AP Physics C: Mechanics · Topic 2.4

Topic 2.4: Newton's First Law

Unit 2: Force and Translational Dynamics20-25% of the multiple-choice section

If the net force on a system is zero, its velocity stays constant. That covers an object at rest and one moving at steady speed in a straight line equally. AP Physics C states this exactly as AP Physics 1 does; what changes is the expected answer, from a number to an expression.

AP Physics: Unit 2 (topics 2.4 Newton's First Law). AP Physics C: Mechanics Unit 2, Topic 2.4. One learning objective, 2.4.A, describe the conditions under which a system's velocity remains constant. Supported by 2.4.A.1 (the net force on a system is the vector sum of all forces exerted on the system), 2.4.A.2 (translational equilibrium is the configuration of forces such that the net force exerted on a system is zero, with the DERIVED equation sum of F_i = 0), 2.4.A.3 (Newton's first law: if the net force on a system is zero, the velocity of that system will remain constant), 2.4.A.4 (forces may be balanced in one dimension but unbalanced in another; the system's velocity will change only in the direction of the unbalanced force) and 2.4.A.5 (an inertial reference frame is one from which an observer would verify Newton's first law of motion). No boundary statement. This topic's content is IDENTICAL to AP Physics 1 Topic 2.4: same single objective, same five statements, same derived equation, no boundary statement in either, and no calculus. Suggested skills are 1.C, 2.B, 2.C and 3.C. The derived equation is NOT printed on the AP Physics C: Mechanics equation sheet; the printed second-law line a_sys = sum F / m_sys is what you set to zero. The CED's sample multiple-choice question 6 aligns to 2.4.A and 2.4.A.1 with skill 2.A, answer B, and is a fully symbolic two-wire equilibrium problem. Sample free-response question 2 also aligns to 2.4.A. NOTE a CED defect: that same alignment table lists 2.4.B for question 2, but no learning objective 2.4.B exists anywhere in the framework and the string appears nowhere else in the document. The exam-conventions box states that the frame of reference of any problem is assumed to be inertial unless otherwise stated.

The content is identical, and the exam question is not

Topic 2.4 is the second of two places in Unit 2 where AP Physics C: Mechanics and AP Physics 1 say exactly the same thing. Both frameworks have one learning objective, 2.4.A, with five essential-knowledge statements numbered 2.4.A.1 through 2.4.A.5, and both print the same derived equation. Neither course prints a boundary statement on this topic. There is no derivative and no integral anywhere in it.

So do not go hunting for a calculus angle. Spend the time on the thing that actually differs, which is what the exam does with the equilibrium condition.

The evidence is in the CED's own sample material. Sample multiple-choice question 6 aligns to learning objective 2.4.A and essential knowledge 2.4.A.1, and the skill it is tagged with is 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway. The question gives a heavy sign of mass $M$ held at rest by two supporting wires between two buildings, each wire making the same angle $\theta$ with the vertical, and asks for the tension in each wire. All four answer choices are expressions in $M$ , $g$ and $\theta$ . There is no number in the question and no number in the answer.

That is the shape to practise. On the AP Physics C: Mechanics exam skill 2.A carries 25 to 30 percent of the multiple-choice section, the largest band the CED lists for any single skill, and equilibrium problems are among the cleanest places to test it, because the two component equations close on themselves and the mass often cancels.

What the CED requires of Topic 2.4

One learning objective, and the suggested skills printed beside it are 1.C, 2.B, 2.C and 3.C.

Objective 2.4.A: describe the conditions under which a system's velocity remains constant.

2.4.A.1: the net force on a system is the vector sum of all forces exerted on the system.
2.4.A.2: translational equilibrium is the configuration of forces such that the net force exerted on a system is zero. Its derived equation is

\sum \vec{F}_i = 0

2.4.A.3: Newton's first law states that if the net force exerted on a system is zero, the velocity of that system will remain constant.
2.4.A.4: forces may be balanced in one dimension but unbalanced in another. The system's velocity will change only in the direction of the unbalanced force.
2.4.A.5: an inertial reference frame is one from which an observer would verify Newton's first law of motion.

No boundary statement. Five statements, one equation, and that equation is labelled a derived equation, which the CED defines on its Required Equations page as one of the equations "provided for reference and guidance, or to demonstrate the final results of derivations expected of students on the exam". Derived equations are not printed on the exam's equation sheet, and $\sum \vec{F}_i = 0$ is not on the AP Physics C: Mechanics sheet. You are expected to get there from $\vec{a}_{\text{sys}} = \sum \vec{F}/m_{\text{sys}}$ , which is printed, by setting the acceleration to zero.

