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Physics HL · Chapter 2: Forces and Newton's Laws

2.2 Equilibrium and Net Force

Translate free-body diagrams into axis equations and solve for unknowns in static and dynamic settings.

Estimated time: 28 minutes

Equilibrium as a Vector Condition

Equilibrium means the vector sum of all forces on an object is zero. This does not mean each force is zero; it means their combined effect cancels. An object in equilibrium can be at rest or moving in a straight line at constant speed. What matters is zero acceleration, not whether velocity is zero.

In one dimension, this is often direct addition with sign. In two dimensions, treat each axis separately: ΣFx = 0 and ΣFy = 0 for equilibrium. The method scales to complex setups with ropes and inclines because component equations preserve vector logic without requiring geometric guesswork.

Choosing Axes to Minimize Algebra

Axis choice is a strategic decision. On an incline, one axis parallel to the slope and one perpendicular usually keeps equations shortest. Along the slope, weight contributes mg sinθ; perpendicular to slope it contributes mg cosθ. This often turns difficult-looking diagrams into two clean equations with one unknown each.

When solving, delay arithmetic until symbolic structure is complete. Writing component equations first reduces sign errors and reveals whether your result is physically plausible. If a normal force solves negative, your assumed contact state is inconsistent and the model must be reconsidered.

Worked Logic: Incline Equilibrium and Breakaway Angle

On an incline, choose +x up the plane and +y normal to the plane. For a resting block, resolve weight into mg sinθ down the plane and mg cosθ into the plane. The normal equation gives R = mg cosθ. The parallel equation gives f_s = mg sinθ, where static friction points up the plane because it opposes impending downward motion.

At the largest angle that still allows rest, friction is at its limit: f_s,max = μsR. Substituting R = mg cosθ and f_s = mg sinθ gives μs = tanθ_max. This gives a practical interpretation of μs: it encodes the steepest no-slip incline. If the surface were frictionless, the same setup would have net force mg sinθ down the plane and acceleration g sinθ.

Fx=max,Fy=may\sum F_x = ma_x,\qquad \sum F_y = ma_y

Set ax or ay to zero only after the diagram and motion constraints justify equilibrium along that axis.

μs=tanθmax,africtionless incline=gsinθ\mu_s = \tan\theta_{\max},\qquad a_{\text{frictionless incline}} = g\sin\theta

The first relation comes from threshold equilibrium; the second from Newton's second law along the slope when friction is absent.

Simulation: Near-Threshold Equilibrium

Use a high-friction setup where static friction can just hold. Nudge applied force around the threshold and watch net force flip from zero to non-zero.

Net Force Lab

Free-body visualization

m = 8.0 kgN = 78.5 NW = mg = 78.5 NApplied: 31.0 NFriction: -31.0 NNet: 0.0 NScene updates from actual force balance, not just sliders.

Max static friction

31.39 N

Net force

0.00 N

Acceleration

0.000 m/s²

State

Static equilibrium

The free-body diagram shows static friction matching the applied pull exactly, so the net-force vector collapses and the block remains at rest.

Interpretation target: with m = 8 kg and μ = 0.40, the static ceiling is about μmg ≈ 31.4 N. At 31 N the model should remain stuck with near-zero net force; a tiny increase above that creates a clear acceleration. This is a direct numerical visualization of the static-friction inequality.

Test Yourself

A 6.0 kg crate is pulled horizontally by 12 N and moves at constant velocity. What is the horizontal friction force magnitude?

Hint: Constant velocity means zero horizontal acceleration, so horizontal forces must cancel.

Test Yourself

A block rests on a rough incline and starts to slip when the angle reaches 30°. Estimate μs.

Hint: Use μs = tanθmax at the threshold of slipping.