Answer:
108.65 N
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
[tex]N_a \times 4 + \mu N_a \times 3 = m g \times 2[/tex]
[tex]N_a \times 4 + 0.17 N_a \times 3 = 25\times 9.8 \times 2[/tex]
[tex]4.51N_a= 490[/tex]
Na= 108.65 N
now,
computing horizontal forces
P = N_a
P = 108.65 N
hence, P is the contact force at point B