One curiosity in the CED worth flagging, because it will confuse anyone cross-checking. The answer key that aligns the sample free-response questions lists question 2 against seven learning objectives, and one of them is written "2.4.B". There is no learning objective 2.4.B in the framework: Topic 2.4 prints 2.4.A and nothing else, and the string "2.4.B" appears nowhere else in the document. Treat it as a typographical slip in the alignment table, not as content you are missing.

Equilibrium is a statement about velocity, not about rest

Statement 2.4.A.3 says that zero net force means constant velocity. Constant velocity includes zero velocity as a special case, and it also includes 30 m/s due north forever. Both are equilibrium; neither needs a force to sustain it.

That is the point the law was written to make. Aristotelian intuition says motion needs a cause, and the first law says only a change in motion needs a cause. A puck sliding on frictionless ice keeps its velocity because nothing is unbalancing it, not because something is pushing it along. A car at a steady 70 km/h on a motorway is in equilibrium: the engine's forward drive exactly balances drag and rolling resistance, and the net force is zero even though the engine is working hard.

Statement 2.4.A.2 defines the configuration rather than the outcome: translational equilibrium is the configuration of forces such that the net force is zero. Notice the word translational. A system can have zero net force and still be angularly accelerating, because torques can sum to something nonzero even when forces sum to zero. Two equal and opposite forces applied at different points do exactly that. Rotational equilibrium is a separate condition and it belongs to Unit 5.

In practice $\sum \vec{F}_i = 0$ is not one equation but one per axis. In two dimensions you get

\sum F_x = 0 \qquad \text{and} \qquad \sum F_y = 0

and two equations mean you can solve for two unknowns. That is why the standard equilibrium problem gives you a weight and two angles and asks for two tensions: the count works out exactly.

Balanced on one axis, unbalanced on another

Statement 2.4.A.4 is the one students skip and then lose marks to: forces may be balanced in one dimension but unbalanced in another, and the system's velocity will change only in the direction of the unbalanced force.

This is the reason you resolve into components at all. Equilibrium is not a property of the whole force set that you can eyeball; it is a property of each axis separately, and one axis can be in equilibrium while the other is not.

The cases where it matters:

A block dragged across a floor by a rope at an angle. Vertically, the normal force, the gravitational force and the rope's upward component balance, so there is no vertical acceleration. Horizontally, they do not, so the block accelerates. The vertical equation is still worth writing, because it is what gives you the normal force and therefore the friction.
A projectile. The horizontal direction has no force at all once the object is in flight, so horizontal velocity is constant, which is the first law doing its work mid-problem. The vertical direction has gravity, so vertical velocity changes. Same object, two different answers, one per axis.
A block on a ramp. Perpendicular to the surface, the components balance and give you the normal force. Along the surface, they generally do not.
Uniform circular motion. The net force is not zero, but its component along the velocity is. Speed is constant while velocity is not, which is Topic 2.10's territory.

The operational rule: write both component equations every time, even when you only need one of them. The one you did not need is usually where the normal force is hiding.

Inertial reference frames, and the exam's default

Statement 2.4.A.5 gives a definition that is circular on purpose: an inertial reference frame is one from which an observer would verify Newton's first law of motion. The first law is the test for the frame, and the frame is where the first law works.

That is not a dodge. It says there is no absolute standard of rest, only frames in which the bookkeeping comes out right. A frame moving at constant velocity relative to an inertial frame is also inertial. A frame that is accelerating, including one that is rotating, is not, and in it objects appear to accelerate with no force you can attribute to a second object, which breaks the 2.2.A.1.i test.

Two places this shows up.

The exam's stated default. The exam-conventions box printed with the AP Physics C: Mechanics Table of Information has three bullets, and the first is: "The frame of reference of any problem is assumed to be inertial unless otherwise stated." The appendix "Vocabulary and Definitions of Important Ideas in AP Physics" repeats it in its own list of what students may assume unless otherwise stated, alongside negligible air resistance, negligible drag forces, negligible edge effects of charged plates, and ideal strings, springs and pulleys.

The equivalence principle. Statement 2.6.C.4, one topic later, states that an observer in a noninertial reference frame is unable to distinguish between an object's apparent weight and the gravitational force exerted on the object by a gravitational field. That is why a person in an accelerating lift feels heavier: in their own noninertial frame the effect is indistinguishable from a change in gravity. Topic 2.6 is where apparent weight is developed.

