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Problem 6Tro blocka are connected by string; shown in the 6ge The smooth inclined Jufface makes angle of 42 Kich the borizontal, aud the block on the ineline has Me...

Question

Problem 6Tro blocka are connected by string; shown in the 6ge The smooth inclined Jufface makes angle of 42 Kich the borizontal, aud the block on the ineline has MeSE of 6.7 kg: Find the mass of tbe hanging block that will cause the system Lo be in equilibrium The pulley frictionlessInbrer :Solur102

Problem 6 Tro blocka are connected by string; shown in the 6ge The smooth inclined Jufface makes angle of 42 Kich the borizontal, aud the block on the ineline has MeSE of 6.7 kg: Find the mass of tbe hanging block that will cause the system Lo be in equilibrium The pulley frictionless Inbrer : Solur102



Answers

Two blocks are connected by a string, as shown in FIGURE 6-44.
The smooth inclined surface makes an angle of $42^{\circ}$ with the horizontal, and the block on the incline has a mass of 6.7 $\mathrm{kg}$ . Find the
mass of the hanging block that will cause the system to be in equilibrium. The pulley is assumed to be ideal.)

Chapter six. Question 40 has the situation. As I've drawn here, you have two blocks, um, wrapped around a pulley on an incline mass. Big M has given his 5.7 kilograms. Master Lim has given us three point two kilograms and the incline is raised to 35 degrees, Um, question states, or ask us to determine the direction and acceleration of. So that's the direction and magnitude of the celebration, which is not often required, Um, in these types of questions where they simply state the blocks air freedom move and was determined, basically, is the magnitude of the acceleration. So this exploration can occur going down or going up? That was part one of what was the chairman that since he's to the direction of magnitude of the celebrations, are very well, interrelated. We'll solve them together, answering and be simultaneously. So what we do, I haven't well again, we're not sure what direction of exploration is. Something was simply just choose one. Um, as we often do, these questions were draw free body diagram. Just examine the forces that were experiencing so this for block em. I have a celebration due to gravity G as well as the tension pulling up on the rope not indicated, but simply pulling up the tension. Mr. Action of Detention Building up in this direction as well on Big M I think it's a my net force in my wind direction is positive tension minus acceleration due to gravity that mass. And finally, I'm going to guess that my A is going to be going up accelerating upward with this mass. But a question, because that's what I'm positing. So then my resulting for us here just be little m Time's A If you look at the X components on block capital and block or block to Maisie accidentally, Look here we have, of course, the tension acting in this direction. If you look at, um this component here, you know the force of gravity would be down and that was a big M. I like that big M times G. If we do like triangles being determined that this Seo here is also theater. Hey, look here It also theater. So the x component of our gravity Here's mg scientific data. So you look at the net force in our ex direction. Of course we have attention minus big and G sign data. And of all of this again, if we're sending blocked too our little M Bloch is going up. Then our block is going down big in block will be going down. Will have negative m A. Yeah, the fragrance, Lee, I'm solving pretension to be and G signed data equal Very close. It's minus Just began times a day. I can substitute this back into my f Why my force of the wind direction equation as we found here, we're just doing that. I have and eliminate tension as an unknown variable mg signed data The M G minus little mg minus big M A and all that equals little in May. Okay, I'm solving for a so I can collect some, like terms here. I could have collected the G's here, but that's okay for the right time. You said a time. Some of the two masses we can rearrange is the salt for acceleration, which I've been here as it falls. We have all these variables given to us Big M little limit data. So you played is in direct happier. We get our emphasize. This is positive. 0.0 76 meters per second squared. There is a very relatively slow acceleration, but it is a positive acceleration nonetheless, which tells us this is positive. That means our initial guests of our acceleration direction to be up is correct because it is a positive value if we chosen it to be negative. So if we said don't do this in green, If he said A is traveling down, we would have resulted in a equals negative 0.76 meters per second squared. Assuming, of course, you kept all of your, um, positive leaders in check and there you have it.

