second squared, like. So, Okay, so after this we can find that the A is actually going to be 2.81 m per second squared. So that will be our final answer, which is 2.1 m per second squared, like. So which is actually going to be answer C. Yeah. All right, So let me know if you guys have any questions and if there's anything that is unclear. And yeah, that's it.