AP Physics 1 · Topic 2.4

Topic 2.4: Newton's First Law

Unit 2: Force and Translational Dynamics18-23% of the multiple-choice section

Newton's first law states that if the net force exerted on a system is zero, the velocity of that system remains constant. That one condition covers both an object at rest and one moving in a straight line at steady speed. Forces can balance on one axis while staying unbalanced on another.

AP Physics: Unit 2 (topics 2.4 Newton's First Law). AP Physics 1 Unit 2, Topic 2.4, covering learning objective 2.4.A, describe the conditions under which a system's velocity remains constant, and essential knowledge 2.4.A.1 through 2.4.A.5. The CED lists four suggested skills for this topic: 1.C, 2.A, 3.B, and 3.C. Unit 2 is weighted at 18 to 23 percent of the multiple-choice section and about 22 to 27 class periods, tied with Unit 3 for the largest share in the course.

What Newton's first law actually claims

The CED puts it in a single sentence: Newton's first law states that if the net force exerted on a system is zero, the velocity of that system will remain constant.

Three words in that sentence carry the load.

Net force. The net force on a system is the vector sum of all forces exerted on the system, so zero net force does not mean no forces. A book resting on a table has two forces on it and a net force of zero.
Velocity. Constant velocity means constant magnitude and constant direction. A skater carving a turn at a steady 6.0 m/s fails the test, because the direction of her velocity is changing, so something has to be pushing her sideways.
System. Topic 2.1 lets you draw the boundary where you like: one block, or two blocks plus the string between them. The law applies to whatever you enclosed.

The contrapositive is the version most questions use. If the velocity of a system is changing, the net force on it is not zero. That is Topic 2.5, and the two learning objectives are written as deliberately complementary sentences: 2.4.A asks for the conditions under which a system's velocity remains constant, and 2.5.A asks for the conditions under which it changes.

Translational equilibrium and the sum that has to vanish

Translational equilibrium is a configuration of forces such that the net force exerted on a system is zero. The CED writes it as a derived equation:

\sum_i \vec{F}_i = 0

Two features of that line are worth pinning down. First, it is derived rather than printed: it is not one of the equations on the AP Physics 1 equation sheet. What the sheet gives you is the second law solved for acceleration, $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$ , and setting the acceleration to zero forces the numerator to zero. You are expected to produce the equilibrium condition from that in one step.

Second, a vector equation is shorthand for one scalar equation per axis:

\sum F_x = 0 \qquad \sum F_y = 0

Both have to hold. A crate whose vertical forces cancel perfectly while a rope drags it sideways is not in translational equilibrium.

Nothing here says the system is at rest. Static equilibrium (sitting still) and dynamic equilibrium (gliding at constant velocity) satisfy the same equation and are solved with the same algebra. The free-body diagram builder is a quick way to check whether a set of arrows really does sum to zero.

Balanced on one axis, unbalanced on another

EK 2.4.A.4 is the statement that turns the first law into a working tool: 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.

So the law is applied axis by axis, not to the object as a whole.

Situation	Balanced axis	Unbalanced axis
Crate dragged across a level floor	Vertical: normal force against gravity	Horizontal: pull against friction
Ball in flight, air resistance negligible	Horizontal: no forces at all	Vertical: gravity alone
Car cruising a straight highway at fixed speed	Both axes balanced	None

The middle row is the whole logic of projectile motion. Because nothing acts horizontally, the horizontal velocity component is protected by the first law and holds its value for the entire flight, while the vertical component changes. Topic 1.5 treats the two axes separately for exactly this reason.

An axis with no forces on it at all counts as balanced: the sum of nothing is zero.

Why constant motion needs no force

The everyday intuition is that motion needs a cause, because in ordinary life you have to keep pushing a box or it stops. The first law says the opposite: a change in velocity needs a cause. The box stops because friction is a real unbalanced force, not because motion runs out.

Test the claim by removing the friction. A puck on smooth ice slides a long way with barely any slowing, and a spacecraft coasting between planets holds its velocity for years. The better the forces are cancelled, the longer the velocity lasts, which is what the law predicts.

That is why zero net force and zero velocity have to be kept apart. All four combinations occur.

