Applied Maths - Mechanics 12

Consider the cone is on the verge of rotating about point A.
http://img207.imageshack.us/img207/5211/slope.jpg
Resolve forces parallel to F & N directions
F = Xcosθ – Wsinθ … (1)
N = Wcosθ + Xsinθ … (2)
Before taking moment about A, let's work out the distances of the forces.
The length of the side of the cone along the steepest slope = r/sinθ
Then c = (r/sinθ)cos2θ
d = rcosθ
Height of the cone = rcotθ
centre of gravity of a cone is 1/4 height from the base =rcotθ/4
b = rcotθ/4(sinθ) = rcosθ/4
Take moment about A,
Xc = W(b + d)
X(r/sinθ)cos2θ = W(rcosθ/4 + rcosθ)
X(r/sinθ)cos2θ = Wr5cosθ/4
X = 5Wcosθsinθ/(4cos2θ)
X = 5Wsin2θ/(8cos2θ) = (5W/8)tan2θ
Now (1) & (2) =>
F/N = (Xcosθ – Wsinθ) / (Wcosθ + Xsinθ)
= (X – Wtanθ) / (W + Xtanθ)
= [(5W/8)tan2θ – Wtanθ]/[W + (5W/8)tan2θtanθ]
= (5tan2θ – 8tanθ)/(8 + 5tanθtan2θ)
If this is greater than tank, then the friction is not sufficient to allow tilt without slipping.
Therefore the critical value for F/N = tank
(5tan2θ – 8tanθ)/(8 + 5tanθtan2θ) = tank
5tan2θ – 8tanθ = tank(8 + 5tanθtan2θ)
5tan2θ – 5tanθtan2θtank = 8tank + 8tanθ
5tan2θ(1 – tanθtank) = 8(tank + tanθ)
5tan2θ = 8 (tank + tanθ)/(1 – tanθtank) = 8tan(θ + k)