Question39. (II) How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 85 m at a speed of $95 \mathrm{~km} / \mathrm{h}$ ?

Studdy Solution

STEP 1

1. The car is rounding a level curve, which implies no banking angle.
2. The car is moving at a constant speed of $95 \, \text{km/h}$ .
3. The radius of the curve is $85 \, \text{m}$ .
4. We need to find the coefficient of static friction ( $\mu_s$ ) necessary to prevent slipping.

STEP 2

1. Convert the speed from km/h to m/s.
2. Use the centripetal force equation to find the required frictional force.
3. Relate the frictional force to the normal force using the coefficient of static friction.
4. Solve for the coefficient of static friction.

STEP 3

Convert the speed from km/h to m/s.
Given speed $v = 95 \, \text{km/h}$ .
Convert to m/s using the conversion factor $1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s}$ .
$v = 95 \times \frac{1}{3.6} \, \text{m/s}$
$v \approx 26.39 \, \text{m/s}$

STEP 4

Use the centripetal force equation to find the required frictional force.
The centripetal force required to keep the car moving in a circle is given by:
$F_c = \frac{mv^2}{r}$
where $m$ is the mass of the car, $v$ is the speed, and $r$ is the radius of the curve.

STEP 5

Relate the frictional force to the normal force using the coefficient of static friction.
The frictional force $F_f$ that provides the centripetal force is given by:
$F_f = \mu_s \cdot F_n$
where $F_n = mg$ is the normal force, and $g$ is the acceleration due to gravity ( $\approx 9.81 \, \text{m/s}^2$ ).
Since $F_f = F_c$ , we have:
$\mu_s \cdot mg = \frac{mv^2}{r}$

STEP 6

Solve for the coefficient of static friction $\mu_s$ .
Cancel $m$ from both sides of the equation:
$\mu_s \cdot g = \frac{v^2}{r}$
Solve for $\mu_s$ :
$\mu_s = \frac{v^2}{rg}$
Substitute the known values:
$\mu_s = \frac{(26.39)^2}{85 \times 9.81}$
$\mu_s \approx \frac{696.6721}{833.85}$
$\mu_s \approx 0.835$
The coefficient of static friction must be approximately:
$\boxed{0.835}$

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