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3 years ago

Required information

Physics

1 answer:

Semmy [17]3 years ago

8 0

Answer:

19,224 N

Explanation:

The given parameters are;

The mass limit of the elevator = 2,400 kg

The maximum ascent speed = 18.0 m/s

The maximum descent speed = 10.0 m/s

The maximum acceleration = 1.80 m/s²

Given that the acceleration due to gravity, g ≈ 9.81 m/s²

The minimum upward force that the elevator cable exert on the elevator car, $F_{min}$ , is given in the downward motion as follows;

$F_{min}$ = m·g - m·a

∴ $F_{min}$ = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N

The minimum upward force that the elevator cable exert on the elevator car, $F_{min}$ = 19,224 N

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2 years ago

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Answer:

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