Any help on a static equilibrium question

A uniform ladder 10.0 m long is leaning against a
frictionless wall at an angle of 60.0° above the horizontal.
The weight of the ladder is 20.0 lb. A 61.0-lb boy climbs 4.00 m
up the ladder. What is the magnitude of the frictional force
exerted on the ladder by the floor?

Draw the Free Body Diagram

What we know:

The wall is frictionless ====> Wall can only exert a horizontal force

Thus: The only other source of horizontal force is the friction from the ground

Therefore: Since the sum of the horizontal forces = 0 for equilibrium, Friction = Wall horizontal reaction

We get the Wall Reaction by summing Moments about the bottom of the ladder. We need the effective arm lengths for each of the forces creating a moment

CG of ladder:

cos 60 = x/5 m
0.5 = x/5
x=2.5 m

M = L*F = 2.5 m * 20 lbs = 50 lbs * m

Boy:

cos 60 = x/4
0.5 = x/4
x = 2

M = 2 m * 61 lbs = 122 lbs*m

Those 2 moments must be balanced by the horizontal force of the wall

Wall

sin 60 = y/10
0.866 = y/10m
y = 8.66 m

Moment

(50 + 122) lbs*m = 8.66 m * F
8.66 m * F = 172 lbs*m
F = 19.86 lbs = Friction

Draw a clear diagram with forces marked

Find the normal horizontal reaction, R, on the top of the ladder by taking moments (torques) about the base of the ladder. From the diagram you should see:

R acts a vertical distance 10sin(60.0° ) from the base of the ladder.
Ladder's weight acts a horizontal distance (10.0/2)cos(60.0° ) from
Boy's weight acts a horizontal distance 4.00cos(60.0° ) from the base of the ladder.

Applying the principle of moments
R x 10sin(60.0° ) = 20.0(10.0/2)cos(60.0° ) + 61.0(4.00)cos(60.0°)

Solve for R (units of lbs, 3 significant figures).

The frictional force = R (because R and frictional force are the only 2 horizontal forces and the ladder is in equilibrium.