Newton's first law states that at rest the sum of the forces acting on a body is zero
The distance S to which the painter can climb is approximately 2.55 meters
The reason the above value for the distance is correct is as follows:
The given parameter are;
The mass of the painter, m₁ = 90 kg
Length of the ladder, l = 4 m
The mass of the ladder, m₂ = 15 kg
The coefficient pf static friction, μ = 0.25
Friction force, [tex]\mathbf{F_f}[/tex] = Normal reaction, N × μ
Normal reaction (flat surface), N = Weight of painter + Weight of ladder
N = m₁·g + m₂·g
Where;
g = The acceleration due to gravity, g = 9.81 m/s²
N = 90 kg × 9.81 m/s² + 15 kg × 9.81 m/s² = 1,030.05 N
∴ [tex]\mathbf{F_f}[/tex] = 1,030.05 N × 0.25 = 257.5125 N
Friction force, [tex]\mathbf{F_f}[/tex] = 257.5125 N (Horizontal force)
At equilibrium, the sum of the horizontal forces = 0
∴ Fₓ + [tex]\mathbf{F_f}[/tex] = 0
Fₓ = [tex]-F_f[/tex] = -257.5125 N (Acting opposite to the direction of the friction force)
The height of the top of the ladder from the ground, h = √(4²-1.5²) = √(13.75)
Taking moment about point A gives;
Clockwise moment = Anticlockwise moment
Fₓ × h = m₁·g·s·cos(θ) + m₂·g·(l/2)·cos(θ)
Therefore;
Fₓ × √(13.75) m = 90 kg × 9.81 m/s² × s × (1.5/4) + 15 kg × 9.81 m × 2 m × (1.5/4)
Fₓ × √(13.75) m = 331.0875 N × s + 110.3625 J
257.5125 N × √(13.75) m = 331.0875 N × s + 110.3625 J
s = (257.5125 N × √(13.75) m - 110.3625 J)/(331.0875 N) ≈ 2.55 m
The distance to which the 90 kg painter can climb without causing the 4 m ladder to slip at its lower end, A, distance S ≈ 2.55 meters
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