Unbalanced forces are exactly the opposite situation from Lesson 2.4's equilibrium: a configuration where the net force on a system is notzero. Whenever that happens, Newton's second law tells you precisely what the system's center of mass does in response.
Try the proportionality yourself below — a single net force acting on a single mass.
A single net force acting on a single mass. Adjust each — watch how acceleration responds proportionally to force and inversely to mass.
Try doubling the force alone — acceleration doubles. Try doubling the mass alone — acceleration is cut in half. That's the proportionality Newton's second law describes.
Newton's second law and first law are really two views of the same idea. A system's center-of-mass velocity changes if — and only if — a nonzero net external force acts on it. Set F(net) = 0 in a⃗ = F⃗(net)/m, and you get a⃗ = 0, which is exactly Newton's first law from Lesson 2.4.
The real payoff of this lesson is turning a free-body diagram into an actual equation — one F=ma equation per axis, per object. When two objects are connected by an ideal string, their accelerations share the same magnitude (the string doesn't stretch), which lets you solve both equations as a system.
Take the incline block (m₁) and hanging mass (m₃) from this unit's header art, connected by a string over an ideal pulley. Writing Newton's second law for each object separately, along its own most convenient axis:
Explore the full system live below — adjust both masses, the incline angle, and the friction coefficient, and watch the acceleration and tension update instantly.
The incline block (m₁) and hanging mass (m₃) from this unit's header, connected by an ideal string over an ideal pulley. Adjust the masses, incline angle, and friction coefficient — the acceleration and tension update from the full system of equations.
m₃ falls, m₁ slides up the incline
m₁ = 4 kg sits on a 30° incline with μₖ = 0.15, connected via an ideal string over an ideal pulley to a hanging mass m₃ = 5 kg. Find the system's acceleration and the string's tension.
Using the same setup, what would happen to the acceleration if the incline were frictionless (μₖ = 0) and the incline angle were 90° (a vertical wall, effectively making this a simple two-mass Atwood machine)?