Same Acceleration, Same Tension: Write One Equation per Body

Treat the system as a whole to find the acceleration, then isolate one body for the tension. Four blocks of 4, 3, 2 and 1 kg pulled along a smooth floor with a tension of 3 N in the cord that pulls the last two (3 kg together): a = 3/3 = 1 m/s², so the whole 10 kg needs F cos 60° = 10 N, and F = 20 N. For two masses joined by a spring, the spring force on one equals the spring force on the other: if the 10 kg mass accelerates at 4 m/s² the spring force is 40 N, so the 20 kg mass pulled by 200 N has a = (200 - 40)/20 = 8 m/s². On an incline of 60° pulled up by F = 120 N, three blocks (2, 4, 6 kg, total 12 kg) have a + g sinθ = F/12 = 10 m/s², so each tension equals the mass below it times 10: 20 N and 60 N.

With a 4 kg block hanging below a 6 kg block on one side and an 8 kg block on the other, a = (10 - 8) g / 18 = g/9 and the tension in the string between the 4 kg and 6 kg blocks is that of the lower block, 4 (g - a) = 320/9 N. A block of mass M on a frictionless incline pulling an equal horizontal block through a pulley accelerates at a = g sinθ / 2 and the tension is Mg sinθ / 2. Pulling the free end of a rope with a force F = 2mg gives acceleration g to a mass m, which is greater than the g/3 produced by hanging a 2m mass on the other end. The force that a clamp exerts on a pulley holding up a block of mass M adds the string tensions (Mg down and Mg sideways) and the pulley's weight mg, giving sqrt((M + m)² + M²) g.

If the ends of a string move down at speed v and the string makes an angle θ with the vertical at the mass M between them, the mass rises at v / cosθ, because only the string-direction component of its velocity equals the string's speed. When a string is cut, the spring force cannot change instantly while the string tension can. Blocks A (m) and B (3m) hang from a spring: the spring pulls up with 4mg, so right after the string between A and B is cut, A accelerates at (4mg - mg)/m = 3g upward and B falls with g.