Elevator dynamics — comparing cable tensions A lift (mass m) moves with the same magnitude of acceleration a in two cases: upward and downward. If the cable tension while moving downward is half of the tension while moving upward, what is the acceleration of the lift?

Introduction / Context:Elevator problems illustrate Newton’s second law with attention to sign conventions. Cable tension depends on both gravity and acceleration direction, and comparing tensions in two scenarios allows solving for the unknown acceleration.

Given Data / Assumptions:

• Lift mass = m (constant).
• Upward motion with upward acceleration magnitude a → T_up = m(g + a).
• Downward motion with downward acceleration magnitude a → T_down = m(g − a).
• Given: T_down = (1/2) * T_up.

Concept / Approach:Apply Newton’s second law along the vertical with the chosen positive direction upward. Write expressions for tension in both cases and use the given ratio to solve for a in terms of g.

Step-by-Step Solution:

Verification / Alternative check:Numerical check with g = 9.9 m/s^2 (approx.): a = 3.3 m/s^2. Then T_up = m(13.2), T_down = m(6.6), indeed T_down = 0.5 T_up.

Why Other Options Are Wrong:

• g/2, g/4, 3g/4: Do not satisfy the given tension ratio when substituted.
• None of these: Incorrect because a consistent value exists: g/3.

Common Pitfalls:Using the same sign for acceleration in both cases or forgetting that T_down decreases when accelerating downward.

Final Answer:g/3