Answer:
0.58
Explanation:
L = length of the ladder = AC = 8.2 m
AB = L/2 = 8.2 /2 = 4.1 m
[tex]F_{g}[/tex] = weight of the ladder = 254 N
[tex]F_{n}[/tex] = Normal force by the floor
[tex]f_{s}[/tex] = static frictional force
F = Normal force by the wall
Using equilibrium of torque about A
F (AC) Sinθ = [tex]F_{g}[/tex] Cosθ (AB)
F (8.2) Sin41 = (254) Cos41 (4.1)
F = 146.1 N
Using equilibrium of force along the horizontal direction
[tex]f_{s}[/tex] = F
[tex]f_{s}[/tex] = 146.1 N
Using equilibrium of force along the vertical direction
[tex]F_{n}[/tex] = [tex]F_{g}[/tex]
[tex]F_{n}[/tex] = 254 N
Coefficient of static friction is given as
[tex]\mu _{s} = \frac{f_{s}}{F_{n}}[/tex]
[tex]\mu _{s} = \frac{146.1}{254}[/tex]
[tex]\mu _{s} = 0.58 [/tex]
The coefficient of static friction between the ladder and the floor is 0.75.
The coefficient of static friction between the ladder and the floor is determined Newton's second law of motion.
F(net) = 0
Force of the ladder perpendicular to the floor - static frictional force of the floor on the ladder = 0
mgcosθ - μmg = 0
μmg = mgcosθ
μ = cosθ
μ = cos(41)
μ = 0.75
Thus, the coefficient of static friction between the ladder and the floor is 0.75.
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