Zero net force, zero velocity: a book on a table.
Zero net force, nonzero velocity: a hockey puck gliding on ice.
Nonzero net force, zero velocity: a ball at the very top of its flight, momentarily at rest with gravity still acting.
Nonzero net force, nonzero velocity: everything else.

The third line is the one worth rehearsing, and it is the reason to look at the forces rather than at the motion when a question asks whether a net force exists.

Inertial reference frames

EK 2.4.A.5 defines the frames in which all of this is allowed: an inertial reference frame is one from which an observer would verify Newton's first law of motion.

The definition looks circular and is not. It is a test you run. Watch an object that has no net force on it. If it keeps a constant velocity as seen from where you are standing, your frame is inertial; if it appears to accelerate on its own, your frame is not. A passenger in a braking bus sees a loose bottle accelerate forward with nothing pushing it, so the bus is a noninertial frame while it brakes.

Two consequences matter. Newton's second law is written for inertial frames, so a frame that fails the first law test invalidates the force analysis built on it. And the AP Physics 1 Table of Information states the exam convention outright: the frame of reference of any problem is assumed to be inertial unless otherwise stated. You may assume it, but you should know what you are assuming.

Topic 1.4 supplies the other half. Acceleration is the same as measured from all inertial reference frames, so inertial observers who disagree about velocities still agree about forces.

What zero net force looks like on a graph

Suggested skill 1.C for this topic is sketching qualitative graphs, so be ready to draw constant velocity three ways.

Graph	Shape when the net force is zero
Position vs time	A straight line, sloping if the system is moving, horizontal if it is at rest
Velocity vs time	A horizontal line at the constant velocity, which may be positive, negative, or zero
Acceleration vs time	A horizontal line lying on the time axis

Read those against the exact graph rules from Topic 1.3. The slope of a position vs time graph is the velocity, the slope of a velocity vs time graph is the acceleration, the area under a velocity vs time graph is the displacement, and the area under an acceleration vs time graph is the change in velocity.

The velocity graph is the useful test. Zero net force means it is flat, at whatever height. A flat line sitting well above the axis still describes a system in equilibrium, which is dynamic equilibrium drawn out. A velocity graph with any slope at all, however slight, means the net force is not zero.

How Topic 2.4 is tested, and what to do next

Four suggested skills are attached to Topic 2.4 in the CED: 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. In the CED's numbering, 1.C belongs to Practice 1, Creating Representations; 2.A to Practice 2, Mathematical Routines; and 3.B and 3.C to Practice 3, Scientific Questioning and Argumentation.

Read the list as a study plan. Sketch the three graphs, derive the equilibrium condition symbolically instead of memorizing it, then state a claim about a system's velocity and defend it with the law. Unit 2, Force and Translational Dynamics, is the joint largest unit in the course: 18 to 23 percent of the multiple-choice section and roughly 22 to 27 class periods, matched only by Unit 3.

From here: how to draw a free-body diagram supplies the picture every equilibrium problem starts from, how to find net force covers the component bookkeeping in detail, and Topic 2.5 picks up where the sum does not vanish.

Finding the force that produces translational equilibrium

Three horizontal forces act on a 5.0 kg puck sliding on frictionless ice, with directions given as seen from above: 42 N east, 27 N west, and 8.0 N south. Take east as positive $x$ and north as positive $y$ . Find the single additional horizontal force that would put the puck in translational equilibrium.

Write the equilibrium condition axis by axis. Translational equilibrium needs $\sum F_x = 0$ and $\sum F_y = 0$ separately, so east-west and north-south are two independent problems.
Sum the given forces along $x$ : $+42 - 27 = +15\ \mathrm{N}$ . The fourth force must contribute $-15\ \mathrm{N}$ , which is 15 N west.
Sum the given forces along $y$ : the only contribution is the 8.0 N south force, $-8.0\ \mathrm{N}$ . The fourth force must contribute $+8.0\ \mathrm{N}$ , which is 8.0 N north.
Combine the two components: $\sqrt{15^2 + 8.0^2} = \sqrt{225 + 64} = \sqrt{289} = 17\ \mathrm{N}$ . Every given force carries two significant figures, so 17 N is the right precision.
Direction: $\tan^{-1}(8.0 / 15) = \tan^{-1}(0.533) = 28^\circ$ , measured north of west.
Notice what never entered the work: the mass. Equilibrium is a condition on forces alone, so the 5.0 kg is not needed and would matter only if the question asked for an acceleration.
Apply the first law. With the fourth force in place the net force is zero, so whatever velocity the puck had, it keeps. Gliding east at 3.0 m/s, it continues east at 3.0 m/s indefinitely; at rest, it stays at rest.

