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MlIn the figure, if the pulley is frictionless and the friction coefficient ifm1 mass of the rope is neglected, find the 10 kg and m2 2 = kg and the masses are movi...

Question

MlIn the figure, if the pulley is frictionless and the friction coefficient ifm1 mass of the rope is neglected, find the 10 kg and m2 2 = kg and the masses are moving at a constant ?speed (ibij 2.5)

ml In the figure, if the pulley is frictionless and the friction coefficient ifm1 mass of the rope is neglected, find the 10 kg and m2 2 = kg and the masses are moving at a constant ?speed (ibij 2.5)



Answers

(III) Two masses $m_A=$2.0 kg and $m_B=$5.0 kg are on inclines and are connected together by a string as shown in Fig. 4-61. The coefficient of kinetic friction between each mass and its incline is $\mu _k=$ 0.30. If $m_A$ moves up, and $m_B$ moves down, determine their acceleration. [Ignore masses of the (frictionless) pulley and the cord.]

So let's assume that the direction off Ah, the emotion off each block is the positive extraction. So here Ah, we're denoting with denoting box A as with gluing and do nothing box be with reading and also why is basically particle to the slope or vertical tow? The XX is now I've already drawn the free body diagram. Is there exactly similar? The only difference is for blockade of friction forces acting Ah, in the opposite direction. Off blocky and four block B. It's again acting in the opposite direction. But the tension force and ah friction force in Block B are in the same different direction, whereas in blockade, the tension force and the friction force are acting in the opposite direction. So as you can see that the questions will be quite similar, the only thing will be the change in sign for the Friction forces. So let's write the questions down for Block A and then we'll do the same thing for plumpy. So I'm using blue ing for block block A so it's gonna be f off in. It's in the wind direction minus and a G cool science Ada Hey, which is equal to zero. And from here we can we see that, if any is equal to M A G co signed did I? And in the extraction we have f off X Day that is equal dough F D minus and a G scientist up. It's actually shrink the picture a little bit so that it would have more room, toe buried equations so and a decide theta minus if off. If, um, let's see. We have f f r a. That's equal to M eight times so similarly in in, we'll write the same similar questions for dog be Israel. So why easy? Will do. Um, if and b minus and b g co sign, they don't be which is equal to zero. And then if off Expedia that's gonna be for big species, it will do and be signed data minus F off f r B minus f t with sequel toe M B times A. So what we can do here is first awfully can substitute this frictional friction force time. So, for the first case we have will have if off f r A, which is equal to new eight times. Um, if off any where Mew is deprivation of kind of friction so we can combine these two equations and substitute that here. Similarly, in Box B, we have f off F R B, which is equal to new B F off N B. So we'll add this question will relate this equation with this and with wybie wybie and then substitute that in the friction equation part. So if we do so, we will end up having to questions. One is in terms off m A and the other is in terms off m b a. So here on the solutions for those questions. So here I am. Hey, this is for the block A and ah for block B, it's m b times eight, so we can add them up and then solve for the ex elation from there. So when we added the questions, we had this league expression and we divided m a place and be on both sides to get rid off limitless and being from the left. And this is what we got here now. Ah, The given values for Amy and and be are two and five cages respectively. Then angles are 51 in 21 cage 21 degrees for its 51 for B. It's 21 mu. And maybe these are the coefficient of planet friction there. Same. Ah, there's essentially the same thing. It's 210.3. So if we use all the numbers we see that ex elation becomes negative. 2.2 Media over second squid. Thank you.

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.

In this problem we have to determine the exploration of this system. So Newton's second law is and the net force is mass into the excavation. So we have from the figure the net force acting on the mass. M one is T one minus M one G. This is a question one. And the net force yeah acting on mass. Everyone is everyone. Mm. This is the equation to So using both the equations we can say Using Creation one and two We can say and one a. is equal to 51- and one G. So from this T one is equal to M. One A plus M one gee. This is a question three. And then that force acting on the mosque two is I am to G -62. This is a question for and the net force acting on them as two is um to a Now using equation Using equation four and 5 we have em too is a call to M two G minus T two. So T two is M two G-. And to desist equation six. And the equation for net torque on the police half mm are square multiplied by a upon. Hi this is equation seven. And the equation for net torque is this is the equation for net torque on literally. Yeah. And the expression for net torque is um RT to minus RT one minus style. This is equation eight. So using equation seven and eight. Mhm. Uh huh. Rt to minus R. T. One minus so is equal to have an eye square multiplied by upon our. So from here the two minus the one minus how are is equal to half M. This is equation nine. So now you think equation three, six and nine we have M two G minus and one g minus tao upon us, he recalled two and one plus M two plus half. I am might be fired by a So from here is a Dorito. I am to G- and one key minus towel upon our of born and one I am too. Let's have Yeah, so We have M to 8.8 Kg. She is 9.8 metaphor 2nd sq my nurse and one is 10.4 kg multiplied by 9.8 metre for second square. My style is point 35 new 10 m upon. It is .15 upon and one is 10.8 K drip plus M two is point for Fiji. Let's have M. S 0.2 G. So from here is 1.2 m for a second square. Therefore the isolation of the system is 1.2 metaphor second square.

Okay, so for this problem, it's important to understand that the tension up of a police is going to be twice or it's going to be equal to both forces pulling down. So if we have this force pulling down across this pulley, this tension and this here is also going to be equal to that force over here. So this right here, Call this tension one, that's going to be two times that force. And so if this is two times, of course it's going to be projected down here. So this is also going to be two F. And so the tension here would call attention to, that's going to be two F plus two F. Or four F. And then that for F is going to be projected around as well. So we're gonna have four F over on this side. And so this tension, the tension of the top part, that's actually gonna be eight times the forest being pulled down here. Now, the weight of this thing MG is going to be being pulled up, um it's going to be being pulled up by seven F. So we have four F. Here to of here in one F. There. So combine that seven F. So this MG Is being held up by seven times the force. And so that forces being pulled down is MG over seven. Except And this tension since it's eight f Tension is going to be eight times this, or 8/7 times the mass times gravity. And we are told the mass is 6.4 kg. So if we multiply 8/7 time 6.4 times 9.8, we're going to get attention for 71.7 Newton's.


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