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So below is the problem and I want to know how to find the location of the center of mass. The book I have is not very clear on the fact of 3-D objects.

Figure 9-27 shows a cubical box that has been constructed from metal plate of uniform density and negligible thickness. The box is open at the top and has edge length 40 cm. Find the coordinates of the center of mass of the box.
______cm (x-coordinate)
______cm (y-coordinate)
______cm (z-coordinate)

Link to image: http://www.webassign.net/hrw/09_33.gif

2007-10-24 17:20:04 · 3 answers · asked by eggyu74 3 in Science & Mathematics Physics

3 answers

advance disclaimer I am not a physics student or teacher, but I do think logically most of the time.

Think of the problem in two steps. Where is the center of mass for a box with no top or bottom and how does it change when you add the bottom?

four sided box would be x=20cm y=20cm z=20cm

the bottom would have the same mass as one of the sides or 1/5 of the total mass. The center of mass should move down by 1/5 or 4 cm

I'm guessing here but I would say x=20cm y=20cm z=16cm

2007-10-24 17:32:10 · answer #1 · answered by don_sv_az 7 · 5 0

The center of mass of each square side is at its center.
Considering the four sides of the box (leaving top and bottom],
the center of mass is at [x/2, y/2, z/2] and the mass = 4m
where m is the mass of each square.

The center of mass of the bottom side is at its center.
[X/2, y/2, 0] and its mass is m.

Now we are having two mass points alone in the vertical line through the center of the base square.

One mass point [= m] is at the center of the base and another mass point [= 4m] is at the distance z/2 = 20 cm from the base.

Let us denote the center of the mass of this system is at a distance r from the base.

Taking moment about the defined center of mass

m r = 4m [20 -r]
r = 80 - 4r
5r = 80
r = 16
Thus the coordinates of the center of mass of the whole box is
[20, 20, 16]

2007-10-24 17:47:05 · answer #2 · answered by Pearlsawme 7 · 0 3

break it down into simpler problem by treating each side of the box as a point mass located at the center of each side.

Then if you break the problem down from the 3D problem into 3 2D problems, you'll be able to combine point masses that are in the 3rd dimention.

As an example, when you are looking at the box from he "front" of the box, you will have 4 points in 2D space, one for the "bottom" of the box, one for the "left" side of the box, one for the "right" side of the box, and one that combines the point masses that make up the "front" and "back" of the box.

2007-10-24 17:42:53 · answer #3 · answered by HooKooDooKu 6 · 0 1

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