ATWOOD MACHINE

An Atwood machine requires treating the two masses as a single system connected by a string. The string is inextensible, so both masses have the same magnitude of acceleration a — one moves up while the other moves down. The net force driving the system comes from gravity acting on both masses, but acting in opposite directions because the string reverses the direction for one of them. The heavier side wins: F_net = (m₂ − m₁)g = (6 − 4) × 9.807 ≈ 19.61 N. Both masses resist the acceleration, so the total inertia is m_total = m₁ + m₂ = 10 kg. Newton's second law for the system gives a = F_net / m_total = 19.61 / 10 ≈ 1.96 m/s². To find the string tension, isolate m₁ (the rising mass). Two forces act on it: gravity m₁g downward, and tension T upward. Net force is upward at m₁a: T − m₁g = m₁a, so T = m₁(g + a) = 4 × (9.807 + 1.961) ≈ 47.1 N. Notice the tension is greater than m₁g (which would be 39.2 N) because it must accelerate the mass upward on top of supporting it against gravity. You can verify using m₂: m₂g − T = m₂a gives T = m₂(g − a) = 6 × (9.807 − 1.961) ≈ 47.1 N — identical, confirming the solution.