Stacking cylinders

Let W be the weight of one of the cylinders. The total weight is then 3W, and so the normal force of the table $F_n$ on a cylinder is 3W/2. Now let us consider the right hand cylinder. There are three points of contact on this cylinder, where the top cylinder touches, where the left cylinder touches, and where the table touches. The top most cylinder touches at an angle of 60 degrees with respect to the horizontal and exerts a force $F_{t,r}$ towards the center of the right hand cylinder and a force $F_{t,h}$ tangent to the right hand cylinder. The table similarly exerts a force $F_{p,r}$ towards the center and $F_{p,h}$ tangent to the cylinder. The left hand cylinder only exerts a force $F_{L,r}$ towards the center of the right hand cylinder by symmetry.

We first note that to balance torque on the right-hand cylinder $F_{t,h}=F_{p,h}=F_h$ . We can then apply a force balance in the x,y directions:

$W+F_h sin 30 + F_{t,r}sin 60=3W/2$ , $F_{L,r} + F_{t,r}sin 30 =F_h + F_h cos 30$ .

If $F_h$ is a minimum, that means $F_{L,r}$ is zero (no extra force needed), so we can solve for $F_h=W/(4+2\sqrt{3})$ and hence $F_h/F_n$ .