A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 21.0° with the horizontal. A 37.0 kg piece of luggage is placed on the carousel, 7.10 m from the axis of rotation. The travel bag goes around once in 44.0 s. (a) Calculate the force of static friction between the bag and the carousel. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to a position 7.88 m from the axis of rotation. The bag is on the verge of slipping as it goes around once every 28.20 s. Calculate the coefficient of static friction between the bag and the carousel.
To solve this problem, we need to understand the forces acting on the luggage and the relationship between these forces. Let's break down the problem step by step:
(a) Calculate the force of static friction between the bag and the carousel:
1. Start by drawing a free-body diagram of the luggage. The forces acting on the luggage are the gravitational force (mg) pointing downward and the static friction force (fs) pointing horizontally towards the center of the circle.
2. The normal force (N) exerted by the carousel on the luggage is perpendicular to the metallic surface and equal in magnitude but opposite in direction to the component of the gravitational force perpendicular to the surface. In this case, the normal force is equal to mg * cos(21°).
3. The frictional force can be represented as fs = μs * N, where μs is the coefficient of static friction.
4. The frictional force fs provides the centripetal acceleration needed for circular motion. Therefore, fs = m * (v^2 / R), where m is the mass of the luggage, v is the linear speed of the luggage, and R is the radius of the circular path.
5. The linear speed v is given by v = 2πR / T, where T is the time it takes for the luggage to complete one full rotation.
6. Combining equations, we have:
μs * N = m * (v^2 / R)
μs * (mg * cos(21°)) = m * [(2πR / T)^2 / R]
μs * g * cos(21°) = 4π^2 * R / T^2
7. Substitute the given values:
m = 37.0 kg
R = 7.10 m
T = 44.0 s
g = 9.8 m/s^2
8. Solve the equation to find the force of static friction (fs):
μs * (37.0 kg * 9.8 m/s^2 * cos(21°)) = 4π^2 * 7.10 m / (44.0 s)^2
(b) Calculate the coefficient of static friction between the bag and the carousel:
1. Repeat steps 1-3 for the new position of the luggage:
N = mg * cos(21°)
fs = μs * N
2. The frictional force fs still provides the centripetal acceleration needed for circular motion:
fs = m * (v^2 / R')
3. The linear speed v is given by v = 2π * R' / T', where R' is the new radius of the circular path, and T' is the new time it takes for the luggage to complete one full rotation.
4. By substituting the values and using the previous equation for fs, we can solve for the coefficient of static friction (μs).
Note: In this case, the luggage is on the verge of slipping, meaning the static friction force is at its maximum and equal to the product of the coefficient of static friction and the normal force.
By following these steps and substituting the values, you can calculate both the force of static friction (a) and the coefficient of static friction (b) between the bag and the carousel.