What is the tension in an elevator cable which must support a 4000 N elevator with maximum upward acceleration of 2.0 g's. What would be the tension if the elevator were allowed to accelerate downward at 0.25 g's?

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Answer 1 · 2017-07-21T09:51:12

The tension for the elevator moving upward #=12000N#
The tension for the elevator moving downward #=3000N#

Elevator moving upward

Let the tension in the cable be #=T_1 N#

The weight of the elevator is #=4000 N#

The acceleration is #a=2gms^-2#

#T-4000=4000/g*2g*#

#T=4000+8000=12000N#

Elevator moving downward

Let the tension in the cable be #=T_2 N#

The weight of the elevator is #=4000 N#

The acceleration is #a=0.25gms^-2#

Applying Newton's Second Law

#4000-T=4000/g*0.25g#

#T=4000-1000=3000N#

Answer 2 · 2017-07-21T10:00:22

#1^"st"# case : #F_"elevator"=12000N#

#2^"nd"# case : #F_"elevator"=3000N#

Let's see the first case, moving upwards with acceleration of #2g# :

From the #2^"nd"# Law of Newton :

#ΣF=ma=>F_"elevator"-W=m*2g=>#

#F_"elevator"-mg=m*2g=>F_"elevator"=3mg=>#

#F_"elevator"=3W=>#

#F_"elevator"=3*4000=12000N#

In the second case the elevator goes down with an acceleration of
#0.25g# :

#ΣF=ma=>W-F_"elevator"=0.25mg=>F_"elevator"=0.75W=>#

#F_"elevator"=0.75*4000=3000N#