1b 35kg block (friction 7ON): Find the The 50 kg was consisted of a 15kg block (friction 3ON) glued onto rope in tension model. tension force inside the glue: You m...

A $2.00-\mathrm{kg}$ block $\left(M_{1}\right)$ and a $6.00-\mathrm{kg}$ block $\left(M_{2}\right)$ are connected by a massless string. Applied forces, $F_{1}=10.0 \mathrm{~N}$ and $F_{2}=5.00 \mathrm{~N},$ act on the blocks, as shown in the figure. a) What is the acceleration of the blocks? b) What is the tension in the string? c) What is the net force acting on $M_{1}$ ? (Neglect friction between the blocks and the table.)

All right now we've got a similar diagram to the previous problem but the five kg block is restrained so let's draw it who it's restrained to here. Then there's f. Pulling in this direction. Um Five and 10. So I'm gonna call this M. One and M. Two M. One is five. M. Two is 10. Okay. Now F. Is 45. Newton's mu for all surfaces is 02 A. Okay so let's draw a diagram of em too. We've got a force acting on it, We've got its weight and the weight of M. One acting downward. So that would be M. One plus M. Two times G. We've got a frictional force in this direction. Uh And I'm gonna that's the one at the bottom. So I'm actually gonna redraw this. Then we've got a frictional force at the top. Okay? Um We've got a normal force at the bottom and also a normal force at the top. Okay. Not sure this one actually needs to be drawn. Um nevertheless. Okay then we got the other block. The other block has mg. It's got a frictional force which is opposing motion. Yeah so it would be going in this direction and we've got a normal force but it's not accelerating at all. So I think I've drawn the friction force in the wrong direction and then here going to have to be the tension. Okay so friction force and tension are opposing each other. Um Normal force and weight are opposing each other um determine the tension in the string and the acceleration of the lower block I believe of the lower block. Okay so f minus force of friction to which would be mu times M one plus M two G minus the other force of friction. Which would be mu times I am one G. I think that would be correct. Uh That minus that minus that is going to equal mass times acceleration but only the lower block is accelerating, so that would be M two. Okay now um let's start using the calculator. M one is five. M two is 10. Mu is 100.2. M one is five. M two is 10. New is 0.2 F. Is 45 G. Is 9.81 m per second squared. Okay so f minus mu and one plus M two G. F minus um You M one plus M two G minus mu M one G minus mu M one G equals M two A. So if I just take this whole thing and I divided by AM M. Two, that gives me a A. Is I can't get my fingers in the right spot. 0.576 um meters per second squared tension. So as far as I can tell the some of the forces would just be tension minus two friction force. So the tension would have to equal the friction force. One friction force, one is mu M one G. Yeah. Yeah. Let me think about this again. The friction force led to mass and acceleration, acceleration on the lower block. So I think it's just mu M one G. Which is 9.81 Really? Okay. I guess it is news. Okay, this is 68 so I really can't check. So thank you for watching.

And this problem, we have a block that's hanging over the police with the rope and then a tension on the other side of the rope. E told the diameter of this poli is half a meter. This, uh, the mass of the poli is 2 kg. The mass of this block is 1.5 kg. The moment of inertia of this holy which we assume is, you know, from disk is 18 times times the mass times the diameter squared. Um, we know that the acceleration of this block is just the radius here, which is the over two times the angular acceleration of police. And we're told that that is minus 1.2 m per second squared. Now what I did again is I cut. This broke this into two pieces, one piece being this block and the other piece being so to Richard bodies here. So let's analyze the block here. I said positive was down. So we have w minus t one equals the mass times acceleration of that block, which you can also write in terms of the angular acceleration of this disc. Um, now we look at the disk here are the polling. We have the torque from this tension here that we exposed when we cut the cable. Um, minus two torque from this attention that were being for the being applied equals the moment of inertia, times the angular acceleration. So again we could solve for Alfa here and plug it in. So what I did is after some algebra, you know, we can solve for t one plug that the here and then that eliminates t one. And we just do a little rearranging. And in the end, we get Let's see, minus m one a plus. The weight of this block minus t equals 1/4. I am the no m Well, minus 1/4. It's actually ones that we minus one half m A. Yeah, so we see that. And then we saw for tea we get, um, the weight minus m one a minus one half m a. Now we can't. The acceleration is is up, we were told. So I said positive was down. So it's a negative acceleration. So plugging in all these values, we wind up with this tension. Just force on this rope. Here is 18 Newton meters, so that would be causing this to rise up, and we can see that it makes sense because, you know, if if the answer we got if if a was zero, then the tension is just the weight, right, it's an equilibrium. And because these things have mass to get them accelerating, we need more attention. And in fact, you could see that where there's a minus sign here, but a is negative. So we're adding more attention now, If they was positive, then we basically be saying, Well, we're going to let the tension be less than the weight. So we basically let this fall, but not as fast as they normally would, just under gravity, because we're gonna hold it here with some tension.

