Question

MY NOTES ASK YOUR TEACHER PRA If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 105 m radius curve banked at 15° m/s (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 25.0 km/h? Additional Materials Reading

Answer 1

Z Ncos e 0 N sin бш

A)

Here, radius of curve, r = 105 m

banking angle, θ = 15º

free-fall acceleration, g = 9.8 m/s²

We have to find out the ideal speed v (the speed for which no friction is required between the car's tires and the surface)

From the free-body diagram for the car:-

Fnet centripetal

mu mg * tana) 7

2 2 _9* tan(0) 7

21 9.8 * tan(15°) = 105

If the car has a speed of about 16.6 m/s, it can negotiate the curve without any friction.

B)

Now if, friction is present then, equilibrium about vertical direction:

$N*cos(\theta) + f*sin(\theta) = mg$

Also, in horizontal direction:

$f*cos(\theta) - N*sin(\theta) = \frac{mv^2}{r}$

Also,

$f = \mu N$, where mu is the coefficient of friction.

putting value of f in vertical direction equilibrium:

$N*cos(\theta) + \mu N*sin(\theta) = mg$

$N = \frac{mg}{cos(\theta) + \mu*sin(\theta)}$

Putting value of N in horizontal direction equilibrium:

$\mu* \frac{mg}{cos(\theta) + \mu*sin(\theta)}*cos(\theta) - \frac{mg}{cos(\theta) + \mu*sin(\theta)} * sin(\theta) = \frac{mv^2}{r}$

$\mu* g*cos(\theta) - g * sin(\theta) = \frac{v^2}{r}*(cos(\theta) + \mu*sin(\theta))$

$\mu*cos(\theta) - sin(\theta) = \frac{v^2}{rg}*(cos(\theta) + \mu*sin(\theta))$

Dividing whole equation by mu * cos θ:

$1 - \frac{tan(\theta)}{\mu} = \frac{v^2}{rg}*(\frac{1}{\mu} + tan(\theta))$

Substituting values of θ, r and g:

$1 - \frac{0.268}{\mu} = \frac{48.1636}{1029}*(\frac{1}{\mu} + 0.268)$

$\mu - 0.268 =0.0468*(1 + 0.268*\mu)$

$\mu - 0.268 =0.0468 + 0.0125424 * \mu$

$\mu - 0.0125424 * \mu=0.0468 + 0.268$

$0.9874576*\mu=0.3148$

$\mu = 0.3188$

The minimum coefficient of friction needed is 0.3188