Answer :
Answer:
Driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.
Explanation:
Let suppose that car is represented by a particle, then we proceed to construct its free body diagram and corresponding equations of equilibrium:
[tex]\Sigma F_{x'} = f - W\cdot \sin \theta = 0[/tex] (Eq. 1)
[tex]\Sigma F_{y'} = N-W\cdot \cos \theta = 0[/tex] (Eq. 2)
Where:
[tex]f[/tex] - Static friction force, measured in newtons.
[tex]N[/tex] - Normal force on the car from the ground, measured in newtons.
[tex]W[/tex] - Weight of the car, measured in newtons.
[tex]\theta[/tex] - Driveway inclination, measured in sexagesimal degrees.
By applying definitions of maximum static friction force and weight, we expand the system of equations presented above:
[tex]\mu_{s}\cdot N -m\cdot g\cdot \sin \theta = 0[/tex] (Eq. 1b)
[tex]N - m\cdot g \cdot \cos \theta = 0[/tex] (Eq. 2b)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]m[/tex] - Mass of the car, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
Then, we substitute normal force and simplify the resulting expression:
[tex]\mu_{s}\cdot m\cdot g \cdot \cos \theta -m\cdot g \cdot \sin \theta = 0[/tex]
[tex]\mu_{s} = \tan \theta[/tex]
[tex]\theta = \tan^{-1}\mu_{s}[/tex] (Eq. 3)
Based on such result, we can conclude that if driveway inclination is greater than value reported, then car shall not be safe in case of parking. Our reference angle is: ([tex]\mu_{s} = 0.15[/tex])
[tex]\theta = \tan^{-1} 0.15[/tex]
[tex]\theta \approx 8.531^{\circ}[/tex]
By direct comparison, we find that driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.
