Elevator dynamics — if the cable tension when a lift moves upward is twice the tension when it moves downward, determine the lift’s acceleration (express your answer as a multiple of g).

Introduction / Context:
Elevator (lift) problems are classic applications of Newton’s second law. Cable tension differs depending on whether the car accelerates up or down. Relating two measured tensions allows you to solve for the unknown acceleration without needing the car’s mass explicitly.

Given Data / Assumptions:

• Tension upward case: T_up.
• Tension downward case: T_down.
• Given T_up = 2 * T_down.
• Lift mass = m, weight W = m * g.
• Same magnitude of acceleration a in both cases (one upward, one downward) over the short comparison interval.

Concept / Approach:

Apply Newton’s second law along the vertical for each motion state, taking upward as positive. When accelerating upward, tension must exceed weight; when accelerating downward, tension is less than weight. Two linear equations in T_up and T_down eliminate T and isolate a in terms of g.

Step-by-Step Solution:

Verification / Alternative check (if short method exists):

Check sign consistency: with a = g/3, T_up = W + m * g/3 = 4W/3 and T_down = W − m * g/3 = 2W/3; indeed T_up = 2 * T_down.

Why Other Options Are Wrong:

g/6 and g/4 produce T_up/T_down ratios different from 2; g/2 doubles the discrepancy; “none” is unnecessary because g/3 satisfies the given relation exactly.

Common Pitfalls (misconceptions, mistakes):

Mixing up direction of motion with direction of acceleration; forgetting that downward acceleration reduces cable tension.

Final Answer:

g/3