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752860Eaku In-am Eoar HES: smcoth &bk obixt which Irn aas ad Epulled Tl 6ee 5 #0 Jon euktr ulfolk felttinn lor clartty of wark, [conalui5"y You uill 'coltatia uanttWCre Uetb ugked 30" abuieimna oblcct showiag all Label all R forces 6t Kntm Ino objccls hlun IrdesEt i2 8r 4+L Ieeten[w:So S 31F : 6-e Sai Fy 2 254 2fueces & Ca 53C8530 =43.3NWhict Bure (4 oleeh OleLe coinpon TOntean elere ) lo 0 A Malwetee Ilnitudes Idueclions F+FEoBz0 DtavtI a1- E2 0 fx - -FEaG Jke Oblle Gv- C

752860 Eaku In-am Eoar HES: smcoth &bk obixt which Irn aas ad Epulled Tl 6ee 5 #0 Jon euktr ulfolk felttinn lor clartty of wark, [conalui5"y You uill 'coltatia uantt WCre Uetb ugked 30" abuie imna oblcct showiag all Label all R forces 6t Kntm Ino objccls hlun IrdesEt i2 8r 4+L Ieeten [w:So S 31 F : 6-e Sai Fy 2 254 2fueces & Ca 53C8530 =43.3N Whict Bure (4 oleeh OleLe coinpon TOntean elere ) lo 0 A Malwetee Ilnitudes Idueclions F+FEoBz0 DtavtI a1- E2 0 fx - -FEaG Jke Oblle Gv- Cllov Hor uocsthe block move ii bpullal fo10? Lx = MA '3E Sefos3o z 2.165 ms 20 Ax-*utat Vo_0 Lx- { ( 2."lS) ( lo) bk = [1ob.25 m Glattad uJtAL



Answers

Let us fix $x-y$ co-ordinate system to the given plane, taking $x$ -axis in the direction along which the force vector was oriented at the moment $t=0$, then the fundamental equation of dynamics expressed via the projection on $x$ and $y$ -axes gives, $$ \begin{aligned} &F \cos \omega t=m \frac{d v_{x}}{d t} \\ &F \sin \omega t=m \frac{d v_{y}}{d t} \end{aligned} $$ and (a) Using the condition $v(0)=0$, we obtain $v_{x}=\frac{F}{m \omega} \sin \omega t$ and $$ v_{y}=\frac{F}{m \omega}(1-\cos \omega t) $$ Hence, $$ v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\left(\frac{2 F}{m \omega}\right)\left|\sin \left(\frac{\omega t}{2}\right)\right| $$

So once again with determining the force vectors the X and Y components for each For Spector shown in the diagram, this diagram would just just it out very quickly. Well, once again, I wish these lines were straight, but it gives you the idea. So we've got a surface upon which there is a particle and then some force factors which will once again draw this one vertically, upwards, this one vertically, downwards. And then one that jumps off of this angle here, what can we know to our first of all? This angle is 25 degrees from the actual vertical access itself. These forces also have labels will give them their appropriate labels. We've got f here, which is dinner. See, with sub script are one and oh, are vertical one upwards f uh, two on. Oh, and finally I last one. This is f Yeah, on. Oh, okay. And now we just need to get the appropriate components for each of these to x and y here on, Let's shut down. The values is are the names of each of these will start at the top one f are two on over, and then we've got if the one opposite E on and finally f our one on Oh, okay. Till I start feeling in this table and we'll look at f r to Ah, no first. So it's going vertically upward on. If we're defining why increasing upwards and ex increasing to the right as his general convention, you can see that the vertical one has no X component. I should add the actual figures from the diagram. That would be a good idea. Sorry. So 40 Newton's upwards here. This little fellow down here has 10 Newtons attributed to it and 30 Newton's attributed to this one. So that's that. That's that. And that's that right? Okay, so given that we also need the 40 mutants to be entirely in the one direction because it's vertical Okay, the next one, the other easy one f e on. Oh, well, clearly. Once again, being vertical has no ex school a component on dhe. Therefore, all of this stuff is concentrated in the white direction being vertically downwards. It's minus 10 mutants. And finally we have the F one on, and once again, we've got an angle associate ID, so we need to do a bit of a minor trigonometry to break in two components for our X component. We have to take the sign off 25 degrees because of how the angle is oriented. So we've got 30 multiplied by sine on 25 is not a special angle. It's a pretty mundane thing. And if we figure it out, we get the answer. Is 13 mutants or thereabout course? Remember, it's actually negative direction because it's moving to the left. That's crucial part for the white component. Off the same. We just take 30 and this time we take the coast time off 25. Now we find out that this is 27 to the nearest Newton and therefore once again acting downwards, minus 27 Newton's, That's what we have. So this is our table once again adding our units, this is a type of you want

