A uniform ladder of mass and length is resting on a wall. A man of mass climbs up the ladder and is in perfect equilibrium with the ladder when he is the way up the ladder. The ladder makes an angle of with the horizontal floor. If the coefficient of static friction between the ladder and the wall is the same as that between the ladder and the floor, which is , what is , expressed to the nearest integer?
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First, draw a free body diagram . We know that F 1 = μ N 1 . Since the system is at translational equilibrium, the net horizontal force is 0 and so N 2 = F 1 . Knowing that N 2 is the normal force of the wall on the ladder, F 2 = μ N 2 = μ 2 N 1 . Again, since there is translational equilibrium, the vertical forces balance out and so N 1 + F 2 = 2 m g : N 1 = 2 m g − F 2 = 2 m g − μ 2 N 1 μ 2 N 1 + N 1 = N 1 ( μ 2 + 1 ) = 2 m g N 1 = μ 2 + 1 2 m g Thus, N 2 = μ 2 + 1 2 μ m g and F 2 = μ 2 + 1 2 μ 2 m g . Now we can use the fact that the system is at rotational equilibrium as well, so Σ τ = 0 . Using the base of the ladder as the pivot point: Σ τ = − m g cos θ ( 2 L ) − m g cos θ ( 3 2 L ) + F 2 L cos θ + N 2 L sin θ = 0 2 − m g cos θ − 3 2 m g cos θ + μ 2 + 1 2 μ 2 m g cos θ + μ 2 + 1 2 μ m g sin θ = 0 μ 2 + 1 2 μ 2 cos θ + 2 μ sin θ = 2 cos θ + 3 2 cos θ = 6 7 cos θ 2 μ 2 cos θ + 2 μ sin θ = 6 7 cos θ μ 2 + 6 7 cos θ μ 2 ( 2 cos θ − 6 7 cos θ ) + 2 μ sin θ − 6 7 cos θ = μ 2 ( 6 5 cos θ ) + 2 μ sin θ − 6 7 cos θ = 0 Plugging into the quadratic formula: μ = 3 7 cos θ − 2 sin θ ± 4 sin 2 θ − 4 ( 6 5 cos θ ) ( 6 − 7 cos θ ) Setting θ = 3 0 ∘ and discounting the negative answer (-2.06), we find that μ ≈ 0 . 6 7 8 and so 1 0 0 0 μ ≈ * 678 *.