Systems & Connected Objects

You’ll be able to

Apply Newton’s second law to systems of objects linked by strings and pulleys
Solve Atwood machines and blocks connected over an ideal pulley
Recognize that objects joined by an inextensible string share the same magnitude of acceleration

Two ways to attack a system

When several objects are connected, there are two complementary tools. Treat the whole system as one body — the internal tension cancels — to find the shared acceleration quickly: a = (net external force)/(total mass). Then isolate a single object with its own free-body diagram to find an internal force such as the string tension. Use the system view for a, the single-body view for internal forces.

An inextensible string enforces a shared acceleration

If a string does not stretch, then whatever length feeds off one side is exactly the length arriving on the other. That geometric fact means the connected objects move with the same speed and the same magnitude of acceleration at every instant — even though their directions may differ (one goes up as the other goes down). An ideal massless string over a frictionless, massless pulley simply redirects the tension without changing its magnitude.

Atwood machine

a = (m₂ − m₁)g / (m₁ + m₂) T = 2 m₁ m₂ g / (m₁ + m₂)

Two masses hang over an ideal pulley. The heavier mass m₂ falls, the lighter m₁ rises, both with the same |a|. The tension is the same throughout the single string.

Worked example

An Atwood machine has masses of 3 kg and 5 kg hanging over a frictionless, massless pulley. Using g = 10 m/s², find the acceleration of the system and the tension in the string.

1.The two masses share |a|. Write ΣF = ma for each, taking the direction of actual motion as positive.
2.Heavier mass (5 kg, falling): m₂g − T = m₂a → 50 − T = 5a. Lighter mass (3 kg, rising): T − m₁g = m₁a → T − 30 = 3a.
3.Add the two equations to cancel T: 50 − 30 = (5 + 3)a → 20 = 8a → a = 2.5 m/s².
4.Back-substitute for tension: T = 30 + 3a = 30 + 3(2.5) = 37.5 N (check: 50 − 37.5 = 12.5 = 5 × 2.5 ✓).

Answer: a = (m₂ − m₁)g/(m₁ + m₂) = 2.5 m/s²; T = 37.5 N

Checkpoint

Two blocks are joined by an inextensible string that runs over an ideal pulley. What do the two blocks necessarily share at every instant?

Checkpoint

In an Atwood machine, masses of 2 kg and 6 kg hang over a frictionless, massless pulley (g = 10 m/s²). What is the magnitude of the acceleration of the system?

Tip

Use the system shortcut a = (net external force)/(total mass) to get the acceleration in one line, then isolate one block to find the tension. Adding the per-block equations is the fast way to cancel the unknown internal tension and solve for a.

Answer the 2 checkpoints as you read.