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Newton" Laws BackwardsThe aPplication of Newton' laws to problcn Lten by components:the following equationsNewton'5 Sccond Lawcos(6u?) sin (60")...

Question

Newton" Laws BackwardsThe aPplication of Newton' laws to problcn Lten by components:the following equationsNewton'5 Sccond Lawcos(6u?) sin (60") 0.40(20 kE) (9.8 m/$ ) (0kg) (-0.50 m/s ) Wotk tlac Newtotan) lysis bckwucls First, €Onsttuct lice boely 4g'4m consisteHt With thesc ccations Muc tel devnse tcalistic [toblem situation (Ot which tlus woull be thc solution (YOu May have t0 Jitele bit €tcative): Be cuelul Speci Iluc iuutcaction Leeli "O [Otces Aued detrtm

Newton" Laws Backwards The aPplication of Newton' laws to problcn Lten by components: the following equations Newton'5 Sccond Law cos(6u?) sin (60") 0.40 (20 kE) (9.8 m/$ ) (0kg) (-0.50 m/s ) Wotk tlac Newtotan) lysis bckwucls First, €Onsttuct lice boely 4g'4m consisteHt With thesc ccations Muc tel devnse tcalistic [toblem situation (Ot which tlus woull be thc solution (YOu May have t0 Jitele bit €tcative): Be cuelul Speci Iluc iuutcaction Leeli "O [Otces Aued detrtmnlitle the values ofAny unknown [ice



Answers

Solve the previous problem quantitatively using Newton's second law.

