**Introduction**

I've included a diagram below.

You will note that, as CaptainBlack correctly said, for equilibrium the weight of the object must be directly over the point of contact.

In the diagram, you can see that

is the point that would be the centre of the flat face of the hemisphere,

is the axis of rotational symmetry of the original hemisphere,

is the point of contact with the inclined plane and

is the location of the centre of mass of the hemispherical object.

**Geometry** is a vertical line and since

is at right angles to the inclined plane we must have that the angle

is

. Also the angle that the flat face of the hemisphere makes with the horizontal is the same as the angle between

and the vertical i.e. we need to find the angle

. Let this angle,

, be

.

Using the sine rule for the triangle

, we find that

since

. Since we know

is acute we can rearrange this so

.

So all we need to do is determine

.

**Dimensions of the cube**
The greatest cube that can be cut out of the hemisphere has its centre on

and so if we imagine expanding a small cube centred directly on top of

we will no longer be able to expand it when its height is such that its four other corners touch the curved surface of the hemisphere. This cube then has one face in the same plane as the flat surface of the hemisphere and the opposite face's corners touching the curved suface. At any one of these corners that meet the curved surface the distance from

must be

so using Pythagoras Theorem we must have

where

is the side length of the cube and

is the distance from

to the corresponding corner of the cube that lies in the plane of the flat surface of the hemisphere.

Again using Pythagoras we must have (for a cube) that

,

hence

,

.

**Centre of mass calculation**
I'm going to assume that you know the result that the centre of mass of a uniform hemispherical solid is on the rotational axis of symmetry at

from

.

Let

be the density of the solid,

the mass and

be the distance of the centre of mass from

on

where the solid referred to is denoted by the subscripts

,

and

for hemisphere (uncut), cube and hemisphere with cube cut out respectively.

Then we must have (in terms of moments about

) that

and thus we can determine

since

,

,

,

,

.

So we have

,

with

.

**Putting it together**

Substituting our value of

into our trigonometric result gives us that

.