Atwood machine (two masses over a smooth pulley) If two bodies of masses M1 and M2 (with M1 > M2) are connected by a light, inextensible string passing over a smooth, frictionless pulley, what is the expression for the tension T in the string?

Introduction / Context:In classical mechanics, the Atwood machine illustrates motion with two different masses M1 and M2 connected by a light, inextensible string over a smooth pulley. The system demonstrates uniformly accelerated motion, tension distribution in a string, and Newton’s second law in connected-body problems. The quantity of interest here is the string tension T under ideal conditions (no pulley friction, massless string).

Given Data / Assumptions:

• M1 > M2 so the system accelerates with M1 moving downward and M2 moving upward.
• String is light (massless) and inextensible; pulley is smooth (frictionless) and massless.
• Acceleration due to gravity is g and is uniform.
• Tension is uniform throughout the string.

Concept / Approach:The connected bodies share the same magnitude of acceleration a due to the inextensible string. Apply Newton’s second law separately to each mass, write equations for forces along the line of motion, and solve simultaneously to eliminate a. The tension T must be less than M1 * g and greater than M2 * g to accelerate the system consistently with M1 > M2.

Step-by-Step Solution:

Verification / Alternative check:By symmetry, exchanging M1 and M2 should not change T except through the product M1 * M2 and the sum M1 + M2. The expression T = (2 * M1 * M2 * g)/(M1 + M2) satisfies this symmetry and fits limiting cases (e.g., if M1 ≈ M2, T → nearly M * g).

Why Other Options Are Wrong:

• (M1 + M2) * g / 2: Ignores dynamics; tension is not a simple average of weights.
• (M1 − M2) * g: This is related to net driving force, not tension.
• (M1 * M2 * g)/(M1 + M2): Misses the factor 2 that arises from combining the two equations.
• M1 * g: Would be the tension only if the system were static with equal masses.

Common Pitfalls:Incorrect sign conventions, assuming T equals the weight of one mass, or forgetting that both masses accelerate together. Also, including pulley inertia or string mass without stating it changes the result.

Final Answer:T = (2 * M1 * M2 * g) / (M1 + M2)