Question

A highway curves to the left with radius of curvature of 36 m and is banked at 26, so that cars can take this curve at higher speeds. Consider a car of mass 1477 kg whose tires have a static friction coefficient 0.61 against the pavement.How fast can the car take this curve without skidding to the outside of the curve?

Answer 1

Answer:

Explanation:

Given

radius of track $r=36\ m$

inclination $\theta =26^{\circ}$

mass of car $m=1477\ kg$

coefficient of kinetic friction $\mu =0.61$

From diagram

$N\cos \theta -f_r-mg=0---1$

Where N=Normal reaction

$f_r=friction$

$N\cos \theta +f_r\cos \theta =\frac{mv^2}{r}---2$

$f_r=\mu N----3$

From 1 2 and 3 we get  maximum velocity without skidding

$v=\sqrt{gr}\times \sqrt{\frac{\tan \theta +\mu}{1-\mu \tan \theta }}$

$v=\sqrt{9.8\times 36}\times \sqrt{\frac{\tan 26+0.61}{1-0.61\cdot \tan 26}}$

$v=\sqrt{551.366}$

$v=23.78\ m/s$

Answer 2

Answer:

23.48 m/s

Explanation:

radius, r = 36 m

angle of inclination, θ = 26°

coefficient of friction, μ = 0.61

mass, m = 1477 kg

Let the velocity is v.

$v=\sqrt{\frac{rg\left ( \mu +tan\theta \right )}{1-\mu tan\theta }}$

$v=\sqrt{\frac{36\times 9.8\left ( 0.61 +tan26 \right )}{1-0.61 tan26 }}$

v = 23.48 m/s