Question

f a curve with a radius of 97 m is properly banked for a car traveling 75 km/h , what must be the coefficient of static friction for a car not to skid when traveling at 100 km/h ?

Answer 1

Answer:

The coefficient of static friction is 0.26

Explanation:

Given;

radius of the road, R = 97 m

banking velocity, V₁ = 75 km/h = 20.83 m/s

velocity of the moving car, V₂ = 100 km/h = 27.78 m/s

Car in a banked circular turn:

$\theta = tan^{-1}(\frac{V_1^2}{gR} )$

where;

θ is the angle between the horizontal ground and road in which the car move on

$\theta = tan^{-1}(\frac{V_1^2}{gR} ) \\\\\theta = tan^{-1}(\frac{20.83^2}{9.8*97} ) \\\\\theta = 24.5^o$

During this type of motion, the body acquires some acceleration which tends to retain the circular motion towards its center, known as centripetal acceleration.

There are two components of this acceleration;

Parallel acceleration,  $a_|_|$ = a*Cosθ

Perpendicular acceleration, a⊥ = a * Sinθ

Parallel acceleration, $a_|_|$  $= \frac{V^2*Cos \theta}{R}$

Perpendicular acceleration, a⊥ $= \frac{V^2*Sin \theta}{R}$

Apply Newton's second law of motion;

sum of perpendicular forces acting on the car;

ma⊥ $= F_N - mg*cos \theta$

$m(\frac{V^2*Sin \theta}{R} ) = F_N - mg*Cos \theta\\\\F_N = mg*Cos \theta + m(\frac{V^2*Sin \theta}{R} )$ --------equation (1)

sum of parallel  forces acting on the car

m$a_|_|$ $= mg*Sin \theta - F_s$

$m(\frac{V^2*Cos \theta}{R} ) = mg*Sin \theta - F_s\\\\F_s = mg*Sin \theta - m(\frac{V^2*Cos \theta}{R} )$ ---------equation (2)

Coefficient of static friction is given as;

$\mu = \frac{F_s}{F_N}$

Thus, divide equation (2) by equation (1)

$\frac{F_s}{F_N} = \frac{mg*Sin \theta - m(\frac{V^2*Cos \theta}{R}) }{mg*Cos \theta + m(\frac{V^2*Sin \theta}{R}) } \\\\\frac{F_s}{F_N} = \frac{g*Sin \theta - (\frac{V^2*Cos \theta}{R}) }{g*Cos \theta + (\frac{V^2*Sin \theta}{R}) }$

V = V₂ = 27.78 m/s

θ = 24.5°

R = 97 m

g = 9.8 m/s²

Substitute in these values and solve for μ

$\frac{F_s}{F_N} = \frac{9.8*Sin(24.5)+ (\frac{27.78^2*Cos (24.5)}{97}) }{9.8*Cos (24.5) + (\frac{27.78^2*Sin (24.5)}{97}) }\\\\\frac{F_s}{F_N} = \frac{4.0641 \ - \ 7.2391}{8.91702 \ + \ 3.299} = -0.26\\\\| \mu| = 0.26$

Therefore, the coefficient of static friction is 0.26