An elevator (cabin mass 600 kg) is designed for a maximum load of 2800 kg, and to reach a velocity of 3.6 m/s in 5 s. For this scenario, what is the tension the elevator rope has to withstand?
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Acceleration is required to be
a = (3.6 m/s)/5 s = 0.72 m/s^2
Total tension in cable when accelerating upward =
Mtotal*(a + g) = 3400*(0.72 + 9.81)
= 35,800 N
a = (3.6 m/s)/5 s = 0.72 m/s^2
Total tension in cable when accelerating upward =
Mtotal*(a + g) = 3400*(0.72 + 9.81)
= 35,800 N
1 answer