How Accurate Is the Center of Mass Calculation for an Open-Top Cubical Box?

[suspenc3]

a cubical box has edges of length 40cm with an open top.

Find the x, y, z coordinates of The Centre Of Mass

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in my head I cut the box along the x-axis to get this

(let m = the mass of one side of the box)

$$COMx=\frac{1}{M}\times(m1x1 + m2x2)$$

$$COMx=\frac{1}{5M}(2.5M\times 40cm)+(2.5M\times 0cm)$$
$$COMx=20cm$$

$$COMy=(2a\times 40cm)+(3M\times 0)$$
$$COMy=16cm$$

$$COMz=\frac{1}{5M}(2.5M\times 40cm)+(2.5M\times 0cm)$$
$$COMz=20cm$$

Therefore $$COM=(20cm,16cm,20cm)$$

 

[kleinwolf]

You could also decompose the problem : the sides are equivalent to a 4m mass on the axis of symmetry. Then the bottom makes lowering of the CM by : l/2/5=l/10.

Which in your case l=40, hence the lowering from the center is 4cm, and because the center is at y=20, you get CM(y)=20-4=16cm which is the same result.