How Much Friction is Needed to Keep the Ladder in Place?

[Kamisama]

Homework Statement

A 5.00-m long uniform ladder leans against a smooth wall and its base rests on a rough floor. The ladder has a mass of 18.0 kg and its base is a distance of 2.30 m from the wall. A person of mass 70.0 kg climbs 2.50 m up the ladder. If the ladder is to remain in place, what frictional force must be exerted by the floor on the ladder?

Homework Equations

t = FR
t =0
N=mgcosθ
ΣF=0
Σt = 0
-m_Lgb_g-m_Pgb_P+F_n,w*a = 0
F_n,w= (m_Lgb_g + m_Pgb_P ) / a

ΣFx = 0
F_s - F_n,w = 0
Fs = F_n,w = (m_Lgb_g + m_Pgb_P ) / a

The Attempt at a Solution

wall (a)= sqrt(30.29)= 5.50m, b_L=2.30m, b_P=cos(67.3)*c_P
c= 5.00m, c_P=2.50m
θ_L= tan-1(5.50/2.30)= 67.3°

Fs = ((18.0)(9.81)(2.30)+(70.0)(9.81)(.965)) / 2.5 = 194 N

I'm not completely sure if I found all forces for this problem, did I find everything for the frictional force?

 

[Let'sthink]

angle of the ladder with the ground = arc cos(2.3/5) = 62.61°. I do not know which angle you find out to be 30.5 and how? You have to label all the possible forces acting on the ladder and take moments about a convenient point. I think it is center of mass of the ladder. You think why and what is the convenience. modify your reply accordingly then I would comment.

 

[haruspex]

b_P=cos(67.3)*c_P

It is rarely necessary to calculate the actual angles from the sides of the right-angled triangle. Just work in terms of the trig functions of the angles.