let-mathbfv_1-ldots-mathbfv_k-be-points-in-mathbbr3-and-suppose-that-for-j1-ldots-k-an-object-with-m

Question

Let $\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}$ be points in $\mathbb{R}^{3}$ and suppose that for
$j=1, \ldots, k$ an object with mass $m_{j}$ is located at point $\mathbf{v}_{j}$ . Physicists call such objects point masses. The total mass of the system of point masses is
$$
m=m_{1}+\cdots+m_{k}
$$
The center of gravity (or center of mass) of the system is
$$
\overline{\mathbf{v}}=\frac{1}{m}\left[m_{1} \mathbf{v}_{1}+\cdots+m_{k} \mathbf{v}_{k}\right]
$$
Compute the center of gravity of the system consisting of the following point masses (see the figure):
figure can't copy

Priyanka Sadarangani · Accepted Answer

okay. And this question, we have to complete the center of gravity center off gravity, gravity off the system, Consistent following point masses. So in the question is even it was 25 minus 43 and masses to Graham and point we do is for three minus two and masses 5 g V three is minus four, minus three minus one. And masses program Okay. And point before is minus 986 and masses, 1 g. Okay, so and the central in the formal of center of gravity is also given is one upon em. And one we one plus and do it too. Plus on em que week. So these are the given in the ocean. So now we have to find the central gravity. Okay, So first we will find on this portion. This was in one upon him. So one upon am it was 21 upon in one place and Douglas entry plus m four. So this will be program now 52 and one, then five, two and one. This will be one upon 10. Okay, Now we have to find this and when we want to do it, okay. And for this when we went in this way, will. So all the same for Eman is too. And we even points are five minus 43 Now I can do it so that I am to his five. And we do is 43 minus two and empty TV three. The empty is do and we tree is minus four, minus three minus one. And am for we for is m four is one and we for is minus 98 and six. Okay. And the resultant will be We have to fight it. Okay, so this will be then minus eight and six on this will be plus 20 plus 15 minus 10 and then minus eight minus six minus two and then minus nine. Plus eight plus six on this will be 13, 9 and zero. Okay, so we have the value. Now we will put these value in the formula. So we central gravity is one upon AM and m one m two plus m m m m money one plus two plus 33 plus emporium. A four is 13, 9 and zero, and that will be 1.30 point nine and zero. That means the central of gravity is 1.30 point nine and zero on. This is our final answer. Thank you.

Runpeng Li · Answer

All right, so for probably 30 Um, we want to know whether the center of a mass is the is in this pan is in the space off eyes in the space spent by the B one B one through. Okay, so so this space spend by, he won to decay. So by our deposition, just 20 the center of the will be, um, one over m times and one times v Juan. Thus two time to be too. Uh, until deal. I&#x27;m Kate. I&#x27;m speaking so we can observe that if we if we have. If we defined the center mass in this way that the V Center Mass would just be a leaner population of B one and b two b three until we say so. I&#x27;m just right. Spend is, uh, question. So we&#x27;ll be in want Wilbraham times V one and two over m times V two. Well, started up plus m k over times a B. K. So this is a linear combination of you want me to and b k. So it is definitely in the space off the inner space penned by V one V. K. So here we we know that I wanna over M, too. It&#x27;s just skater and them two over em over AM is that he&#x27;s another skater, so we just take a different skaters to find out. The center messed, so that means to seven Center for Mass. They see the spent. He&#x27;s in the span of 31

