To get up on the roof, a person (mass 85 kg) places a 6.82 m aluminum ladder (mass 12.6 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. At the top ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2.0 m from the bottom. The person is standing 3.0 m from the bottom of ladder. What are the magnitudes of the forces on the ladder at the top and at the bottom

 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