This question. We are giving this for us with this time diagram, Uh, on a 1.5 kg particle. So, uh, you want to calculate the impulse in certain timetables and the final speed of the object? Uh, some thanks. Okay, so in Friday, I want to calculate the impulse, uh, from T equals zero to teach with the three seconds. Okay, So the impulse is equal to the aerial and, uh, false. That's the time graph. Okay, so for part, A is from 0 to 3 seconds, so it's only, uh, during this area. Okay, so it's four times stream is equal to 12. 10 seconds. Okay. He then be you want to calculate the impulse from over the whole time interval from T equals 02 teaspoons of five seconds. So, again, this is the area under the graph, but we from 3 to 5 seconds. The area here is negative. Okay, so this area here is negative. So we have, uh, just now, Charles, uh, plus negative area, which is minus two times two. And we get it new 10 seconds in overall. Okay. Then in policy, you want to find, uh, particle speed at close to three seconds when it is initially at rest. So be, uh, using. Okay. Impulse? Yes. You go to don't have p, which is equal to end times Delta V. Right. And then, um uh, final velocity. Well, yes, equal to the I class that Harvey, which is the I plus Delta p over em. Well, so we are zero. So we only look at the top, Um, in this case and yeah, only we want to find a particle speed at close to three seconds. So we only need to look at the impulse from 0 to 3 seconds. So 12 divided by 1.5. There, we get 8.0 m per second. So this is the final speed. Okay? Of the political. Okay. Then in property again, we want to find a final speed of the particle. At T goes in five seconds. So uh huh. Okay. It's the same formula. Same step as part C. So you only look at Delta p over em. But this delta P will be from 0 to 5 seconds. So it's eight divided by 1.5. You get 5.3 tree. You know this for a second. Okay, so this is the answer for party, and that's all

Hello. So for this question Oh I have the diagram on the left, we have F. One and then we have F. Two which has two components right? One year is the one to south. So the diagram on the right represents only only F. Two. So you see the national F. Two and the husband distinct component in the southern component. So because the body is unrest is in a room so we can resolve the forces in the X. And Y. Direction. So if you do for the X. Rather for the X. The force is going to the right will be 1 50 30 right and 30 from the F two members to 1 80. There is no force to the left. Us. When I sit we'll talk about why little force when the wind direction is 40 nothing the negative Y direction, it was negative 40. So if you re draw the diagram you realize that there is a force not going this way to the right and that's from 1 80 right? That there is another one downwards which is friday. Mhm. So if you want to trump the box at rest between that we have to try and balance all these forces. It means that we need a force that is growing up was which is also for the we need a four status in the opposite direction to the 1 80 which is also the same 1 80. If you have something like this that the body will remain in a kilogram. Okay, so now we're going to focus on just this one. So these are the forces, we need secure the body in the hologram. So uh if you're looking for a single force to do that, then it has to be the result of this too. Right? So the result of this to be in this direction, This is the second part. Right? So that's the second program. And that is where the new force or after it should be, inevitably body will be addressed. So, if you want to find the value of a single force which is going to use photographs here. All right. So that force listen us F train just be discouraged of 1 80 squared. It was funny sweat. And I shall give us approximately 1 84 uh, Charles. So this is the direction of F tray. Or is the value of after it? And it should be in this direction. Right. That's uh northeast, not worst, or west of north, whatever. Yeah. So what does that? That's the second record. Right here. Thank you very much.

For this problem on the topic of equilibrium and elasticity, We have shown a structure which has three forces acting on it, F one, F 2, and F three. And we wish to place the structure and equilibrium Like by applying a 4th force at a point p and the fourth force has components F H N F V. For the horizontal and vertical components, respectively. With the data that we are given, we want to find F H F V. And lastly we want to find the distance D. Now, firstly, by analyzing horizontal forces which need to add to zero for equilibrium, we can see that the horizontal force Component of the 4th, 4th of H must equal F three. And so this is simply five newtons. Now we need to find the vertical component of this unknown force V. And again we can use the equilibrium of the vertical forces, which says that every Must balance with the sum of F one and F two. And so F one plus F two is equal to F. V. And this gives us a total of 30 Newton's. Now we know the force that needs to be applied to put the structure into equilibrium. So, lastly, to find the we can compute talks about zero. and competing talks about how we get if the time's D Is equal to F two times B plus F three times A, meaning we can rearrange and been solved 40. And so we just divide both sides by a. V. And we get D. To be 10 newtons, which is F two times B, Which is three m plus F three, which is five newtons times A, Which is two m divided by F. D. Which we calculated to be 30 Newton's. And so computing this, we get the distance D to be 1.3 m.


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