Eso question 11. It's really related to Question 10 Very much, it says the right. Newton's second law in component form for each of the situations described a problem. 10. Eso. You've seen this as I've done before. If you've watched all of these videos, you've seen me do a similar thing with problems. Eight, which were forced diagrams. And nine which, with the appropriate equations, we're gonna do a similar thing for 10 where we describe force diagrams for situations on these air, the appropriate equation. So let's just remind ourselves there are four situations in this question. I would start with Part A, which is just asking us to describe a cinderblock sitting on at the ground, and so are to remind for the fourth diagram we had normal reaction force from the surface on the force of gravity acting downwards. No other forces acting left or right, and we're doing this in component form for the first time ever. I don't draw a direction emotion. This thing really just isn't moving, Um, and so we don't have two months to go on in terms of motion. There is no X component. Eso If I wanted to start with X. Nothing to do. There's no forces left or right, so we'll go straight to the white component. Where will define our accedes as exits? Horizontal on. Why is vertical? So I want component. Newton's second law separation and why is equal to the force, the Net force and why divided by em and being the mass off the cinder block? And furthermore, we can say the Net force on the cinderblock is equal to the pick a direction, I guess, the force of gravity minus the normal reaction force. But of course, the net force is equal to zero because it's just sitting on the ground. And so really, what we can say from what this is, the force of gravity is equal to the normal reaction force off the surface situation is very straightforward. Okay, party A rope pools at an angle of 30 degrees relative to the horizontal step moving on a horizontal surface. Uh, that is moving an increasing speed. That right, the surface is not smooth. Onda Cero proportions that parallel to the incline slope. Sorry, I've moved straight under question. See, I should reiterate Part B, as it should stand would be a rope fools at an angle of 30 degrees relative, the horizontal on a sled moving on a horizontal surface. The sled it moves in. The increasing speed to the right on the surface is not smooth. So to remind ourselves off this force diagram, we've got a force of gravity acting downwards, a normal reaction force acting upwards as we've got a rough surface. We got a force of friction opposing motion and we're putting with a rope. So we're gonna deny a T to me. Attention forced on this angle is 30 degrees and so in component form. Once again, let's just get a view. Motion is horizontally to the right As I've drawn it, Andi X and Y will donate as horizontal and vertical as before, So in component form in the X Direction three acceleration in the X direction is equal to the Net force in the X direction over M, which is the mass off the sled on the F Net X. Remember that F Net X, therefore force the net force in the X direction is equal to the tension. We might as well use the positive part first tens, but not the full tension need to take the co sign of 30 because, of course, we're in considering the horizontal in this and opposing that motion. Is the force off friction okay? And then, furthermore to that, we have to go onto the Y component in the Y component is simply the acceleration in the Y. Direction is equal to the force. The net force in the Y direction divided by the mess on DA. This time, of course, we're not taking off. We're not going up or down. This is zero. We're not descending into the surface, so I could probably use a bit of information from that. Anyway, The net Force in Why, Yeah, Just considering hot diagram. Well, we're moving up or down, so don't really matter which direction it chooses. Positive and negative. Let's go with tension was positive last time. It's positive this time, So tensions signed 30. Ah, because we got a force pulling us up in the Y Direction is component attention and we'll ask that in the normal reaction force and was attract from that the force off gravity. But we know that all of this is gonna be equal to zero because our sled isn't either taking off or descending into the surface. We're not pulling a sled on clouds said last time something that house to go through the surface. So the force of gravity is equal to attention. Sign 30 added to the normal reaction force. And so this is the situation for this. Let okay support, see a game Fairly simple. A rope pulls a sled parallel to and inclined Slope inclined at an arbitrary angle on the set Moves in increasing speed up the slope, All straightforward. So once again we're model this in a particle as a particle on with the force diagram Onda as I sort of said in part, uh, Ponte, Um, Question 10 is really quite useful for us not to always consider Axum. Why? Although we can redefine our coordinates as much as we'd like X and y being, um, defined as we like them. But for the sake of our because personal consistency, you can talk about forces parallel and perpendicular to slip, so we know that our sled is not falling into the slope. So I got a normal reaction force we're pulling up the slope was some main force that we don't know So we do know that just as an f we've got a force of friction opposing that motion and good old gravity ex straight down and doesn't care. One jot for what we're doing here and as we're dealing with parallel and perpendicular, then will define. First of all, our direction of motion is up the slope and the axes that were considering donated like this where we've got a parallel direction on a perpendicular direction and this is just it's good to start off, measure out what we're looking at. So first of all, before we still will break gravity into components, and then we can see how acts parallel and perpendicular on this angle that I've done it. Hear it to know here is feet up as the same theater that we've written in black here and now components up the slope. We've got a parallel direction and will say the acceleration in the parallel direction has got to be equal to the net force in the parallel direction. Newton's second law, or divided by the Mass on the Net force in the parallel direction is equal to the main force that we've drawn the force that is pulling us up, Sled up the slope. Andi, From that we've got to take a few things. So the main force was subtract from that the force of friction. But of course, we're also opposing gravity. So we're gonna take from that Also the force of gravity. Uh, not the full thing. We'll take the force of gravity sign feature. OK, so that's our parallel, Doris. And now we definitely know it's accelerating. The question specifically says so. So there's no more We can write about this. We can't have inconvenient. Zero Simplify anything for the perpendicular direction. The acceleration perpendicular direction is equal to the net force in the perpendicular direction divided by the Mass. Newton's second law gonna drill at home as much as we can. On the Net. Force in the perpendicular direction is defined to be the normal reaction force from which we subtract the force of gravity, not the full force of gravity. Once again, it's a component off. This time is the co sign of theater. Andi Thisted Time acceleration perpendicular is definitely here. The net force is definitely zero. So a normal reaction force it is definitely equal to the force of gravity. co sign feta NASA situation. See? So it's the final hurdle here. Ah, situation D A. Situation D is talking about a sky Doha falling downwards at a constant terranova velocity. So it's a constant velocity. And if we're going a constant velocity, our forces balance. So a force of drag from the air is equal to our force off gravity. And once again we do have a direction. Emotion. This time it's downwards. But this is a constant velocity. Will be, didn't let's see here and for axes, we might as well go back to X and y. Considering wise, the vertical people are very comfortable with that as an idea. So there's no horizontal forces. So for X, we have very little to do. Might as well not talk about it and for why, in this situation, well, force of gravity minus the force of drag is equal to zero. Well, I accelerations equals zero. We should probably go into as we've done for all the previous examples. Acceleration in the Y direction is accorded the net force in the Y direction ever M, which is equal to zero, which means our net force itself is equal to zero net force is equal to force of gravity minus the force of drag MassEquality zero. So, of course, in this situation are force of gravity is equal to our forces, drag due to the air. And that's the situation. As it stands this how to answer that question, these the appropriate

