Question

An empty elevator weighing 100 kgs, stops at the third -level of a building and got 3 passengers weighing 200 kgs. Considering a single cable supporting the elevator cart, what is the
(a) tension of the cable if it is going up at an acceleration of 1m/s2.
(b) tension of the cable if it is going down with the same acceleration of 1m/s. ​

Answer 1

Explanation:

in order for the cable to not break the tension force on the cable must be equal to the weight it supports . apparent weight of the 3 passengers plus the elevator itself must be equal to the tension force. mathematicaly:

Total apparent weight = Tension force

(a) tension of the cable if it is going up at an acceleration of 1m/s2

elevator App weight=m(a + g)

elevator App weight=100kg(10m/s2 + 1m/s2)

elevator App weight=1100N

pass. App weight=m(a + g)

pass. App weight=200kg(10m/s2)+ 1m/s2)

pass. App weight=2200N

tension F= total apparent weight

tension F=2200N+1100N

tension F=3300N

(b) tension of the cable if it is going down with the same acceleration of 1m/s.

in this case ,due to the downward movement of the elevator the acceleration is assigned a negative sign(-1m/s2)

elevator App weight=m(a + g)

elevator App weight=100kg(10m/s2 +( - 1m/s2))

elevator App weight=900N

pass. App weight=m(a + g)

pass. App weight=200kg(10m/s2)+(- 1m/s2))

pass. App weight=1800N

tension F= total apparent weight

tension F=1800N+900N

tension F=2700N