Answer:
Fₓ = 123.8 N and F_{y} = 0
Explanation:
This is a static balance exercise, to see the forces, see the attachment. In this W is the weight of the ladder and W1 the weight of the man.
locate a reference system at the bottom of the ladder with the horizontal x axis, it will assume that the counterclockwise turns are positive
Σ τ = 0
x + Fₓ y - W₁ d₁ -W d₂ + N 0 + fr 0 = 0
use trigonometry to find the distances
cos θ = x / L
θ = cos⁻¹ (x / L)
θ = cos⁻¹ (2 / 6.82)
θ = 72.9º
in 72.9 = y / L
y = L sin 72.9
y = 6.82 sin 72.9
y = 6.519 m
Center of mass ladder
cos 72.9 = d₂ / 2
d₂ = 2 cos 72.9 = 0.588 m
x coordinate of man
cos 72.9 = d₁ / 3
d₁ = 3 cos 72.9 = 0.882 m
we substitute
2 + Fₓ 6.519 - W₁ 0.882 -W 0.588 = 0
as they indicate that the plastic basket has no friction
F_{y} = 0
Fₓ 6.519 - W₁ 0.882 -W 0.588 = 0
Fₓ 6.519 = M₁ g 0.882 + m g 0.588
we calculate
Fₓ = 85 9.8 0.882 + 12.6 9.8 0.588) / 6.519
Fₓ = 123.8 N
