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

Answer:

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

Explanation:

We need to remember that to keep an object moving in a circle, this object should meet as following:$F_{c}=m*a_{c}=m*\frac{v^{2}}{R}$. Now analysing the free body diagram for tires car, we can find:$a_{x} =\frac{v^{2} }{R}*Cos\beta$, and $a_{y} =\frac{v^{2} }{R}*Sin\beta$. 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:$F=m(\frac{v^{2} }{R}*Cos\beta-gSin\beta)$ and$N=m(\frac{v^{2} }{R}*Sin\beta +gCos\beta)$. 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:$m(\frac{v^{2} }{R}*Cos\beta-gSin\beta) \leq u*m(gCos+\frac{v^{2} }{R}*Sin\beta)$. Finally replacing the values and solving for vmax we can find the higher speed as:$v_{max}=\sqrt{Rg(\frac{Tan\beta+u }{1-u*Tan\beta } )}=\sqrt{(31*9.8)*(\frac{Tan(24)+0.76}{1-0.76*Tan(24)})} =23.52(m/s)$.

Answer 2

The speed of car to  curve without skidding to the outside of the curve is 23.52 m/s.

Given data:

The radius of curvature of highway is, r = 31 m.

The banking angle is, $\theta = 24^{\circ}$.

The mass of car is, m = 1029 kg.

The coefficient of static friction against the pavement is, $\mu=0.76$.

The given problem resembles the condition of banking of road. Banking means the driving path is inclined by making some angle with respect to the horizontal. And the standard relation for the speed of vehicle on the Banked road with given amount of frictional coefficient is,

$v = \sqrt{rg \times \dfrac{tan\theta+\mu}{(1-\mu)tan \theta} }$

here, g is the gravitational acceleration.

Solving as,

$v = \sqrt{31 \times 9.8 \times \dfrac{tan24+0.76}{(1-0.76)tan24} }\\\\v=23.52 \;\rm m/s$

Thus, we can conclude that the speed of car to  curve without skidding to the outside of the curve is 23.52 m/s.

Learn more about the Banking of road here:

https://brainly.com/question/20385799