Regina Hays · Answer

okay. What we want to dio is we want to find the center of mass of the given system of point masses. And I&#x27;m actually going to draw a picture in the cornet playing so, um, to help us out. So here is our coordinate playing. And at at 22 we have a point mass of five, and that is at coordinates 22 Um, and then at negative 31 we have a point mass of one, and that is that Negative 31 and then we have 1/3 1 at one negative four. And so that point mass is of three. And he is at, um, one negative for okay, and we want to find the center mass of the system. So the first thing we need to do is we need to find the total mass, and so we&#x27;re just gonna add one. We&#x27;re just add up all the masses and we get a value of nine. Then we need to find the moment about the why axis. And so it is going to be all of the masses, all of the masses times the respective X&#x27;s value, because we&#x27;re doing the moment about the y axis or the vertical. And so this is going to be one times negative. Three plus, um, one times three plus five times two. And so this is gonna be equal to 10. Um, and then worry, Do the moment about the X axis, which is gonna be all of the respective masses. And since we&#x27;re doing the moment about the X axis were actually saying okay with a distance basically the distance of each point, mass is from the X axis, so that would be based off of the Y values and so whips. So this would be, um, point mass of one times one plus the point mass of five times two plus the point mass of three times negative for and so this will be equal to negative one. And so the X I, you, um, of Thesent er of mass is gonna be given by the moment about the y axis over mm over the mass. So that is gonna be 10 over nine. And of course we would. We would reduce, if possible. The y value your center mass is given by the moment about the, um, X axis over the mass, which is gonna be equal to negative one night. So my center of mass is 10 nights comma, negative one night. And so where that is gonna occur, go ahead and put a red dog in there. Where occurs? It&#x27;s a little over one and down here at negative one night. So it&#x27;s about right. There&#x27;s where that center of masses

Shiva Gopalakrishnan · Answer

let us find a send off. Most off the given dado morning in cagey located at X equal to Did he meet death? Yeah. Do you goto three kg located at X equal to minus one? We did the former force and rough Mosses exporting Would do you have more next? Warrantless? Um do we extrude? You were in vain. You have one less m two send it does sketches given system on the number line. You want you going to do 10 kg? You said three Yeah, until you go to three k g He said minus one Let us a place given value in the in this form Gloves and Ralph Moss ex money could do then in country unless, you know minus one Do it anyway then. Less thing The golden 30 minus treaty with waiter Dean So the center of Mosses its body could do the sound You&#x27;re better, Dean Approximately I knew approximately 20. Sound your weight. 13 decoder 2.8 Study on this On your bettered Any goto who 0.8 So the central mosque is located at this hour You&#x27;re a waiter

Samuel Hannah · Answer

So this problem We&#x27;ve been given this to muss free, spink free spring coupled equation of drawn up on the screen here. We&#x27;ve been asked to try and determine what the differential equations governing the this system. So hey, we&#x27;re just written out. So when this is an equilibrium position, so we have that the extensions X and Y or equal to zero. But let&#x27;s suppose that we move this spring. This would so X is non zero. The change in X is simply Delta Rex, so it&#x27;s simply the extension of this left hand spring. It&#x27;s about still to Rex, but now the length of the middle spring has automatically got smaller. So the change if this spring the middle spring was go to the right hand side. It&#x27;s not just depended on the position of why it&#x27;s also dependent on position of X. So the change in why is Belt X is Delta y minus Delta Rex? Because if the position of the left hand spring moves more than why the Springs crossover and so for math M two was actually moved to the left. So here we see that, but seven to spurt the change of the 1st 2 springs Now simply for the last spring. The spring K free. This is only dependent on the last Mass, and this changes with length minus y, and that&#x27;s important to now. We know that for the first spring it&#x27;s dependent only on the position of X Middle. Spring kits, dependent only on the visit, is dependent on both positions. Andi For the last spring, it&#x27;s again dependent only on the Y position. So now we can use hooks along and you&#x27;ve been second Lol to get some differential equations. So we know that from Newton&#x27;s second law, let&#x27;s look at the first mass that m one times the acceleration so he sled thanks. He t squared is equal to the sum of forces. F one is going in the opposite direction to what we&#x27;re traveling. That&#x27;s minus F one, and then we get plus F two. Now F one is the force of the first spring Onda. We know that that is dependent only on the X extension. So using hooks, little F one is equal to minus K one thanks and so as to again just hook. So we know that spending on both X and Y and that&#x27;s positive. So that gives is plus K two Holmes. Why minus X, which is what we found here. And this is absolutely differential equation governing Be first mess. I&#x27;m looking at this. We have to say it&#x27;s actually exactly the same as to walk with come across in the textbook when the third spring or when there isn&#x27;t assert spring. But that&#x27;s easy to see because if we just restrict ourselves so looking, get Spring one, Mass one and spreading to. That&#x27;s exactly the same system that we have, like even when we don&#x27;t have the last spring, so back to the surprises. But now let&#x27;s look at how the second mass has changed. So the new and 2nd 2nd law again, we have that and to d squared. Why DT Sward is equal to the sum of the forces of this time. After is in the opposite direction. So it&#x27;s minus F two plus F free. Using Hook&#x27;s Law, we find that F two it&#x27;s simply minus the F two we found before, so that&#x27;s minus K two times Y minus sex. After Jeffrey, this is simply plus the third spring coefficient K free. Now it&#x27;s times the extension of the spring. That extension is dependent on minus wine that goes to minus wife and it&#x27;s retire. Leave yourself a bit. This is simply minus K two. Why mind sex minus K free. Why on this is different to what we have before, because we now have to take into account how the third spring resists the, um, the movement of the mast of right hand side. But overall, that is the differential those off the differential equations governing this system.

