Conveyor belt Question? - suitcase belt
The driver left the bag containing 8.80 kilograms baggage on a treadmill running at 2.10 m / s. The materials are such that, according to U.S. = 0.440 and K = 0.290 or secondary.
How far is your suitcase dragged before it surfs smoothly on the tape?
() In meters
Friday, January 22, 2010
Suitcase Belt Conveyor Belt Question?
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Mass = 8.80 kilograms bags
ReplyDeleteWeight of the bag = 8.8 * 9.8 = 86.24 N
Force of friction f = 0.290 * 86.24 N = 25.01 N
The bag pull in the direction Belt Movement.
Acceleration of a bag of 25.01/8.80 = m / s ^ 2
a = 2.84 m / s ^ 2
After the bag is dropped,
Initial velocity u = 0
V is the terminal velocity after moving smoothly on the belt = belt speed
O, V = 2.10 m / s
Let d = distance
v ^ 2 = u ^ 2 + 2AS
O, 2.10 ^ 2 = 0 + 2 * 2.84 * s
O, S = 2.10 ^ 2 / (2 * 2.84) = 0.776 m
Answer: 0776 m