friction

(i) There are 4 forces acting on the block:
- the weight of the block, acting vertically downward;
- the normal reaction, acting perpendicularly away from theb plane surface'
- the applied force, acting horizontally;
- firctional force, acting downward along the plane.

(ii) Along the plane:
F.cos(30) = 10g.sin(30) + Ff --------------------- (1)
where F is the horizontally applied force, g is the acceleration due to gravity, and Ff is frictional force

Normal to the plane:
10g.cos(30) + F.sin(30) = R ------------------ (2)
where R is the normal reaction

(iii) When the block just starts to move upward, frictional force acting on the block equals to limiting friction. Hence,
Ff = 0.22.R
Using equation (1); P.cos(30) = 10g.sin(30) + 0.22R ---------------- (3)
Substitute R from (2) into (3):
P.cos(30) = 10g.sin(30) + 0.22.(10g.cos(30) + P.sin(30))
solve for P