A rotating frame is the case worth naming because it is where invented forces come from. Sitting in a car going round a bend, you feel thrown outward, and in the car's rotating frame you would have to invent an outward force to explain it. From the road, an inertial frame, nothing is pushing you outward at all; the door is pushing you inward, and your body is going as straight as the door will let it. The centripetal versus centrifugal force comparison covers that distinction in full.

Traps this topic sets

Reading "no net force" as "no forces". A book on a table has two forces on it and is in equilibrium. Statement 2.4.A.2 says the configuration of forces sums to zero, not that the set is empty.

Assuming equilibrium means at rest. Constant velocity is equilibrium. A question that says "moving at constant speed in a straight line" has just told you the net force is zero, and that is usually the whole solution.

Treating equilibrium as a single equation. It is one equation per axis. A problem with two unknowns needs both.

Forgetting that constant speed on a curve is not equilibrium. The velocity vector is changing direction, so it is not constant, so the net force is not zero. The first law is about velocity, and velocity is a vector.

Assuming the tensions in two support wires are equal. They are equal only if the angles are equal. The first worked example below derives the general case, and the tensions come out in the ratio of the sines of the opposite angles, which is the opposite of most people's first guess.

Adding magnitudes instead of components. Two 10 N forces at $60^\circ$ to each other do not give 20 N. Resolve, add per axis, then recombine.

If you want the algebra-based version of this topic

The physics is the same in both courses, so pick by which exam you are sitting. If you are taking AP Physics 1, read AP Physics 1 Topic 2.4: Newton's First Law, which covers the same five statements with algebra-based worked examples. If you are taking AP Physics C: Mechanics, stay here for the symbolic framing and the exam context.

	AP Physics 1 Topic 2.4	AP Physics C Topic 2.4
Objectives	2.4.A	2.4.A
Essential-knowledge statements	2.4.A.1 to 2.4.A.5	2.4.A.1 to 2.4.A.5
Derived equation	$\sum \vec{F}_i = 0$	$\sum \vec{F}_i = 0$
Boundary statement	none	none
Calculus involved	none	none
Typical expected answer	a number	a symbolic expression

This is a genuinely shared topic and there is no point pretending otherwise. What is worth your time in the C course is finishing every equilibrium problem in symbols before you touch a calculator, because that is what the CED's own sample multiple-choice question asks for.

Related on this site: the Newton's first law glossary entry, the equilibrium glossary entry, the net force guide and the tension guide for the routine, and the net force calculator for checking a vector sum.

How Topic 2.4 is tested

Unit 2 is 20 to 25 percent of the multiple-choice section across about 15 to 25 class periods, the highest minimum weighting of any unit in AP Physics C: Mechanics (only Unit 3, at 15 to 25 percent, reaches the same 25 percent ceiling). Topic 2.4 contributes one of the unit's nineteen learning objectives, and it is one of the two the CED's sample multiple-choice set actually uses.

The suggested skills are 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of the physical system; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.C, compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Skill 1.C is the graph-sketching one, and it points at the representation this topic owns. On a velocity-versus-time graph, equilibrium is a horizontal line, at any height including zero. On a position-versus-time graph it is a straight line of any slope. On an acceleration-versus-time graph it is the line $a = 0$ . A question that shows you a velocity graph with a flat section is telling you the net force is zero over that interval, and a question that asks you to sketch the motion of an object with no net force wants a straight line, not a flat one, unless it also started at rest.

The CED's sample free-response question 2, Translation Between Representations, worth 12 points, aligns to 2.4.A among its seven learning objectives, and its skills are 1.A, 3.B, 2.A, 1.C, 2.D and 3.C. Combined with sample multiple-choice question 6, that puts Topic 2.4 in both halves of the CED's sample exam.

The unit's "Preparing for the AP Exam" note ties Unit 2 to question four of the free-response section, the Qualitative/Quantitative Translation, and warns that students exposed primarily to numerical problem solving often struggle with it because it requires expressing a conceptual understanding of course content and representations. Equilibrium arguments are close to the centre of that question type. Practice on the surrounding material is in the forces and Newton's laws practice set.

Two support wires at unequal angles, done symbolically

A sign of mass $M = 12$ kg hangs at rest from two wires attached to buildings on either side. The left wire makes an angle $\theta_1 = 35^\circ$ with the vertical and the right wire makes $\theta_2 = 50^\circ$ with the vertical. Derive expressions for the two tensions, then evaluate them with $g = 9.8\ \mathrm{m/s^2}$ .

