How to find Torque without Coefficient of Friction?

[Tyler Riley]

Hello! I am an AP Physics student and had a quick question for validity. I was prompted with this question in class today:
Example 7: An 8-meter ladder of weight 355 N leans at an angle of 50° to the floor against a smooth vertical wall. Assume the center of weight (gravity) for the ladder is at its linear center. Find the force exerted by the wall and the floor upon the ladder and the friction between the floor and the ladder.

I wasn't really paying attention and missed a class day of notes so I really don't understand anything really and was hoping you could answer my few questions.
Here is what I copied:

f = floor
r = lever arm
F_f = Friction force
N_f = Normal force or Floor
W_⊥ = Weight
N_w = Normal force of Wall

r_L = (4m)cos(50) = 2.57m
r_W = (8m)sin(50) = 6.13m

∑F_y = 0
N_f - W_⊥ = 0
N_f = W_⊥
355 N = 355 N

∑F_x = 0
F_f - N_w = 0

∑τ = 0
-W_⊥ * r_L - N_w * r_W = 0
(-355 N) * (2.57) - N_w * (6.13) = 0
-912.35 = N_w * (6.13)
N_w = -148.833

Here are my questions:

Why isn't it (4m)sin(50) instead of (8m)sin(50)?

And why does she plug in
-W_⊥
in
-W_⊥ * r_L - N_w * r_W = 0
instead of
N_f * μ
in
N_f * μ * r_L - N_w * r_W = 0

 

[haruspex]

Why isn't it (4m)sin(50) instead of (8m)sin(50)?

I assume you have a diagram.
When taking moments, you have to choose an axis. Any axis will do, usually, but it helps to pick one which several forces act through. That eliminates them from the torque balance equation since they have no torque about that axis.
The teacher has chosen the point of contact with the floor as the axis. The only forces with moment about that point are the weight of the ladder and the normal force from the wall. What is the lever arm of the force from the wall about the point of contact with the floor?

And why does she plug in
-W_⊥
in
-W_⊥ * r_L - N_w * r_W = 0
instead of
N_f * μ
in
N_f * μ * r_L - N_w * r_W = 0

Same reason.