Physics question about grade and static friction. Pls help, thanks!

a.
b.
c.
A grade is the ratio of the vertical rise to horizontal run of a road
The coefficient of static friction is a number which shows the amount of friction between two surfaces.
Since grade = rise/run
= tan θ
So, θ = tan⁻¹(grade)
Since grade for Telegraph hill = 31.5%
= 31.5/100
= 0.315
θ = tan⁻¹(grade)
θ = tan⁻¹(0.315)
θ = 17.48⁰
So, the angle for telegraph hill is 17.48⁰
For no sliding the coefficient of static friction, μ ≥ tanθ
≥ tan17.48⁰
≥ 0.315
So, the coefficient of static friction has to be no less than 0.315
Since grade = rise/run
= tan θ
So, θ = tan⁻¹(grade)
Since grade for Canton Ave. = 37%
= 37/100
= 0.37
θ = tan⁻¹(grade)
θ = tan⁻¹(0.37)
θ = 20.3⁰
So, the angle for Canton Ave. is 20.3⁰
For no sliding the coefficient of static friction, μ ≥ tanθ
≥ tan20.3⁰
≥ 0.37
So, the coefficient of static friction has to be no less than 0.37
Since grade = rise/run
= tan θ
So, θ = tan⁻¹(grade)
Since grade for Bradford Street = 41%
= 41/100
= 0.41
θ = tan⁻¹(grade)
θ = tan⁻¹(0.41)
θ = 22.3⁰
So, the angle for Bradford Street is 22.3⁰
For no sliding the coefficient of static friction, μ ≥ tanθ
≥ tan22.3⁰
≥ 0.41
So, the coefficient of static friction has to be no less than 0.41
Learn more about grade of a road here:
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