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.

 

[quicknote]

where M is the total mass of the box.

This approach is not entirely accurate as the x, y, z coordinates will vary depending on how the box is cut. A more precise method would involve dividing the box into smaller sections and calculating the center of mass for each section, then finding the overall center of mass using the weighted average of these smaller centers. Also, it is important to note that the center of mass for a cubical box will always be at the exact center, so the coordinates would be (20cm, 20cm, 20cm) regardless of how it is cut.