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

Step-by-step 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 =(mv^2)/(r)---2`

`f_r=\mu N----3`

From 1 2 and 3 we get maximum velocity without skidding

`v=√(gr)* \sqrt{(\tan \theta +\mu)/(1-\mu \tan \theta )}`

`v=√(9.8* 36)* \sqrt{(\tan 26+0.61)/(1-0.61\cdot \tan 26)}`

`v=√(551.366)`

`v=23.78\ m/s`

Answer 2

23.48 m/s

Step-by-step 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{(rg\left ( \mu +tan\theta  \right ))/(1-\mu tan\theta )}`

`v=\sqrt{(36* 9.8\left ( 0.61 +tan26  \right ))/(1-0.61 tan26 )}`

v = 23.48 m/s