Problem

$\begin{array}{|c|c|c|c|}\hline m_{i} & {8} & {1}…

02:19

Problem 11 Easy Difficulty

Center of Mass of a Two-Dimensional System In Exercises $11-14,$ find the center of mass of the given system of point masses.
$\begin{array}{|c|c|c|c|}\hline m_{i} & {5} & {1} & {3} \\ \hline\left(x_{i}, y_{i}\right) & {(2,2)} & {(-3,1)} & {(1,-4)} \\ \hline\end{array}$

Answer

$\left(\frac{10}{9},-\frac{1}{9}\right)$

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus of a Single Variable

Chapter 7

Applications of Integration

Section 6

Moments, Centers of Mass, and Centroids

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Video Transcript

okay. What we want to dio is we want to find the center of mass of the given system of point masses. And I'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're just gonna add one. We'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's value, because we'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'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's a little over one and down here at negative one night. So it's about right. There's where that center of masses

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