17 N at 28 degrees north of west. With that force added the puck is in translational equilibrium, and Newton's first law then says its velocity, whatever value it had, stops changing.

Balanced on two axes, unbalanced on one

A 0.60 kg puck glides east at 4.0 m/s across frictionless ice. A steady 1.2 N horizontal force pointing due north is applied for 3.0 s. Take east as positive $x$ , north as positive $y$ , and up as positive $z$ . Find the puck's velocity at the end of the 3.0 s.

Sort the forces by axis. Vertical: the normal force from the ice balances the gravitational force, so $\sum F_z = 0$ . East-west: nothing acts, so $\sum F_x = 0$ . North-south: the applied force is unbalanced, so $\sum F_y = +1.2\ \mathrm{N}$ .
Apply the first law to the two balanced axes. Zero net force along $z$ means no vertical velocity develops. Zero net force along $x$ means $v_x$ holds at its initial $+4.0\ \mathrm{m/s}$ for the whole interval: a force pointing due north has no eastward component and cannot slow the eastward motion.
Apply the second law to the unbalanced axis: $a_y = \frac{F_y}{m} = \frac{1.2\ \mathrm{N}}{0.60\ \mathrm{kg}} = 2.0\ \mathrm{m/s^2}$ north. The force is constant, so the acceleration is constant on this axis and the kinematic equations apply to it.
Find the new component: $v_y = v_{y0} + a_y t = 0 + (2.0)(3.0) = +6.0\ \mathrm{m/s}$ .
Recombine: $v = \sqrt{(4.0)^2 + (6.0)^2} = \sqrt{16 + 36} = \sqrt{52} = 7.2\ \mathrm{m/s}$ , directed $\tan^{-1}(6.0 / 4.0) = \tan^{-1}(1.5) = 56^\circ$ north of east.
This is the structure of every projectile problem. In Topic 1.5 the horizontal axis carries no force, so its velocity component holds, while the vertical axis is unbalanced, and the two are recombined only at the end.

7.2 m/s at 56 degrees north of east. The eastward component never changed, because the forces along east-west stayed balanced. Only the north-south component grew, which is EK 2.4.A.4 in action: the velocity changes only in the direction of the unbalanced force.

Frequently asked questions

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

That is half of it. The law says that if the net force on a system is zero its velocity remains constant, and zero is just one constant value. The other half matters more on the exam: a system already moving keeps moving in a straight line at the same speed, and no force is needed to maintain that.

What is translational equilibrium?

A configuration of forces such that the net force exerted on a system is zero. In practice it means the force components sum to zero on every axis separately. It does not mean the system is stationary: a sled gliding at constant velocity is in translational equilibrium just as much as a book resting on a table.

Is Newton's first law just the second law with zero acceleration?

Algebraically yes. Put zero acceleration into the second law and the sum of the forces has to vanish. Physically the first law does something extra, because it defines which reference frames the second law may be used in. The CED puts it this way: an inertial reference frame is one from which an observer would verify Newton's first law of motion.

What is an inertial reference frame?

One in which Newton's first law holds when you check it. Watch an object with no net force on it; if it keeps a constant velocity as seen from your frame, your frame is inertial. A braking bus is not, since a loose bottle inside accelerates forward with nothing pushing it. The AP Physics 1 Table of Information states that the frame of reference in any exam problem is assumed to be inertial unless otherwise stated.

What do motion graphs look like when the net force is zero?

Position versus time is a straight line, velocity versus time is a horizontal line at any height including zero, and acceleration versus time lies on the time axis. The velocity graph is the test: flat means the net force is zero, and any slope at all means it is not, because the slope of a velocity-time graph is the acceleration.