Draw the free-body diagram of the sign: three arrows on one dot, the gravitational force straight down and the two tensions along their wires. Following the Topic 2.2 boundary statement, no components go on the diagram.
Declare axes: $x$ positive to the right, $y$ positive up. The acceleration is zero in both directions, so use 2.4.A.2 twice.
Horizontal equation. The left wire pulls up and to the left, the right wire up and to the right, and both angles are measured from the vertical, so the horizontal components are $T_1\sin\theta_1$ and $T_2\sin\theta_2$ : $T_2\sin\theta_2 - T_1\sin\theta_1 = 0$ .
Vertical equation: $T_1\cos\theta_1 + T_2\cos\theta_2 - Mg = 0$ .
Solve in symbols. From the horizontal equation, $T_2 = T_1\dfrac{\sin\theta_1}{\sin\theta_2}$ . Substituting into the vertical equation gives $T_1\left(\cos\theta_1 + \dfrac{\sin\theta_1\cos\theta_2}{\sin\theta_2}\right) = Mg$ .
Put the bracket over a common denominator: $T_1\dfrac{\cos\theta_1\sin\theta_2 + \sin\theta_1\cos\theta_2}{\sin\theta_2} = Mg$ . The numerator is the sine addition identity, so it collapses to $\sin(\theta_1 + \theta_2)$ .
The symbolic answers, which is where an AP Physics C question stops: $T_1 = \dfrac{Mg\sin\theta_2}{\sin(\theta_1+\theta_2)}$ and, by the same route, $T_2 = \dfrac{Mg\sin\theta_1}{\sin(\theta_1+\theta_2)}$ .
Read the structure before computing. Each tension is proportional to the sine of the other wire's angle, so the wire nearer to vertical carries more load. Set $\theta_1 = \theta_2 = \theta$ and both reduce to $Mg/(2\cos\theta)$ , the familiar symmetric result. Let $\theta_1 + \theta_2 \to 180^\circ$ , meaning both wires approach horizontal, and the denominator goes to zero and both tensions blow up, which is why a clothesline cannot be pulled straight.
Now the numbers. $Mg = (12)(9.8) = 117.6$ N, $\theta_1 + \theta_2 = 85^\circ$ , $\sin 85^\circ = 0.9962$ , $\sin 50^\circ = 0.7660$ , $\sin 35^\circ = 0.5736$ .
$T_1 = 117.6(0.7660)/0.9962 = 90.09/0.9962 = 90.4$ N. $T_2 = 117.6(0.5736)/0.9962 = 67.45/0.9962 = 67.7$ N.
Check both equations. Horizontal: $90.4\sin 35^\circ = 51.9$ N and $67.7\sin 50^\circ = 51.9$ N, equal. Vertical: $90.4\cos 35^\circ + 67.7\cos 50^\circ = 74.1 + 43.5 = 117.6$ N, which is $Mg$ .

$T_1 = \dfrac{Mg\sin\theta_2}{\sin(\theta_1+\theta_2)} = 90.4$ N in the left wire and $T_2 = \dfrac{Mg\sin\theta_1}{\sin(\theta_1+\theta_2)} = 67.7$ N in the right wire. Each tension goes with the sine of the other wire's angle, so the wire closer to vertical carries the larger share.

Balanced vertically, unbalanced horizontally

A $5.0$ kg block sits on a frictionless horizontal floor. A rope pulls it with tension $T = 20$ N at $\theta = 30^\circ$ above the horizontal. Find the normal force and the acceleration. Then find the tension that would be needed at that same angle to lift the block off the floor.

This is statement 2.4.A.4 in one picture: the vertical direction is in equilibrium and the horizontal direction is not. Handle each axis separately.
Declare axes: $x$ horizontal, positive in the direction of the pull; $y$ vertical, positive up. Three forces: gravity down, normal force up, tension at $30^\circ$ .
Vertical. The block stays on the floor, so $a_y = 0$ and the vertical components sum to zero: $F_N + T\sin\theta - mg = 0$ , giving $F_N = mg - T\sin\theta$ .
Numerically, $F_N = (5.0)(9.8) - 20(0.500) = 49 - 10 = 39$ N. Note that this is not $mg$ : the rope is carrying 10 N of the block's weight, so the floor carries only 39 N.
Horizontal. There is no friction and no other horizontal force, so the tension's horizontal component is the net force: $ma_x = T\cos\theta$ , giving $a_x = \dfrac{T\cos\theta}{m}$ .
Numerically, $a_x = 20(0.8660)/5.0 = 17.32/5.0 = 3.46\ \mathrm{m/s^2}$ , horizontal and in the direction of the pull.
Notice what would go wrong without splitting the axes. The magnitude of the net force is 17.32 N, all of it horizontal, and if you had used the full 20 N you would have got $4.0\ \mathrm{m/s^2}$ , about 15 percent too large.
Now the lift-off condition. The block leaves the floor when the normal force reaches zero, so $T\sin\theta = mg$ , giving $T = \dfrac{mg}{\sin\theta}$ .
At $\theta = 30^\circ$ that needs $T = 49/0.500 = 98$ N, nearly five times the tension in the original problem. With the original 20 N rope it is impossible at any angle, because even straight up, $\sin\theta = 1$ , gives only 20 N against a 49 N weight.
One more reading of the same expression. If a rope could supply 60 N, lift-off would come at $\sin\theta = 49/60 = 0.8167$ , so $\theta = 54.8^\circ$ . Above that angle the block leaves the floor; below it, the floor still pushes.

