16-18 An elevator car is suspended by a single steel cable. The car has a mass o

Question

16-18 An elevator car is suspended by a single steel cable. The car has a mass of 520 kg and it contains a passenger whose mass is 80 kg. The mass of the cable can be ignored. 16. If the elevator is traveling vertically downwards at a constant speed of 2.00 m/s, calculate the tension in the cable.

17. If the elevator is accelerating downwards with a constant acceleration of magnitude 1.00 m/s2, calculate the tension in the cable.

18. If the maximum tension that the cable can safely withstand is 9000 (N), and the maximum upward
acceleration of the elevator is 1.00 m/s2, how many 80 kg passengers may be safely carried in the elevator?

Explanation / Answer

the elevator car mass m1 = 520 kg and passenger mass m2 = 80 kg. (16) if the elevator traveling constant speed then the acceleration is zero.    i.e a = 0 (since v = constant)      Applying the Newton's 2nd law of motion,     the tension in cable T = (m1+m2)g +0                                T = 5880 N (17) If the elevator is accelerating downwards with a constant acceleration of magnitude 1.00 m/s2.          a = 1.00 m/s62 the tension T = (m1+m2)g - (m1+m2)a                  T = 5880 - 600                T = 5280 N (18) If the maximum tension that the cable can safely withstand is 9000 (N),    therefore T = (m1+m2)g + (m1+m2)a                     = m1(1+9.8)+80(1+9.8) 9000 = m1(10.8) + 864    m1 = 753.33 kg the maximum mass carried is 753.33 kg.

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