A sports car is accelerating up a hill that rises 17.0° above the horizontal. The coefficient of static friction between the wheels and the road is µs = 0.79. It is the static frictional force that propels the car forward.
(a) What is the magnitude of the maximum acceleration that the car can have?
m/s2
(b) What is the magnitude of the maximum acceleration if the car is being driven down the hill?
m/s2
acc up:
mu g cosθ - g sinθ = a
acc down:
mu g cosθ + g sinθ = a
To find the maximum acceleration of the car in both cases, we need to analyze the forces acting on the car.
Let's start by drawing a free body diagram of the car on the hill.
In both cases, there are two main forces acting on the car: the force of gravity pulling it downward and the force of friction between the wheels and the road.
(a) When the car is accelerating up the hill:
The force of gravity can be split into two components: a component parallel to the hill, pulling the car downhill, and a component perpendicular to the hill.
The perpendicular component of gravity (mg * cos θ) is balanced by the normal force from the road, and it does not affect the motion of the car in this case.
The parallel component of gravity (mg * sin θ) acts downward and opposes the motion of the car. The static frictional force between the wheels and the road must overcome this component to propel the car forward.
The maximum static frictional force that can be generated is given by:
fs = µs * N
where fs is the static frictional force and N is the normal force.
The only force that can provide the necessary forward acceleration is the static frictional force. So, the maximum acceleration is equal to the maximum static frictional force divided by the mass of the car:
(a) a_max = fs / m = (µs * N) / m
Since N = mg, we can substitute it into the equation:
(a) a_max = (µs * mg) / m
We are given that θ = 17.0° and µs = 0.79.
Substituting these values into the equation gives us:
(a) a_max = (0.79 * mg) / m
(b) When the car is being driven down the hill:
In this case, the force of gravity acts along the direction of motion, aiding the movement of the car. The static frictional force will act in the opposite direction, opposing the motion.
The maximum static frictional force in this case can be calculated in the same way as before.
(b) a_max = fs / m = (µs * N) / m
But now, since the force of gravity aids the motion, the component of gravity acting parallel to the hill (mg * sin θ) and the maximum static frictional force will add up.
(b) a_max = (µs * mg) / m + (mg * sin θ) / m
Again, substituting θ = 17.0° and µs = 0.79 into the equation gives us:
(b) a_max = (0.79 * mg) / m + (mg * sin θ) / m
Now, we have the expressions for both cases to find the maximum acceleration (a_max) of the car.
To calculate the magnitude of the maximum acceleration in each case, we need to consider the forces acting on the car and use Newton's laws of motion.
(a) When the car is accelerating up the hill:
First, let's calculate the gravitational force acting on the car (Fg). The component of the gravitational force parallel to the hill is given by:
Fg_parallel = m * g * sin(θ)
where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the hill (17.0°).
Next, we calculate the maximum static frictional force (Ffriction_max) that can propel the car up the hill. The maximum static frictional force is given by:
Ffriction_max = µs * N
where µs is the coefficient of static friction and N is the normal force.
Since the car is on an incline, the normal force (N) is equal to the gravitational force acting perpendicular to the hill:
N = m * g * cos(θ)
Now, we can substitute these values and calculate the maximum acceleration:
Ffriction_max = µs * m * g * cos(θ)
Using Newton's second law (F = m * a), where F is the net force acting on the car, we have:
Fnet = Ffriction_max - Fg_parallel
Setting Fnet equal to m * a, we can solve for the maximum acceleration (a):
a = (Ffriction_max - Fg_parallel) / m
Now, substituting the values:
a = (0.79 * m * g * cos(θ) - m * g * sin(θ)) / m
Simplifying the equation:
a = g * (0.79 * cos(θ) - sin(θ))
Finally, we can calculate the magnitude of the maximum acceleration:
a = g * (0.79 * cos(17.0°) - sin(17.0°))
(a) Calculated Result:
a = 9.8 m/s^2 * (0.79 * 0.96 - 0.29)
a ≈ 8.25 m/s^2
Therefore, the magnitude of the maximum acceleration that the car can have when accelerating up the hill is approximately 8.25 m/s^2.
(b) When the car is being driven down the hill:
The maximum acceleration when the car is being driven down the hill can be calculated using the same method. However, in this case, the direction of acceleration is opposite to the direction of the parallel gravitational force.
Using the same equation and values as before, the magnitude of the maximum acceleration when the car is being driven down the hill will be:
a = g * (0.79 * cos(17.0°) + sin(17.0°))
(b) Calculated Result:
a = 9.8 m/s^2 * (0.79 * 0.96 + 0.29)
a ≈ 15.87 m/s^2
Therefore, the magnitude of the maximum acceleration that the car can have when accelerating down the hill is approximately 15.87 m/s^2.