Question

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

Answer 1

We have given acceleration when lift goes upward and downward, and also given the weight in elevator i.e., mg. When lift goes upward, then total tension equals to the sum of weight and force due to acceleration. When lift goes downward, then total tension equals the difference of weight and force due to acceleration.

Complete step-by-step solution:
First take the case when elevator is moving upward: -
Let the tension in the cable =$T_{1} N$
The weight of the elevator, $mg = 4000N$
The acceleration is a=$2.0g\;\; ms^{-2}$
Applying Newton's Second Law:
$T – mg = ma$
$T -4000 = \dfrac{4000}{g} \times 2g$
$T=4000+8000=12000N$
Now take the case when elevator is moving downward: -
Let the tension in the cable =$T_{2} n$
The weight of the elevator =$4000N$
The acceleration is a=$0.25 g\;\; ms^{-2}$
Applying Newton's Second Law:
$mg – T =ma$
$4000 - T = \dfrac{4000}{g} \times 0.25g$
$T=4000-1000 = 3000N$
Hence, the tension for the elevator moving upward is $12000N$ and the tension for the elevator moving downward is $3000N$.
Additional Information: - Elevator has no acceleration means the acceleration of the elevator is zero then the elevator can be at stationary position, zero velocity or moving with a steady speed. When lift goes upward, then the apparent weight of the body is more than the actual weight. When lift goes downward, then the apparent weight of the body is less than the actual weight.

Note: The perpendicular force is equivalent to our possible weight. We feel slightly heavier than expected when the elevator accelerates up and lighter than expected when the acceleration is downward. When the elevator has zero acceleration then total force is equal to the weight of the body.