Find the direction and the magnitude of Frictional force

[chwala]

Homework Statement

See attached Question 5.

Relevant Equations

Mechanics

In my working i have, the attached. My question is Why is the Reaction force not being considered here,

My equations are;

Frictional force = $30\cos 50^0 - 20\cos 40^0 = (19.28 - 15.32)N = 3.96N$
The direction will be downwards.

 

[Lnewqban]

The reaction force is perpendicular to the direction of freedom of movement of the block.

 

[chwala]

The reaction force is perpendicular to the direction of freedom of movement of the block.

Ok a bit confusing...then if we were to talk of Limiting friction that's when the Reaction force comes into play...
like in this example;

aaaargh i think okay! It's only the parallel forces being resolved in both scenarios.

 

[Lnewqban]

The key phrase here is "the block... which is at rest".
Meaning all forces acting along the direction of freedom of movement cancel each other.
Not much to do with the normal forces, because the value of the coefficient of friction is not given.