Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 11 (Q.No. 37)
37.
A car negotiates a curve on a slippery road at a constant speed of 10 m/s. If the coefficient of friction is 0.5, what is the minimum radius of the arc in which the car can turn ?
Discussion:
2 comments Page 1 of 1.
Rahul verma said:
2 years ago
e = 0.
f = V^2/gR,
R = V^2/gf,
R = 10^2/9.81 * 0.5.
R = 20m.
f = V^2/gR,
R = V^2/gf,
R = 10^2/9.81 * 0.5.
R = 20m.
Daud unnadpotra said:
7 years ago
V = √(μrg).
Given μ = 0.5,
v = 10,
10 = √(0.5 * r * 10),
10^2 = 5r,
r = 20.
Given μ = 0.5,
v = 10,
10 = √(0.5 * r * 10),
10^2 = 5r,
r = 20.
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