In an Atwood machine with masses m1 and m2 connected by a rope over a frictionless pulley, the acceleration is given by which expression?

In an Atwood machine with masses m1 and m2 connected by a rope over a frictionless pulley, the acceleration is given by which expression?

• A. The acceleration depends on the sum m1 + m2 and g
• B. a = (m1 - m2) g / (m1 + m2)
• C. a = (m1 + m2) g / (m1 - m2)
• D. a = g / (m1 - m2)

Answer: (B)

When two masses are connected by a rope over a frictionless pulley, the system accelerates because the heavier mass pulls harder due to gravity. Apply Newton’s second law to each mass with a single rope and no friction: for the heavier mass, the downward gravitational force minus the rope tension equals m1 a; for the lighter mass, the rope tension minus its weight equals m2 a. Adding these two equations eliminates the tension and gives (m1 − m2) g = (m1 + m2) a. Solve for the acceleration: a = (m1 − m2) g / (m1 + m2). The direction is downward for the heavier mass and upward for the lighter mass. This shows the acceleration depends on the weight difference over the total mass, not on the sum in the numerator or on any other form.

When two masses are connected by a rope over a frictionless pulley, the system accelerates because the heavier mass pulls harder due to gravity. Apply Newton’s second law to each mass with a single rope and no friction: for the heavier mass, the downward gravitational force minus the rope tension equals m1 a; for the lighter mass, the rope tension minus its weight equals m2 a. Adding these two equations eliminates the tension and gives (m1 − m2) g = (m1 + m2) a. Solve for the acceleration: a = (m1 − m2) g / (m1 + m2). The direction is downward for the heavier mass and upward for the lighter mass. This shows the acceleration depends on the weight difference over the total mass, not on the sum in the numerator or on any other form.