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 $= 12000 N$ $=12000N$
The tension for the elevator moving downward $= 3000 N$ $=3000N$

Elevator moving upward

Let the tension in the cable be $= T_{1} N$ $=T_1 N$

The weight of the elevator is $= 4000 N$ $=4000 N$

The acceleration is $a = 2 g m s^{- 2}$ $a=2gms^-2$

$T - 4000 = \frac{4000}{g} \cdot 2 g \cdot$ $T-4000=4000/g*2g*$

$T = 4000 + 8000 = 12000 N$ $T=4000+8000=12000N$

Elevator moving downward

Let the tension in the cable be $= T_{2} N$ $=T_2 N$

The weight of the elevator is $= 4000 N$ $=4000 N$

The acceleration is $a = 0.25 g m s^{- 2}$ $a=0.25gms^-2$

Applying Newton's Second Law

$4000 - T = \frac{4000}{g} \cdot 0.25 g$ $4000-T=4000/g*0.25g$

$T = 4000 - 1000 = 3000 N$ $T=4000-1000=3000N$

Answer 2 · 2017-07-21T10:00:22

$1^{st}$ $1^"st"$ case : $F_{elevator} = 12000 N$ $F_"elevator"=12000N$

$2^{nd}$ $2^"nd"$ case : $F_{elevator} = 3000 N$ $F_"elevator"=3000N$

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

From the $2^{nd}$ $2^"nd"$ 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
$0.25g$ :

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

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