Eso. I've said this before and I'll say it again. I'm not the biggest fan of equation Jeopardy questions, but I've said this before as well. They do have their uses. The principle is that we are given the horizontal and vertical that is excellent. Why components on those forms of Newton's second law applied to some physical process. So for the unknowns, and then work backwards to construct a force diagram for the object interest. And then we invented programs which the equations might provide, an answer for which there are many possibilities. The first things First, we're going to write out the equations as the book questions us. And they say simply that five kilograms multiplied by the acceleration in the X direction is equal to the following summer. Forces on we have 50 Newtons multiply by the co signer 30 well, and to that end, which is not Newton's, it's an unknown variable em multiply by the co sign of 90 from that would take no 0.5 and to sign zero, and we'll answer that five 0.0 kilograms, multiplied by 9.8 Newtons per kilogram, was squeeze a listen here, and that is most blood by the care sign of 90 of the second equation. Because we're going to need to is equal to the five kilograms. Once again, it's the same object. Doesn't matter if we're the horizontal vertical. And this time it's most blood, not by an acceleration in the Y direction. It is much blood by zero. Cabinets were not moving in the Y direction, and this time we take minus 50 Newtons or negative 50 Newtons multiplied by this sign up 30 to wish we will add and sign of 90. Well, answer. That's no 0.5 and sign off zero. And finally we was attract 5.0 kilograms more supply by 9.8 Newtons per kilogram on that is most blood by the sign of 90. All right, some bookkeeping festival. We've got a lot of signs and co signs which are very simplify rubble, uh, signing cause of 30 Suncoast under 30. We can come to those later, but the coastline of 90 0 and a co signer zero is one the co sign 90. Once again it's zero. Likewise this side of nineties one on the sun, 00 certain 90 being one and so This whole thing simplifies down immensely, and so are modified. Equations, as we would like to treat them, would give us phone times. The acceleration X direction is equal to 50 co sign 30 minus no 0.5 in on. And equivalently, we have zero equal to minus 50. Sign 30 plus n minus five times 9.8, which very quickly we can see a few things with. And that is that sign of 30 is 1/2. It's one of our special angles, leaving us with minus 25 at the front, plus 10 minus five times 9.8. And I said this before as well. What you have done Enough of these questions with the gravitational constant, you start getting quite good at your multiplication by 9.8. This is 49 and so consequently, you rear end everything to have an equal. Teoh 74 Neumann's It's a force, and you'll notice immediately that the only substitution we then have to make is for the n equals 74. To go into here on day quickly ranging tells us that a X is going to be equal to 10 curso in 30. I was subtract from that nor plate warm times n and is known. So it's minus nor 0.74 Okay, Next we simply say that we can write this as five through three. What is that? Well, because of 30 coastline of 30 is the square root of 3/2 cancer. The square root of three attitude most blood by 10 is simply five multiplied by the square it above three. I was subtract from that 7.4, as we have on the line above and sleeps is with 1.3 meters per second squared, so we solve for the unknown. Now we simply have to draw force diagram and are forced. Diagram is based in the horizontal vertical axes, and what we have here is pretty clear. All of these forces between the two lines correspond to each other, and we should have taken the hint quite often So, first of all, we have to incredibly vertical forces just purely for vertical forces. Here in hand, 9.8 tells us that we're dealing with a gravitational constant, and so we have a weight acting down and this is our weight off 49 Neumann's We also have a normal reaction force acting upwards. We've determined the value of that normal reaction force and it is 74 Newtons. Now this chapter is on friction. And so we have no 0.5 multiplied by the normal reaction force. Here is our frictional force, Andi, as it's half of 74 Newtons. It's fairly easy to figure out that our original force, that is the force opposing motion is gonna be half of 74 which is 37 mutants. And finally we have one force that is acting in the positive X direction but the negative Y direction with an angle of 30 degrees. So we're looking at a force that looks like this. Now we can see other front that forces magnitude is 50 Newtons where this angle is 30 degrees. And of course you can come up with lots of examples. For example, anything, anything with a constant force that is giving an acceleration. Except this time, of course, it has to have friction. Voice that have an acceleration will that we should very quickly note that the horizontal forces are not balanced. That is that if we take out 50 Newtons on, we multiplied by the co sign of 30. Well, if we put this into a calculator to see that this is 43 Newtons, which is greater than 37 Newtons, which is opposing motion, and so the block is definitely accelerating. So anything where you're pushing down into the right, uh, would be an option or anything. Where you pulling from down to the right would also be an option. Perhaps you're pulling closed. A set off I can only describe is extremely heavy curtains or pushing on quite a light shopping cut. You can come up with many examples for this on inventing the possible situation is really no quite the point. This is how to do the mathematics here. Fast question 36 1 of the equation. Jeopardy questions.

All right. So this question is pretty straightforward playing which no. So, for the unknown variable in this case, trying to find a church which is the amount of time it takes for are given temperature to reach the final comfortably wanted to reach, which in this case is freezing. So start out by simplifying this this expression and will soon we'll start by adding Tween of both sides will have 52 equals will combine the two numbers in here, which is 95 times you to negative 0.12 each. Um, some extra the by both sides, but 95. This gives us 0.55 equals negative. 0.12 hitch. All right. And now we have to get rid of this exported. So we'll take the natural lock of both sides because that's what undoes basically exponents. Exponential operator. This thesis was natural. August 0.55 because negative 0.12 h We can divide both sides by negative 0.1 tube. So we have gridlock of 0.55 over the top 0.0. Want you equals h when you sell for H you get h equals 50 takes about 15 minutes for the desired object to the center of 30 degrees.

