A uniform cube of edge length $l$ sits on a horizontal table. The four horizontal edges in contact with the table are cut into at $45^0$.
Find the position of the center of gravity of the new solid when the length (and height) $x$ has been sliced away.
Also, show that the eroded block falls when $x=\frac{5}{9}l$ (approx).
Here we need the centroid of the frustrum, that is created, but it is upside down. The equation for the centroid of the frustrum of a square based pyramid is:
$$y = \frac{h(A_1 + 2√A_1A_2 +3A_2)}{4(A_1 + √A_1A_2 + A_2)}$$
where $A_1$ is the bottom base area and $A_2$ is the top base area (which will also be where the cuboid surmounts the frustrum), if we model this as a frustrum surmounted by a cuboid. We can take $O$ to be at the base of the cuboid at $\frac{l}{2}$.
The mass of the frustrum is $\frac{1}{3}(A_1 + √A_1A_2 + A_2)$.
Would you use these equations or would you use alternative ones?
You could not just say that the center of gravity for the frustrum is $\frac{h}{4}$, as this would be the center of gravity for a 'complete' pyramid, not a truncated one. Also, the fact that the frustrum is 'upside down' i.e. the smaller edge on the horizontal table, could change the center of mass compared to one the right way up.
An answer:
After cutting off the four corners at a height $x$ above the horizontal but below $\frac{l}{2}$ we have:
Taking $O$ to be at the apex (base) of the truncated pyramid (with the cuboid on top of it), after cutting,
Centroid of frustrum, with $A1=l$ and $A2=(l-2x)$ is
$$\frac{15xl^2 - 42 lx^2 + 36x^3}{12l^2 - 24lx + 16x^2}$$
Volume of frustrum is
$$xl^2 - 2x^2l + \frac{4}{3}x^3$$
Centroid of cuboid is
$$x+\frac{(l-x)}{2}$$
Volume of cuboid is
$$l^3 - l^2x$$
Total volume of solid is
$$l^3-2x^2l+\frac{4}{3}x^3$$
So the centroid of the whole (truncated) solid is:
$$y= \frac{(\frac{(15xl^2 - 42 lx^2 + 36x^3)}{12l^2 - 24lx + 16x^2}*(xl^2 - 2x^2l + \frac{4}{3}x^3)) + ((x+\frac{(l-x)}{2})*(l^3 - l^2x))}{l^3-2x^2l+\frac{4}{3}x^3}$$ so $$y=\frac{6l^4+9x^2l^2-42x^3l+36x^4}{12l^3-24x^2l+16x^3}$$
For $x=0$ this gives $y=\frac{l}{2}$ as we would expect and for $x=\frac{l}{2}$ we get $\frac{21l}{32}$
Now, when we arrive at $x=\frac{l}{2}$, we have cut at the halfway point from the horizontal to the top of the original cube (which was at a height $l$).
So at this halfway point we have a square based pyramid of height $x=\frac{l}{2}$ and a cuboid of height $x=\frac{l}{2}$.
So we can work out the centroid here as well and compare it to the result we just got for the solid before we reached the halfway point at $x=\frac{l}{2}$.
Again taking the point $O$ to be on the horizontal:
Centroid of pyramid,
$$\frac{3h}{4}=\frac{3l}{8}$$ (as it is 'upside down')
Volume of pyramid is
$$\frac{l^3}{6}$$
Centroid of cuboid is
$$\frac{3l}{4}$$
Volume of cuboid is
$$\frac{l^3}{2}$$
Total volume of solid is
$$\frac{2l^3}{3}$$
So the centroid of the whole solid is:
$$y=\frac{\frac{3l}{8}\frac{l^3}{6} + \frac{3l}{4}\frac{l^3}{2}}{\frac{2l^3}{3}}$$ So $$y=\frac{21l}{32}$$ which agrees with our previous result.
Now for the part of the question which asks us to 'show that the eroded block falls when $x=\frac{5}{9}l$ (approx).'
If we continue to cut the solid above $x=\frac{l}{2}$ i.e. $x>\frac{l}{2}$ we still have a square based pyramid of height $\frac{l}{2}$ but we have a cuboid of dimensions $l*l*(l-x)$ so we can now find the centroid for this 'new' solid as we have just done for $x<\frac{l}{2}$
Again taking the point $O$ to be on the horizontal:
Centroid of pyramid,
$$\frac{3h}{4}=\frac{3l}{8}$$ (as it is 'upside down')
Volume of pyramid is
$$\frac{l^3}{6}$$
Centroid of cuboid is
$$\frac{3l-2x}{4}$$
Volume of cuboid is
$${l^3-xl^2}$$
Total volume of solid is
$$\frac{7l^3}{6}-l^2x$$
So the centroid of the whole solid is:
$$y=\frac{(\frac{3l}{8}\frac{l^3}{6}) + \frac{(3l-2x)}{4}*(l^3-l^2x)}{\frac{7l^3-6l^2x}{6}}$$
So, for $x=\frac{l}{2}$, we have
$$y=\frac{15l}{32}$$
which is approx
$$\frac{4l}{9}$$
If we subtract this result from $l$ we get $$\frac{5l}{9}$$
Can we show that the solid will topple at this point?
