Answer:
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