In this problem, we have given a diagram like this. This is an inclined plane and hear a mass of 5.7 kg is connected by a road via fully and hear another mass of mass and is connected with this room. And this inclined plane is friction less. That means there is no friction here. And this angle from the horizontal is 35 degrees Mhm. Mhm. Yeah, it is given the mass of this hanging block is 3.2 kg. Thank you. So we have to find the acceleration. So we know that here m G force will be acting in the downward direction. So there will be two components mg signed five sign 35 and MD costs 35 will be there and hear this M mhm do you will be here. I suppose this is capital and don't can get confused. Sure, capital and is equal to 5.7 kg and the small lamp is equal to 3.2 kg. So suppose acceleration of please follow are in this direction. So we will write Newton's second law equation for a small hanging block. So we will write net force in that war direction T minus small M G is equal to 7 to 8. This is a question number one. And for this vlog, we will right net force in this direction. You were right. Mg sine 35 minus t should be equal to two. This is a question over to We will add this to a question. We get energy sign 35 minus small energy is equal to small and capital and into a so from here, exploration will be equal to this is 5.7 in two, 9.8 into sign 35 minus 3.2 until 9.8 the world away 5.7 plus 3.2 So this acceleration is equal to if we saw this is equal to 5.7 multi private, 9.8 multi private sign 35 minus 3.2 multiply but 9.8 divided by 5.7 plus 3.2. This is equal to 0.76 m per second squared. This is the magnitude of the exhalation of two blocks and we can say the direction of hanging block the direction of acceleration of hanging vlog in the up our direction. Yeah, Yeah, right

Okay, so to begin with, let's write the equations of motion, the net forces acting on each of these individually. So let's start with mass a. Which is on this 30° incline Um Assay is a mask 3.7 kg. Um So its mass times acceleration, some mass times acceleration. The net forces acting on this, It's got its weight force down. So there's going to be a perpendicular, sorry, a parallel component to that. It's going to have some tension this way and there's not going to be any friction. This is a frictionless plane. So those are the only two forces that we have to consider. The why are the parallel component of its weight force and the tension force. So we can pick either up or down to be positive in this case because we don't know which direction the acceleration is going to be in just yet. I'm going to choose up the slope to be the positive direction. So I'm going to say I'm going to assume that we're going to be accelerating up um and then if we end up with a negative value, that means that we're actually going down. So so we've got a positive tension and then we're going to subtract the weight force from that. So I. M A G sign of photo where that data is that 30 degree angle. So this is for the first mess on the incline right here, this is in the parallel direction, parallel to the some of the forces in parallel direction, the 2nd mass this mass beat, we'll have the same acceleration because these are tied together. Um But in this case because I chose this way to be positive, I'm going to say that down is positive for this, which means it's it's weight force. M P G is going to be the positive direction, and I'm going to subtract tension like that. So the Mhm. I've got two equations here and I've got two unknowns, my acceleration in my tensions now, the first part, I'm supposed to find the magnitude of the acceleration. So to do that and notice that I've got a positive tension appear and the negative tension down here. So I'm just going to have these two equations together and I will get on this side, some of the masses and a plus MB times their acceleration. Well equal this value minus this values M G M B G minus M A G Sang data. And then to find acceleration, you just take this whole value here and be G minus M a g ST peter, and I can factor out my G if I wanted to, but I don't need to and then you divide by this coefficient here and then we get our value for acceleration. And so if we plug our values in here, M B is 2.3. M a is 3.7 And then sign fatal, which is 30°. We should get a value of 0.735 meters per second square. And this is a positive value. So that tells me that yes, I picked my direction correctly, have these will the mass on the platform will be accelerating up the slope. And that means that for a letter B, which direction is this going to be accelerating? It's going to be accelerating town. So that 2nd mass mass B is going to be falling and then we get to part C. C. We need to find the attention. So to do that, I can just plug in this acceleration into either of these two equations and solve for T. So let's just go ahead and do this second equation. I think that would be a little bit, but they're about the same. So I'm just going to add T to both sides and subtract this term from both sides. And so for part C, I get tension is equal to this value minus this value and be G minus. And the hey, Where a is .735. I'm going to get attention value of 20.8 Newton's.

