Atwood-type system with unequal masses Two loads of 50 kg and 75 kg are hung at the ends of a light inextensible rope passing over a smooth, frictionless pulley. What is the tension in the rope (expressed in kilogram-force terms)?

Introduction / Context:This is a classic Atwood machine problem with unequal masses. With a massless rope and frictionless pulley, both masses share the same magnitude of acceleration, and the rope tension is the same on both sides. Working in kilogram-force (kgf) units simplifies numbers for many exam problems.

Given Data / Assumptions:

• m1 = 75 kg (heavier), m2 = 50 kg (lighter).
• Light inextensible rope; smooth, frictionless pulley.
• Use g consistently; when giving tension in kgf, effectively we quote T/g in kg units.

Concept / Approach:Apply Newton’s second law to each mass along the line of motion, taking the heavier mass moving downward and the lighter moving upward. Solve for the common acceleration first, then back-substitute to obtain the tension.

Step-by-Step Solution:

Verification / Alternative check:From the lighter side: T − m2 * g = m2 * a → T = m2 (g + a) = 50 g (1 + 0.2) = 60 g, confirming the same result.

Why Other Options Are Wrong:

• 50 kgf: Equals the lighter weight but tension must exceed it to accelerate upward.
• 75 kgf: Equals the heavier weight; not possible since the heavier mass accelerates downward.
• 25 kgf: Confuses net driving force with tension.
• 65 kgf: Numerical distractor; does not satisfy both equations of motion.

Common Pitfalls:Mixing kg, N, and kgf inconsistently; forgetting that the same tension acts on both sides; sign errors when writing equations.

Final Answer:60 kgf