Answer:
the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping is 3.3424 m
Explanation:
Given the data in the question and as illustrated in the images below;
without the ladder immediately slipping, the net torque and the net force mus all balance out.
from the first image;
In the x, the force is;
[tex]F_{f}[/tex] = N₂
mg = N₁
the torque about the ground contact point gives the following equation
N₂Lsin∅ = mgcos∅[tex]\frac{L}{2}[/tex]
solving for ∅
tan∅ = mg / 2N₂ 55
∅ = tan⁻¹ ( mg / 2N₂ )
we already know that N₂ = [tex]F_{f}[/tex] = μN₁ = μmg
so,
∅ = tan⁻¹ ( mg / 2μmg )
∅ = tan⁻¹ ( 1 / 2μ )
given that; The coefficient of static friction between the level ground and the foot of the ladder μ = 0.35
we substitute
∅ = tan⁻¹ ( 1 / (2×0.35 ) )
∅ = tan⁻¹ ( 1.42857 )
∅ = 55°
now to get the required distance;
from the second image; cos∅ = d / L
d = Lcos∅
given that; length of the ladder = 5.2 m
we substitute
d = 5.2cos(50)
d = 3.3424 m
Therefore, the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping is 3.3424 m
