x (max) = 0,25*L*tanα
Explanation:
Letá call
A: point at the top of the ladder
B: the point at the bottom of the ladder
C: point where the man is up the ladder
L the length of the ladder
α angle between ladder and ground
"x" distance between C and B
Description
The following forces act on the ladder
Point A: horizontal (to the right) reaction (T) of the wall against the ladder
Point B : Vertical (upwards) reaction (R) of ground against the ladder
frictional horizontal ( to the left ) force (F)
Point C : Weight (vertical downwards)) of the man mg
mg = 75 Kg * 9,8 m/s² mg = 735 [N]
Then in static conditions:
∑Fx = T - F = 0 ⇒ T = F
∑Fy = mg - R = 0 ⇒ 735 - R = 0 ⇒ R = 735
∑Torques(b) = 0
Note: In point B the torque produced by forces R and F are equal to 0
Then:
∑Torques(b) = 0
And the arm lever for each force is:
mg = 735
d₁ for mg and d₂ for T
cos α = d₁/x then d₁ = x*cosα
sin α = d₂ / L then d₂ = L*sinα
Then:
∑Torques(b) = 0 ⇒ 735*x*cosα - T*L*sinα = 0
735*x*cosα = T*L*sinα
T = F then 735*x*cosα = F*L*sinα
F = (735)*x*cosα/L*sinα cos α / sinα = cotgα = 1/tanα
F = (735)*x*cotanα/L or F = (735)*x/L*tanα
When F < μ* R the ladder will stars slippering over the ground
μ(s) = 0,25 0,25*R = 735*x/L*tanα
x = 0,25*R*tanα*L/735
But R = mg = 735 then
0,25*L*tanα = x
Then x (max) = 0,25*L*tanα