Question

A highway curves to the left with radius of curvature of 31 m and is banked at 24 ◦ so that cars can take this curve at higher speeds. Consider a car of mass 1029 kg whose tires have a static friction coefficient 0.76 against the pavement. How fast can the car take this curve without skidding to the outside of the curve? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.

Answer 1

The fastest that car can take the curve is:23.52(m/s)

Step-by-step explanation:

We need to remember that to keep an object moving in a circle, this object should meet as following:
`F_(c)=m*a_(c)=m*(v^(2))/(R)`. Now analysing the free body diagram for tires car, we can find:
`a_(x) =(v^(2) )/(R)*Cos\beta`, and
`a_(y) =(v^(2) )/(R)*Sin\beta`. Then applying Newton's second law we get:
`m*a_(x)=F+mgSin\beta` and
`m*a_(y)=N-mgCos\beta`, and replacing ax and ay we can find F and N as:
`F=m((v^(2) )/(R)*Cos\beta-gSin\beta)` and
`N=m((v^(2) )/(R)*Sin\beta +gCos\beta)`. But the car has to take the curve without skidding so
`F\leq u*N` so that replacing F and N in this equation we get:
`m((v^(2) )/(R)*Cos\beta-gSin\beta) \leq u*m(gCos+(v^(2) )/(R)*Sin\beta)`. Finally replacing the values and solving for vmax we can find the higher speed as:
`v_(max)=\sqrt{Rg((Tan\beta+u )/(1-u*Tan\beta ) )}=\sqrt{(31*9.8)*((Tan(24)+0.76)/(1-0.76*Tan(24)))} =23.52(m/s)`.