Asked by Luke
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 23.0 degrees with the horizontal. A piece of luggage having mass 31.0 kg is placed on the carousel, 6.75 m from the axis of rotation. The travel bag goes around once in 48.0 s. Calculate the force of static friction between the bag and the carousel.
Answers
Answered by
MathMate
r=6.75
θ=23°
m=31 kg
ω=2π/48 radians/s
g=9.8 m/s/s
μ = coefficient of static friction
Assuming the bag does not slide due to friction, then
force due to gravity downwards along the incline plane, Fg
= mgsin(θ)
downward force along inclined plane due to rotation, Fr
= mrω²cos(θ)
Normal reaction due to gravity less component due to rotation, N
= mgcos(θ)-mrω²sin(θ)
μ=(Fg+Fr)/N
Can you take it from here?
Also check my logic.
θ=23°
m=31 kg
ω=2π/48 radians/s
g=9.8 m/s/s
μ = coefficient of static friction
Assuming the bag does not slide due to friction, then
force due to gravity downwards along the incline plane, Fg
= mgsin(θ)
downward force along inclined plane due to rotation, Fr
= mrω²cos(θ)
Normal reaction due to gravity less component due to rotation, N
= mgcos(θ)-mrω²sin(θ)
μ=(Fg+Fr)/N
Can you take it from here?
Also check my logic.
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