Answer:
108.65 N
Step-by-step explanation:
given,
mass of road = 25 Kg
static friction at A = 0.17
contact force at B
there will a normal force acting at point A i.e Na and a friction force in the vertical direction at Point A i.e. μNa
A horizontal force is acting along point B i.e N_b.
At the middle weight of of the rod will be acting i.e mg at 2 m
diagram attached below shows the forces,
now, taking moment about point B
computing horizontal forces



Na= 108.65 N
now,
computing horizontal forces
P = N_a
P = 108.65 N
hence, P is the contact force at point B