Okay. Question 37. Sadly, another equation. Jeopardy. I'm not a fan. As I've said before, I should always reiterate on these in case you're just checking out this one video. But they do have their uses. Um, eso Essentially, the idea is that we were given three the x and y component form of Newton's second law, in this case, applying to a physical process for an object on an incline. And we have to Seoul for the unknowns on, then work backwards and constructive force diagram for the object. Then, finally, we have to invent a problem for which the equations might provide an answer for which there may be many possibilities. Um, I think possibly that is one of the reasons that I'm not a huge fan of the equation. Jeopardy questions. It's because the inventing a problem seems quite contrived. But we'll put that aside. So what have we got here? Well, first of all, we have 5.0. I'm just copying out the equation from the book. It's quite important to talk it through. Go got the mass. Times zero. This is in the X direction way Soon is acquitted, plus some force F most blood by the co sign zero degrees. Well, after that, some unknown in and this end is the variable end, not Newtons. This is multiplied by the carousel of 90 degrees. From this we were subtract nor 0.5 in not supplied by the co sign zero degrees. Now finally, from the whole thing we were subtract five in brackets, 4.0 kilograms. No, supplied by 9.8 Newtons per kilogram. You should recognize the gravitational constant now almost play that by co sign of 60. This tells us we're on things like on the second equation. Somewhat similar 5.0 kilograms multiplied by zero is equal to the positive force multiplied by the sign of zero plus in the variable n sign 90 from the source Attractive No 0.5 and sign zero Andi finally minus five point Is your kilograms multiplied by 9.8 Newtons per kilogram on the whole thing Must blood by the sign No. 60. So onto our bookkeeping I always like to do this. Anything that we can get rid off should be thought off very quickly to get rid off, Even if it's eliminating that we don't have to write it. So, for example, the co signer zero is one the care son of 19 0 Once again, the case size here is one. Now, because I'm a 60 will leave till later. But the sign of zero is zero and the sign of zero Because one signs your so nineties what on the sign of zero is zero. So we're in a place to simplify these equations. Let's write the 1st 1 first. Well, it doesn't matter what mass you are if you're not accelerating. There is zero on this side of the equation, and that's going to be equal to U F, which would take half times one. Uh, we ignore this because this has his most Popeye's here, and we take it to B minus no 0.5, and and also my ass five by 9.8 Care sewing 60. Good stunt. Uh, the second equation zero once again. This time it is just equal to this time that normal reaction force and minus five most God bar 9.8 and the sign or 60 Good. Well, the first thing to note is that this equation the second question is very easy to solve precipitate the five multiplied by the night. Poor Kate on the sign. 60 No to the other side. Leading us with five cents Night for Hay is 49. 49. Sign 60 is equal to. And now putting this into a calculator, we simply figure out that N is equal to 42.4. Oh, that's Newton's because it's a force. And now we can substitute into the first equation where we figure out that F is simply equal to no 0.5 times en plus five times 9.8 care sign 60 s O F. Is equal to 21.2, which is half of 42.4, and will and that 49 which is five by 9.8 care sewing 60 and I calculated work. It was the rest, so there are no forces 45.7 Newtons, and now we simply have draw fullest diagram on the question did absolutely specify that we are on an incline. And so it's very important that in this case we have a parallel direction and a perpendicular direction to our slip. Uh, just from the angles, we can tell that our forces pulling us directly up the slope and friction is always going to be acting parallel. So that is gonna be pulling us down the slope. Ah, Next we have a normal reaction force which is perpendicular to a slope. And obviously our mass acts downwards with energy, which has two components on this angle. Here we simply have to turn to our equations to look at it on our parallel. Comparing two days, 60 co sign Sorry, are parallel comparing is talking about the co sign of 60. So if our parallel component is talking about the co sign of 60 it's actually making this angle 30 and therefore we simply write that fi 45.7 mutants is pulling us up the slope. In the meantime, on normal reaction force here is 42 went full MZ acting downwards. Basically tell immediately that this MGI is this making this equal to 49 and our frictional force which is opposing our machin. We've already seen this earlier. He's just here 21.2. So now what we have is a century full start, Graham, and we have to come up with with the scenario that Dick takes Want this force diagram is on. And I would say if I have kilogram block being towed up a hill providing this is a constant velocity. Friction is acting. Uh, but it could be moving at a constant velocity. That way there will be no acceleration on this side of the equation would still be valid. But you could come up with lots of different examples. This is just simply one, and that is questioned 37.


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