Yaqub Khan · Answer

the gravitational potential energy between the two masses is minus capital de on one I am too divided by are there&#x27;s r is the distance between that too. Bodies So we can write. This is minus G and one I&#x27;m toe divided by is our is equal to extra occur for less. Why soccer? How to the power off now the X component off force exerted by and one on em too physical toe minus. Why should revert you off potential energy We respect to X that is equal toe minus partial divided by partial X and the potential energies minus G and one I am too divided by extra er less Why soccer hall to the power Ah, this can be written as X com brand of force equal to Capital G and one I am too in tow. Partial upon partial X ex occur Bless why soccer hold to the power minus half. So this gives us capital G and one. I am too minus half in tow. Ex occur plus why sugar minus half minus one that is minus three, divided by two and the video off Prescribe use two x and driven to a wife Securities zero. So this gives us X component off force equal to minus D and one I am too divided by I am too X divided by extra care. Plus why sugar for to the power she upon to Similarly, we can calculate why com print off force exerted by a one on em too as y component of force equal toe minus positivity off potential energy. We respect to why that is equal toe minus partial upon partial y in tow minus G and one I am too divided by extra Kurtulus. Why so girl, hold to the power off. So this gives us why Complaint of force equal toe minus G and one I am too. Why divided by ex girl less Why should you care? Hold the power trade right by two we can is the calculate magnitude of the force exerted by and one on em too. So magnitude of force he called toe so carried off So care of X competent for us So get off Why complaint s x com print off the forces minus gm one I am Toe X divided by extra care plus y soccer hold the power half And why complained off the forces minus G on one AM toe Why? Divided by extra Kurtulus vice sugar off So taking g m one m two common from both Trump&#x27;s. So this gives us my computer force equal to G and one I am too in tow Scheduled off extra ca divided by ex occur Nice! Why secure? Sorry, I diddle little mistake here Here we have extra careful advice occur Hold the power three right by two Here again we have extra terrible advice You can hold the power she upon to So we have here Execute divided by X squared Plus why soccer or to the power She less Why should care? Do I did buy X squared plus y soccer all to the power three in the next time we have a magnitude of four sequel Toe gravitational constant into And one I am too in tow by taking extra care Plus why should get hold of the poetry Elsom, we have ex occur Plus why is a girl in numerator s extra care? Plus why soccer is equal to Are you okay? So by putting the value here Capital de I am one AM toe in tow Our SCA, divided by our CICA to the power tree. This gives us capital G and one I am toe in tow. One divided by r to the power for this gives us magnitude a force equal toe the Putin g m one m two divided by So get off our So this is the many Utah force exerted by and one on I am too. So here this is I am too This is. And what so x component off force exacted by and one on em too is negative. This is effects and why component off force exerted by and one on him too is also negative like this. So we have there is a 10 force. So we have the net force. That is the magnitude of force exerted by one on em too will be like this. So here the directional force exerted by and one on and two is towards and one So this gives us and one and one exerts and attractive force on em too.

Problem

Let $\mathbf{v}$ be the center of mass of a syste…

Problem 29 Easy Difficulty

Answer

$( 1.3,0.9,0 )$

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

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Absolute Value - Example 1

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$( 1.3,0.9,0 )$

Linear Algebra and Its Applications

Grace H.

Catherine R.

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Absolute Value - Example 1

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David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Grace H.

Catherine R.

Caleb E.

Michael J.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications