Asked by CodyJinks
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
Answered by
Damon
.......but the mass cancels
Answered by
Damon
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
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