Reaction forces 
If a circular lamina (in equilibrium) is supported from two points that are of equal distance from the line joining the centre and the lowest point of the circle (so the arrangement is symmetrical about a vertical line), then the reaction forces at the two supporting points are equal (and their sum is the weight of the lamina). But what if the supporting points are not symmetrical about this line? Is the ratio of the magnitude of reaction forces the same as the ratio of the cosine of the angles made by the line connecting the point of support to the centre of the circle and the vertical.
 A: Just draw the free body diagram yourself and work out the equations for force balance at the center.

Notice that the sum of the horizontal components must be zero (since there is not applied horizontal force) and that sets up the ratio of forces on the two supports.
A: I assume that there is no friction. Then one finds the horizontal equilibrium condition 
$$
N_1\sin\alpha-N_2\sin\beta=0,
$$
and thus clearly
$$
\frac{N_1}{N_2}=\frac{\sin\beta}{\sin\alpha},
$$
so it is the ratio of sines that you need. This survives a sanity check -- if $\beta\ll \alpha$, then almost all the weight is going to be on the second support, and we will have $N_1\ll N_2$, which is what follows from the relation above.
One can find the reaction forces explicitly using the vertical equilibrium
$$
N_1\cos \alpha+N_2\cos\beta = mg.
$$
Multplying by $\sin\beta$ and adding a $\cos\beta$ multiple of the first equation we find 
$$
N_1(\sin\alpha\cos\beta+\cos\alpha\sin\beta)=mg\sin\beta,
$$
giving
$$
N_1=\frac{\sin\beta}{\sin(\alpha+\beta)}mg,
$$
and analogously
$$
N_2=\frac{\sin\alpha}{\sin(\alpha+\beta)}mg.
$$
Interestingly, this problem is way more subtle with friction. Friction adds two unknown variables -- friction force at each the supports. However, we have not yet used only one equation -- the zero net torque relative to the center of the cylinder. It then follows that one cannot actually determine all the forces if friction is allowed. This has the following explanation: imagine the cylinder is made of very firm rubber; you would then be able to "push" it in between the supports a little bit, dramatically increasing the friction and reaction forces. Therefore, the solution in presence of friction depends on the precise way the configuration was reached.
