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?

Question

9607 views around the world

You can reuse this answer
Creative Commons License

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#