Center of Mass for a cubical box with a missing top

[jdawg]

Homework Statement

A cubical box has been constructed from uniform metal plate of negligible thickness. The box is open at the top and has edge length L=40cm. Find (a) the x coordinate (b) the y coordinate, and (c) the z coordinate of the center of mass of the box.

Ok, so my professor already worked this problem for us, but I have no clue what she did.

Homework Equations

The Attempt at a Solution

L=40 cm=0.4 m

Then she divided that by two for some reason?
Front and Back center of mass: (0.2, 0.2, 0.2)
Right and Left center of mass: (0.2, 0.2, 0.2)
Top and Bottom (top is missing): (0.2 ,0.2 , 0)

x_cm=[itex]\frac{1}{5m}[/itex](2m(0.2)+2m(0.2)+m(0.2))=0.2m
y_cm=[itex]\frac{1}{5m}[/itex](2m(0.2)+2m(0.2)+m(0.2))=0.2m
z_cm=[itex]\frac{1}{5m}[/itex](2m(0.2)+2m(0.2)+m(0))=0.16m

I am so confused! Can someone please explain to me what she did? Is there a formula that she used? I looked in my book and the only one I found needed the mass to use it.

 

[jackarms]

That method involves mass moments -- so you find the x, y, and z components for the center of mass first. So for example the x moment you do:
$$\Sigma x_{i}r_{x,i} / M$$

You don't need the mass since it ends up cancelling out -- you can still use it if you call the mass of each side ##m##, so then the total mass is ##5m##. That looks like the ##\frac{1}{5m}## in what you put down.

Also, you can reduce each side to their own center of masses by saying the center of mass is at the center, and then using that to find the center of mass of all of the sides.

 

[jdawg]

Ok, thanks so much! Great explanation :)