A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is the same as that between the ladder and the wall. If this coefficient of static friction is μs = 0.495, determine the smallest angle the ladder can make with the floor without slipping.
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Ответ:
45.3°
Explanation:
Given,
Length of ladder = l
Weight of ladder = w
Coefficient of friction = μs = 0.495
Smallest angle the ladder makes = θ
If we assume the forces in the vertical direction to be N1, and the forces in the horizontal direction to be N2, then,
N1 = mg and
N2 = μmg
Moment at a point A in the clockwise direction is
N2 Lsinθ - mg.(L/2).cosθ = 0
μmgLsinθ - mg.(L/2).cosθ = 0
μmgLsinθ = mg.(L/2).cosθ
μsinθ = cosθ/2
sin θ / cos θ = 1 / 2μ
Tan θ = 1 / 2μ
Substituting the value of μ = 0.495, we have
Tan θ = 1 / 2 * 0.495
Tan θ = 1 / 0.99
Tan θ = 1.01
θ = tan^-1(1.01)
θ = 45.3°
Ответ:
Bobby’s interquartile range is 113, while Amelia’s is 179. A lower interquartile range means Bobby’s scores are closer to the median.
Step-by-step explanation:
A lower interquartile range is clustered closer to the median than a higher interquartile range.