So no, sir. Here in this the problem that ex friction coefficient is mentioned. So we're not going to deal with the static friction of the static frictional force. And we're simply going to only deal with the kinetic virtual force so we can apply Newton's second law to the X axis on. We're going to say that the sum of forces in the X axis for the mass sub one would be equal to the mass of one times the acceleration of the system. Because the masses are both connected here. This would be equal to the tension force minus the force of the kinetic force of friction. Ah, we can apply then. Ah, some of forces for the second clock. This would be equal to EMS of two G sinus data minus detention force. This is equaling m sub two times the acceleration of the system. We're going to add the two equations and by adding the two equations, we find that the acceleration is that equal to M sub to g sign of data. Essentially, the tensions cancel out on, so this will be minus the kinetic force of friction divided by the sum of the masses m sub one. Plus I'm some too. And at this point, we can solve, uh, knowing that force of friction kinetic would simply be equal. Toothy coefficient of kinetic friction multiplied by, um, sub one G um, this is given that the sum of forces in the wind direction equals zero for the second mass, and this would equal force normal minus am someone G. So we can essentially assume that the normal force force blocks block one would be equal to m sub one g. Given that the and we know that kinetic frictional forces equaling the coefficient of kinetic friction times the force normal sub one plug that into the friction off the formula for the frictional force. And again we solved that the kinetic force of friction is equaling to the coefficient of kinetic friction times the weight of the first mass. And so the acceleration well then be equal to m sub too. G sign of Fada minus Musa Kay. I'm someone g divided by again that some of the mess so at this point, we can solve for a a will be equal to 3.0 kilograms multiplied by 9.8 meters per second squared multiplied by sine of 30 degrees. Our angle of multiplied by the coefficient of kinetic friction multiplied by 2.0 kilograms times 9.8 meters per second squared. So we find that difference and then this will be divided three plus two, 5.0 kilograms and we find that the acceleration is equaling one point 96 meters per second squared. If we wanted to find force tension, we can simply say that the force, the tension force, would be equal. We're taking this equation now and we can say the tension force would be equal to M sub to G sign of fada minus m sub two times the acceleration this would be equal to I'm some too. In this case, uh, rather we know that himself to 3.0 kilograms times 9.8 meters per second squared time sign of 30 degrees. This would be minus 3.0 kilograms multiplied by the acceleration that we just found 1.96 meters per second squared and we find that the tension force is equaling 8.8. The Newmans This would be acceleration and you're forced tension. That is the end of the solution. Thank you for watching

Here for the solution. The net force acting on blog is zero. That is sigma ethical to F. One plus two plus three which is equal to zero. And we consider it as the equation one. Hello everyone is the force applied by the lady on block. After is the force applied on the man on block by the man on block and after is the additional force now we re arrange occassion. Want to obtain an expression for F. Three. So by rearranging we get after equal to minus and break it at one plus F. Two. And we consider it as a question too. Now for the calculation the total work done by the lady is that Afghan factory equal to 5.5. And and break it cost 30° except minus signed 30 degree white cap. Or it will be equal to 4.76 and X cap plus minus 2.75 White Cap. So now the total work done by the man is that it's happened factory called to 3.5. And and bracket cost 37 degree. Xscape minus sign 37 degree. Why Cap all? It will be equal to mm hmm 2.80 and X. Caplis 2.11 N. Y. Cap. Now with the substitute 4.76 and X. Caplis minus 2.75 And white cap for F. One and 2780 and X. Caplis 2.11 And wake up for F. Two in decoration too. two. When the additional force keeps the block at rest, that is f. three. Oh sorry. So it will become after equal to minus in bracket 4.76 and Xscape blessed In Bracket -2.75 and white cap plus and back at 2.8 and X cap plus 2.11 and white cap. Or by simplifying this, we get equal to minus 7.6 and X cap plus 0.64 newton white cap. Therefore the additional force keeps the block at rest is -7.6 X capital, us 0.64 and white cap. So this is a complete solution. Please go through this. Thank you.