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United StatesPhysics C: MechanicsSyllabus dot point

What does it mean for an object to be in translational equilibrium, and how do we use the condition of zero net force to solve for unknown forces?

Topic 2.4 Newton's First Law: state the law of inertia, define translational equilibrium as zero net force, and apply the equilibrium conditions to find unknown forces.

A focused answer to AP Physics C: Mechanics Topic 2.4, covering Newton's first law and inertia, the meaning of translational equilibrium as zero net force, the difference between mass and weight, and applying the equilibrium conditions axis by axis to find unknown forces, with worked examples.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. The law of inertia
  3. Translational equilibrium
  4. Mass versus weight
  5. Try this

What this topic is asking

The College Board (Topic 2.4) wants you to state Newton's first law (the law of inertia), to define translational equilibrium as the condition of zero net force, and to apply that condition to find unknown forces. Equilibrium problems, hanging signs, blocks held on inclines, objects at constant velocity, are a staple of the exam, and they all reduce to setting the net force to zero on each axis.

The law of inertia

The first law overturns the everyday intuition that motion needs a sustaining force. A puck sliding on frictionless ice keeps going at constant velocity with no forward push; it slows on a real floor only because friction supplies a net backward force. The law also defines an inertial reference frame as one in which it holds, the frames used throughout the course. Mass is the measure of inertia: a more massive object resists a given change in velocity more strongly.

Translational equilibrium

When the net force on an object is zero, it is in translational equilibrium and its acceleration is zero, so it is at rest or moving at constant velocity. The condition is a vector equation that splits into one scalar equation per axis:

∑F⃗=0⟺∑Fx=0   and   ∑Fy=0.\sum \vec{F} = 0 \quad\Longleftrightarrow\quad \sum F_x = 0 \;\text{ and }\; \sum F_y = 0.

This is the engine of equilibrium problems. Draw the free-body diagram, choose axes, resolve every force into components, and set the sum on each axis to zero. The result is a pair of equations you solve for the unknowns (cable tensions, the normal force, an applied force). Equilibrium does not mean motionless: an elevator rising at constant speed and a car cruising on the highway are both in equilibrium.

Mass versus weight

A recurring distinction is between mass and weight. Mass mm is the amount of inertia, measured in kilograms, and is the same everywhere. Weight W=mgW = mg is the gravitational force on the object, measured in newtons, and changes with the local gravitational field strength gg: an astronaut's mass is unchanged on the Moon, but her weight is about one sixth of its Earth value because the Moon's gg is smaller. In equilibrium on Earth, the weight is the downward force you balance against tensions and normal forces.

Try this

Q1. A 5050 N crate slides at constant velocity across a floor under a horizontal pull. State the friction force on it. [2 points]

  • Cue. Constant velocity means equilibrium, so the friction equals the applied pull in magnitude and opposes it; if the pull is, say, 5050 N, friction is 5050 N backward.

Q2. Explain the difference between the mass and the weight of a 1010 kg object taken to the Moon. [2 points]

  • Cue. Its mass stays 1010 kg (intrinsic inertia); its weight W=mgW = mg drops because the Moon's gg is about one sixth of Earth's.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)4 marksSection II (short FRQ). A 2020 kg traffic light hangs from a junction of two cables. One cable runs horizontally to a wall; the other runs up to a support at 40∘40^\circ above the horizontal. Take g=9.8g = 9.8 m/s squared. (a) Draw a free-body diagram of the junction. (b) Apply the equilibrium conditions. (c) Determine the tension in each cable.
Show worked answer →

A 4-point equilibrium FRQ.

(a) Diagram (1 point): three forces at the junction - the weight of the light W=mg=196W = mg = 196 N straight down, the horizontal cable tension T1T_1, and the angled cable tension T2T_2 at 40∘40^\circ.
(b) Equilibrium (1 point): ∑Fx=0\sum F_x = 0 and ∑Fy=0\sum F_y = 0 since the junction is at rest.
(c) Solve (2 points): vertical: T2sin⁡40∘−196=0⇒T2=196sin⁡40∘=305T_2\sin 40^\circ - 196 = 0 \Rightarrow T_2 = \dfrac{196}{\sin 40^\circ} = 305 N. Horizontal: T2cos⁡40∘−T1=0⇒T1=305cos⁡40∘=234T_2\cos 40^\circ - T_1 = 0 \Rightarrow T_1 = 305\cos 40^\circ = 234 N.

Markers reward applying both equilibrium equations and solving the vertical one first to get T2T_2.

AP 2020 (style)1 marksSection I (multiple choice). An object moves in a straight line at constant velocity. Which statement must be true? (A) a single force acts on it (B) the net force on it is zero (C) no forces act on it (D) it is at rest. Justify your reasoning.
Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

By Newton's first law, constant velocity (including zero velocity) means zero net force, but not necessarily zero forces: several forces can act and balance to a zero resultant. A book sliding at constant speed has gravity, normal force and friction all acting, summing to zero. Choice (C) is wrong because balanced forces are allowed; (D) is wrong because constant velocity need not be rest. The trap is to equate zero net force with no forces.

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