At a speed of 90 km / h, the radius of the turning safety arc is at least 220 m. Find the coefficient of friction.
| September 11, 2021 | education
V = 90 km / h = 25 m / s.
R = 220 m.
g = 9.8 m / s2.
μ -?
In order for the motorcyclist not to fly off the road when turning, the friction force Ffr must be equal to the centripetal force.
Let’s write 2 Newton’s law in vector form: m * a = m * g + N + Ftr.
OH: m * a = Ftr.
OU: – m * g + N = 0.
F = Ftr.
N = m * g.
The friction force Ffr is determined by the formula: Ffr = μ * N = μ * m * g.
m * a = μ * m * g.
a = μ * g.
The centripetal acceleration a is expressed by the formula: a = V2 / R.
V2 / R = μ * g.
μ = V2 / R * g.
μ = (25 m / s) 2/220 m * 9.8 m / s2 = 0.29.
Answer: the coefficient of friction should be μ = 0.29.
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