Question 75 part of a series of questions all about a block on an inclined plane with various different attributed forces and facts to go with it on a few things that we have to work out. So we haven't no 0.2 kilogram block on inclined plane, and it's connected by a string going over a pulley to a no 0.6 kilogram hanging book. We simply have to determine the acceleration of the system. If there's no friction between the block on the surface of the implying plane eso with no friction, this becomes really quite a nice, straightforward question that's model are blocks as particles because we don't need to consider too much. That's complicated about this. And so we have a situation here where we should just model our forces on the weight of ah block. Here is no 0.6 most blood by G. Onda. Of course, our weight off this block is no point to multiplied by G. Please don't read the size of the arrows to me, anything in terms of their after scaled forces are merely drawing this arrow bigger so that I can better express it is in the diagram in a moment cause we have tension that's putting us in both directions on a normal force is acting this way. Um, this question specifies this is a 30 degree angle. So will pop that in. And we should also say that this this force that weight can be split into two components. One that's parallel one that's perpendicular to sleep, which would make this angle here also 30 degrees. She's quite nice. So you've got body one on 22. So we call this 11 and this one, too. For a start. We want to consider the forces sort of generally balancing out. Certainly this body one is not descending into the slope. And so it's normal force, especially when we don't have friction. It doesn't really matter to us. We're only concerned about motion parallel to the slope. And so these are the only equations with to take, so we'll just simply take the idea that power out. It's like we have attention. Uh, um was attract from that, uh, no points to G sign 30 because, of course, we need to take the appropriate parallel component, and that has got to be equality. North Point to okay and likewise a body to weaken. Draw a quick set of equations, which is no 0.6 g minus. T is equal to no 0.6 pay. Um, on. Of course we can do a bit of rearranging. We tell. Do in a second. I want to explain something for us, which is that I have chosen very specifically the body to is the body descending on body one is the one a sending. It's going up the slope. Uh, so there is a reason for this, and that is basically that well, we don't consider friction, which simplifies the problem incredibly, but the more massive block is going to be wanting to dissent quicker than the less massive block on. If it's on a slope, it's downwards forces its forces down the slope, parallel to the slope, or even less wanting to descend than they are. If it's just perpendicular, we going. So it's obvious in this situation, the body to is the one descending okay, s. So we can certainly rearrange these equations a little bit. Let's get both in terms of attention. So let's say the tea is equal to on rather than using theatrical points or no points. Six. Let's go for a general case because it's always nice to see what's happening in the general case. So simply say, T is equal to M one. A. Plus g son 30. We'll keep the angle is 30. We might as well, and in this one t is simply equal to em to I mean in brackets G minus a. Okay, so equating these two tensions, we can simply say that M two G minus a is equal to and one a plus g sign feta course at 30. In this question, Andi, it's probably best to expend everything out because we're about to rear end for acceleration. So we said we want, uh, something like T G minus end to a or AM, too, as I've just written it there. And that's gonna be equal to M one. A. Plus m. One g. It's sign 30. Lovely, very lovely on. So hence getting all the cheese until one side and all the days until one side we now have in expressing G into minus m one sign 30. I think I got that right. Absolutely have is equal to m one plus m to all in parentheses. A. That's a rearrange they on. This is also straightforward. It's simply the left hand side. G two minus m. One sign 30 divided by the sum of the two masses and one plus m two. Andi, these air nice numbers. Simply put, the M one is no part in the empty is no 10.6 but the same 30. But that's just 1/2. So actually, we can, from our substitution, simply say that acceleration is equal to G. Apply the numbers now, and two is 9.6 and M one is no point to. But Sign 30 is 1/2 so good 1/2 that for no port one that's good. And then on the bottom, we have m one plus Empty said no 10.6 plus door point to it is no 0.0.8, and so the acceleration is simply equal to five takes G, which we put into a calculator, and we reasonably quickly establish that our acceleration is 6.12 uh, five or 6.13 You should probably express it so few fewer significant figures 6.13 meters. A second squared on. That is all the question is asking us. It's not asking about any tensions or any other forces, so we'll leave it at that. And of course, we'll carry on what these slow questions Emily next.


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