For translational equilibrium,

T + f = mg sin60$$^\circ$$ ..... (1)

For rotational equilibrium,

Torque about center of mass should be zero.

$$ \therefore $$ $$\tau $$_com = 0

$$ \Rightarrow $$ f $$\times$$ R $$-$$ T $$\times$$ R = 0

$$ \Rightarrow $$ f = R ..... (2)

Putting value of (2) in (1),

f + f = mg sin60$$^\circ$$

$$ \Rightarrow $$ f = $${{mg\sin 60^\circ } \over 2} = {{\sqrt 3 mg} \over 4}$$ = 0.433 mg

$$ \therefore $$ cylinder need minimum 0.433 mg friction force to stay in equilibrium.

But here maximum friction force possible,

f_max = $$\mu$$_sN

= $$\mu$$_smg cos60$$^\circ$$

= 0.4 mg $$\times$$ $${1 \over 2}$$

= 0.2 mg

As maximum possible friction force is less than 0.433 mg. so cylinder will not stay in equilibrium. It will slide on the surface. So friction will be kinetic friction.

f_k = $$\mu$$_k N

=$$\mu$$_k mg cos60$$^\circ$$

= 0.4 mg $$\times$$ $${1 \over 2}$$

= $${mg \over 5}$$