$F_N = mg - T\sin\theta = 39$ N and $a_x = T\cos\theta/m = 3.46\ \mathrm{m/s^2}$ horizontally. Lift-off needs $T = mg/\sin\theta = 98$ N at $30^\circ$ , so a 20 N rope cannot lift this block at any angle.

Frequently asked questions

What is Newton's first law in AP Physics C?

Essential knowledge 2.4.A.3 of the AP Physics C: Mechanics course and exam description states that if the net force exerted on a system is zero, the velocity of that system will remain constant. Constant velocity includes the case of zero velocity, so the law covers an object at rest and one moving in a straight line at steady speed with the same sentence. Essential knowledge 2.4.A.2 gives the associated configuration, translational equilibrium, and its derived equation, the sum of all forces equals zero. The statement is word for word the same as the AP Physics 1 version of 2.4.A.3.

Is the equilibrium equation on the AP Physics C equation sheet?

No. The sum of all forces equals zero is labelled a derived equation in essential knowledge 2.4.A.2, and the CED's Required Equations page explains that derived equations are provided for reference and guidance, or to demonstrate the final results of derivations expected of students on the exam, and that not all equations in the framework appear on the sheet. Checking the AP Physics C: Mechanics Table of Information confirms it is absent. What is printed is the acceleration of a system equals the net force divided by the system mass, from which you get equilibrium by setting the acceleration to zero.

Does Newton's first law mean an object at rest stays at rest?

That is only half of it. Essential knowledge 2.4.A.3 says the velocity remains constant, and constant velocity covers both an object at rest and one moving in a straight line at steady speed. A hockey puck sliding on frictionless ice is in exactly the same state as a book on a table, as far as the first law is concerned. This matters on the exam, because a question that describes an object moving at constant speed in a straight line has told you the net force is zero, and that is usually the fact the whole question turns on.

Can forces be balanced in one direction and unbalanced in another?

Yes, and essential knowledge 2.4.A.4 says so explicitly: forces may be balanced in one dimension but unbalanced in another, and the system's velocity will change only in the direction of the unbalanced force. A block dragged by a rope at an angle across a floor is the standard case. Vertically, the normal force, the weight and the rope's vertical component sum to zero, so there is no vertical acceleration. Horizontally they do not, so the block accelerates. This is why translational equilibrium is one equation per axis rather than a single equation.

What is an inertial reference frame?

Essential knowledge 2.4.A.5 defines an inertial reference frame as one from which an observer would verify Newton's first law of motion. In practice that means a frame that is not accelerating: any frame moving at constant velocity relative to another inertial frame is also inertial, while an accelerating or rotating frame is not. The AP Physics C: Mechanics exam conventions state that the frame of reference of any problem is assumed to be inertial unless otherwise stated, so unless a question says you are in a rotating or accelerating frame, ordinary Newtonian bookkeeping applies.

Are the tensions in two support wires always equal?

Only when the two wires make equal angles. For a sign of mass M hung from wires at angles theta one and theta two from the vertical, applying translational equilibrium in both directions gives the tension in the first wire as Mg times the sine of theta two divided by the sine of the sum of the two angles, with the roles swapped for the second wire. Each tension goes with the sine of the other wire's angle, so the wire closer to vertical carries more of the load. When the angles are equal, both reduce to Mg divided by twice the cosine of that angle.

Is there a learning objective 2.4.B in AP Physics C Mechanics?

No. Topic 2.4 in the AP Physics C: Mechanics course and exam description prints exactly one learning objective, 2.4.A, with five essential knowledge statements. The string 2.4.B does appear once in the document, in the table that aligns the sample free-response questions to the framework, where question 2 is listed against seven objectives including 2.4.B. Since no such objective exists anywhere in the course framework, that entry is best read as a typographical slip in the alignment table rather than as content that is missing from the unit.