An elevator (mass 4025 kg) is to be designed so that the maximum acceleration is 0.0900g.?

Question

An elevator (mass 4025 kg) is to be designed so that the maximum acceleration is 0.0900g.
What is the maximum force the motor should exert on the supporting cable?
N
What is the minimum force the motor should exert on the supporting cable?
N

Answer 1

Consider an FBD of the elevator moving upward

m*a=T-m*g
if a=0.0900 g
4025*9.81*(1+0.0900)=T
T=43039 N

in the downward direction for the minimum T:

m*a=T-m*g
this is a=-0.0900*g

T=m*g*(1-0.0900)
T=4025*9.81*(1-0.0900)
T=35932 N

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