Example 1 - Discrete Masses

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Suppose the there room three allude masses i ordered it as presented in the number at right. Where is the center of fixed of this 3-object system?

Then:

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So, the center of massive of the system is in ~ the allude (2.0 m, 1.7 m).

You are watching: Center of mass of a rectangle

Note: the would have actually been quicker and also easier to notice that the masses in the diagram in ~ left room symmetric around x = 2 m, so the x-coordinate that the center of mass needs to be 2.0 m.

Note: Given this setup of masses, girlfriend would alleviate the amount of calculation by put the origin of the coordinate device at the location of among the points. Ns didn"t carry out that for reasons you will quickly see.

Example 2 - A constant Object

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Suppose the a 2 m through 3 m rectangle is cut from a square item of plywood i beg your pardon originally had actually sides of length 4 m (as shown at right). What is the center of massive of the resulting "U-shaped" piece of plywood?

Well, the center of mass of the the a homogeneous rectangle is in the geometric center of the rectangle (by symmetry). Why no divide the U-shaped piece right into three rectangles as displayed in the diagram at appropriate - 2 2m x 3 m rectangles and one 4 m x 1 m rectangle. (This is not the only way to division the U-shape right into rectangles, and any of the other ways will work just as well.)

The centers of massive of the three rectangles are indicated in the figure at right. (Does it look familiar?) We deserve to now change the rectangular sheets by your centers of mass, and using the area the the rectangle together a "stand-in" because that its mass, girlfriend get:

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Note: to compare this to the example above.

Moral: If you can "chop up" a continuous object into pieces whose centers of fixed are basic to find, you have the right to reduce the trouble to finding the facility of fixed of a system of discrete points.

Example 3 - An alternative Method

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Perhaps chopping perfectly-good objects into tiny pieces does not appeal to you. Here"s an alternative, based upon the idea the the center of mass of an object is in the same ar no matter exactly how you calculate it.

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The facility of massive of the originall 4 m x 4 m item of plywood is in ~ its geometric facility (middle dot at right), therefore the y-coordinate that the facility of fixed of the original square is 2 m.

On the various other hand, you can find the center of massive of the initial square by detect the facility of massive of the 2-point mechanism consisting that the centers of mass of the U-shape and the facility of mass of the plywood that initially fit in the "hole."

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where Y is the y-coordinate of the initial square, au is the area of the U-shape, yu is the y-coordinate of the facility of fixed of the U-shape, ah = area that the "hole", yh is the y-coordinate that the plywood that initially filled the hole, and also A is the area that the original square (which amounts to au + ah). A little algebra gives: