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The masses $ m_i $ are located at the points $ P_i $. Find the moments $ M_x $ and $ M_y $ and the center of mass of the system.

$ m_1 = 5 $ , $ m_2 = 4 $ , $ m_3 = 3 $ , $ m_4 = 6 $ ;$ P_1 (-4, 2) $ , $ P_2 (0, 5) $ , $ P_3 (3, 2) $ , $ P_4 (1, -2) $

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Calculus 2 / BC

Chapter 8

Further Applications of Integration

Section 3

Applications to Physics and Engineering

Applications of Integration

Campbell University

University of Michigan - Ann Arbor

Idaho State University

Lectures

01:03

The masses $m_{i}$ are loc…

01:39

The masses $ m_i $ are loc…

01:30

04:33

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03:44

$23-24$ The masses $m_{i}$…

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01:02

Find the moments $M_{x}$ a…

In Problems $11-14$, find …

Find the center of mass of…

in this problem were given four messes and accordance that those messages are located. Whereas to find X moment by moment as well as the central myself to system, we know that my moment and wise Eagle to submission Off Mass multiplied by exposition, cited as five times in a four plus four times zero plus three times three plus six times, one that is equal to negative five. And we also know that ex moment is a call to summation off mass multiplied by y position. So that is five times two plus four times five plus three times two plus six times negative too. And that is equal to 24. Hear those points are these points. So those are the X coordinates the first ones and second was R Y cordons and the Notre exports is equal to em. Why divide by summation of mess that is negative. Five be wanted by five plus four plus three plus six that is negative. Five divided 18. And why bar is MM X divided by summation off mass on a total mess. And Max here in the first part of you founded as 24. You know what a total mess is 18 and we can simply find us for six, and that is equal to for I three sort of central mess off the system will be a negative five over 18 and four over three.

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