Question
Dana loads luggage into an airplane using a conveyor belt tilted at an angle of 20°.She places a few pieces of luggage on the belt before it starts to move, then she turns the belt on. It takes the belt 0.70 s to reach its top speed of 1.2 m/s. Does the luggage slip? Assume a static friction coefficient of 0.5 between the luggage and belt
I really just would like to know what to find i.e. in what order. I have all the formulas. I have found the accel. of the belt and found what the normal force would be equal to but dont know the mass of the luggage
I really just would like to know what to find i.e. in what order. I have all the formulas. I have found the accel. of the belt and found what the normal force would be equal to but dont know the mass of the luggage
Answers
.......but the mass cancels
a = acceleration up slope = 1.2/0.7 = 1.714 m/s^2
so net force up slope, m * a, must be 1.714 m
now how
much a do we get maximum
weight force down slope = m g sin 20
normal force = m g cos 20
so max friction force up slope = 0.5 m g cos 20
0.5 m g cos 20 - m g sin 20 = m a
m g ( .470 - .342) = m a
a = .128 g = 1.25 m/s^2
we only have enough friction to get a = 1.25 but we need 1.714
so net force up slope, m * a, must be 1.714 m
now how
much a do we get maximum
weight force down slope = m g sin 20
normal force = m g cos 20
so max friction force up slope = 0.5 m g cos 20
0.5 m g cos 20 - m g sin 20 = m a
m g ( .470 - .342) = m a
a = .128 g = 1.25 m/s^2
we only have enough friction to get a = 1.25 but we need 1.714