To find the minimum coefficient of static friction (μ) needed for a car to make a 23.0-meter radius turn at 7.85 meters per second on flat ground, we need to balance the centripetal force required to keep the car moving in a circle with the frictional force that keeps the car from sliding.
Step-by-step, we follow these key steps:
1. Identify the centripetal force:
The centripetal force ([tex]\(F_c\)[/tex]) required to keep an object moving in a circle is given by:
[tex]\[
F_c = \frac{m \cdot v^2}{r}
\][/tex]
where [tex]\(m\)[/tex] is the mass of the car, [tex]\(v\)[/tex] is the velocity, and [tex]\(r\)[/tex] is the radius of the turn.
2. Identify the frictional force:
The frictional force ([tex]\(F_f\)[/tex]) that acts between the tires and the road for static friction is given by:
[tex]\[
F_f = \mu \cdot m \cdot g
\][/tex]
where [tex]\( \mu \)[/tex] is the coefficient of static friction and [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\(9.81 \, m/s^2 \)[/tex]).
3. Set the centripetal force equal to the frictional force:
For the car to successfully navigate the turn without slipping, the frictional force must at least equal the centripetal force:
[tex]\[
F_c = F_f
\][/tex]
Substituting the formulas for each force, we get:
[tex]\[
\frac{m \cdot v^2}{r} = \mu \cdot m \cdot g
\][/tex]
4. Simplify the equation:
Notice that the mass [tex]\(m\)[/tex] of the car appears on both sides of the equation, so it can be cancelled out:
[tex]\[
\frac{v^2}{r} = \mu \cdot g
\][/tex]
5. Solve for the coefficient of static friction ([tex]\(\mu\)[/tex]):
Rearrange the equation to solve for [tex]\( \mu \)[/tex]:
[tex]\[
\mu = \frac{v^2}{r \cdot g}
\][/tex]
6. Plug in the given values:
We have [tex]\(v = 7.85 \, m/s\)[/tex], [tex]\(r = 23.0 \, m\)[/tex], and [tex]\(g = 9.81 \, m/s^2\)[/tex]:
[tex]\[
\mu = \frac{(7.85)^2}{23.0 \cdot 9.81}
\][/tex]
7. Calculate the result:
[tex]\[
\mu = \frac{61.6225}{225.63}
\][/tex]
[tex]\[
\mu \approx 0.2731
\][/tex]
Therefore, the minimum coefficient of static friction necessary for the car to make the turn safely is approximately [tex]\(0.2731\)[/tex].
