Sunday, June 5, 2016

An elevator weighing 3830 N is pulled upward by a cable with an accleration of 2.22 m s^-2. What is the tension in the cable?

When accelerating upwards, the cables must support the weight of elevator and also compensate for the acceleration. Hence the tension in the wire support is given as:


F = ma + mg


where, mg is the weight of the elevator, m is elevator mass and a is elevator acceleration. 


Mass of elevator = weight/g = 3830/9.81 = 390.42 kg


Hence, F = (390.42 x 2.22) + 3830 = 4696.73 N


Thus, the tension in the...

When accelerating upwards, the cables must support the weight of elevator and also compensate for the acceleration. Hence the tension in the wire support is given as:


F = ma + mg


where, mg is the weight of the elevator, m is elevator mass and a is elevator acceleration. 


Mass of elevator = weight/g = 3830/9.81 = 390.42 kg


Hence, F = (390.42 x 2.22) + 3830 = 4696.73 N


Thus, the tension in the elevator cable (when accelerating upwards) is 4696.73 N. This is the reason, we feel that our weight is greater, when elevator goes up. The elevator has to compensate for our weight (and its own weight) and provide acceleration, on top of the weight. In comparison, when going down, we feel our weight is less, since the cable support have lower tension (F= mg-ma).


Hope this helps.

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