A ladder rests in limiting equilibrium against a rough vertical wall and with its foot on rough horizontal ground, the coefficient of friction at both points of contact being 0.5. The ladder is uniform and weighs 300N. Find the angle which the ladder makes with the horizontal ground. Find also the normal reactions at the wall and the ground.

θ = 36.9°

N = 240 N

R = 120 N

Explanation:

Draw a free body diagram.  There are five forces on the ladder:

Weight force mg pulling down at the center of the ladder,

Normal force N pushing up at the ground,

Friction force Nμ pushing right at the ground,

Reaction force R pushing left at the wall,

and friction force Rμ pushing up at the wall.

Sum of forces in the x direction:

∑F = ma

Nμ − R = 0

R = Nμ

Sum of forces in the y direction:

∑F = ma

N + Rμ − mg = 0

N + Nμ² − mg = 0

N (1 + μ²) = mg

1 + μ² = mg / N

Sum of torques about the base of the ladder.

∑τ = Iα

mg (L/2 cos θ) − R (L sin θ) − Rμ (L cos θ) = 0

mg (L/2 cos θ) = R (L sin θ) + Rμ (L cos θ)

mg (1/2 cos θ) = R (sin θ + μ cos θ)

mg / 2 = R (tan θ + μ)

mg / (2R) = tan θ + μ

Substitute:

mg / (2Nμ) = tan θ + μ

(1 + μ²) / (2μ) = tan θ + μ

(1 − μ²) / (2μ) = tan θ

Given μ = 0.5:

tan θ = (1 − 0.5²) / (2 × 0.5)

tan θ = 0.75

θ = 36.9°

The reaction forces are:

N = mg / (1 + μ²) = 240 N

R = Nμ = 120 N

not too sure thinking all accelerations are pointed down