Newton’s Laws of Motion: From Interaction to Equation

1) Force as an Interaction (Not Something an Object “Has”)

A force is a push or pull that arises from an interaction between two objects (or an object and its environment). This habit matters: instead of saying “the object has a force,” train yourself to ask: “What is interacting with it?”

Forces come from identifiable sources

• Contact interactions: push by a hand, friction from a surface, tension from a rope, normal force from a floor.
• Non-contact interactions: gravity from Earth, magnetic forces, electric forces.

Free-body thinking: focus on one object

To connect interactions to equations, you will repeatedly do this: choose a system (often a single object), then list all external forces acting on that system. This list is the bridge from a physical situation to Newton’s laws.

Practical habit: always name forces by their source: “force of the table on the book,” “force of the rope on the crate,” “force of Earth on the ball.” This automatically reminds you that forces come in interactions.

Concept check (quick)

• If a box is sliding on a floor, does it “carry” friction with it, or does friction exist only because the floor is interacting with it?
• When you let go of a stretched spring, what object is exerting a force on the spring (and what object is the spring exerting a force on)?

2) Newton’s First Law and Inertial Frames

Newton’s First Law (N1): If the net external force on an object is zero, its velocity remains constant (it stays at rest or moves in a straight line at constant speed).

N1 is not saying “objects naturally stop.” It says: changes in motion require a net force. If you observe an object slowing down, turning, or speeding up, you should immediately suspect a nonzero net force.

Continue in our app.

• Listen to the audio with the screen off.
• Earn a certificate upon completion.
• Over 5000 courses for you to explore!

Or continue reading below...

Download the app

Inertial frames: where N1 works in its simplest form

An inertial reference frame is one in which an object with zero net force moves at constant velocity. Frames that accelerate or rotate are non-inertial and require extra (fictitious) forces to use Newton’s laws in their usual form.

Example A: Sliding puck (low friction)

Imagine a hockey puck gliding on very smooth ice.

• Observation: it keeps moving nearly straight at nearly constant speed.
• Interpretation: the net horizontal force is close to zero (friction is small).
• Key habit: do not invent a “force of motion” to keep it going. Motion continues without a force; only changes in motion require net force.

Example B: Car making a turn

When a car turns left at constant speed, its velocity changes direction, so it has acceleration (centripetal acceleration). Therefore, the net force is not zero.

• Source of force: typically static friction from the road on the tires provides the sideways force.
• Passenger feeling: you may feel “thrown outward.” In the ground (approximately inertial) frame, what happens is your body tends to keep moving straight while the car turns underneath you; the car door exerts an inward force on you to make you turn with the car.

Concept questions

• A puck slides at constant velocity. What can you say about the net force on it?
• A car moves in a circle at constant speed. Is the net force zero? Why or why not?

3) Newton’s Second Law as a Vector Equation

Newton’s Second Law (N2) connects interactions to measurable motion:

ΣF⃗ = m a⃗

It is a vector equation: direction matters. The left side is the net external force (vector sum of all forces on the chosen object). The right side is mass times acceleration.

Step-by-step method (the core workflow)

1. Choose the object (system) you are analyzing.
2. List all external forces acting on it (by source).
3. Choose axes (often x horizontal, y vertical; or align one axis with motion).
4. Write N2 in components: ΣF_x = m a_x, ΣF_y = m a_y (and z if needed).
5. Use constraints (e.g., “no vertical acceleration” means a_y = 0).
6. Solve for unknowns (forces or acceleration).

Component form (what you actually compute)

In 2D:

ΣF_x = m a_x  (1)
ΣF_y = m a_y  (2)

This is where Newton’s laws connect to the motion you can measure: if you can measure acceleration components, you can infer net force components, and vice versa.

Link to kinematics (without re-deriving it)

Kinematics tells you how to relate acceleration to changes in velocity and position. Newton’s second law tells you why the acceleration is what it is: because of a net force. In practice, you often do:

• Use forces to find a via ΣF = m a, then use that acceleration in motion predictions.
• Or measure a (with video analysis or a phone sensor) and compute ΣF = m a to infer the net force.

Numeric exercise 1: Measured acceleration → net force

A 0.50 kg cart on a track is measured to accelerate at 1.6 m/s² to the right. Assume horizontal is the only relevant direction.

• Given: m = 0.50 kg, a_x = +1.6 m/s^2
• Compute: ΣF_x = m a_x = (0.50)(1.6) = 0.80 N
• Interpretation: the net horizontal force on the cart is 0.80 N to the right.

Numeric exercise 2: Two forces in 1D

A 2.0 kg block is pulled to the right by 9.0 N while friction from the floor acts to the left with 3.0 N. Find the acceleration.

• Net force: ΣF_x = 9.0 - 3.0 = 6.0 N
• Acceleration: a_x = ΣF_x / m = 6.0 / 2.0 = 3.0 m/s^2 to the right.

Numeric exercise 3: Components in 2D (force at an angle)

A 1.5 kg object experiences a net force of magnitude 6.0 N at 30° above the +x direction. Find a_x and a_y.

• Resolve force: F_x = 6.0 cos 30° = 5.20 N, F_y = 6.0 sin 30° = 3.00 N
• Apply N2: a_x = F_x/m = 5.20/1.5 = 3.47 m/s^2, a_y = F_y/m = 3.00/1.5 = 2.00 m/s^2

Concept questions

• If the net force on an object points east, can the object’s acceleration point north? Explain using ΣF⃗ = m a⃗.
• If you double the net force on the same object, what happens to its acceleration?
• If two different objects experience the same net force, which one has the larger acceleration?

4) Newton’s Third Law: Paired Forces on Different Objects

Newton’s Third Law (N3): If object A exerts a force on object B, then object B exerts a force on object A that is equal in magnitude and opposite in direction.

Write it as a pair:

F⃗_{A on B} = - F⃗_{B on A}

The most important rule: action–reaction forces act on different objects

This is the key to avoiding a common mistake. Third-law pairs do not cancel on a single free-body diagram because a free-body diagram is for one object, and the pair is split across two objects.

Example: Hand pushes a box

• On the box: force of hand on box (push forward), plus friction/normal/gravity as applicable.
• On the hand: force of box on hand (push backward).

The box can accelerate forward even though the forces are paired, because the backward force is on the hand, not on the box.

Example: Book resting on a table

• Forces on the book: gravity (Earth on book) downward; normal force (table on book) upward.
• Third-law pairs: the upward normal on the book pairs with a downward force of the book on the table; the downward gravity on the book pairs with an upward gravitational pull of the book on Earth.

Notice: the normal force and gravity on the book are not a third-law pair (they act on the same object). They can balance to give zero net force, but that is a first/second-law statement about net force, not a third-law pairing.

Misconception to confront: “Action–reaction forces cancel, so nothing moves”

If you treat action–reaction forces as if they act on the same object, you would incorrectly predict zero net force in many situations (walking, swimming, pushing off a wall). The correction is always the same: identify which object each force acts on.

Example: Walking forward

• Your foot pushes the ground backward (force of foot on ground).
• The ground pushes your foot forward (force of ground on foot). This forward force on you can accelerate you forward.

Concept questions

• When a truck collides with a small car, which force is larger: the force of the truck on the car, or the force of the car on the truck?
• If the forces are equal and opposite in a collision, why do the truck and car often have very different accelerations?

Numeric exercise 4: Third law + second law together

During a collision, a car exerts a force of 12,000 N on a truck. The truck’s mass is 6000 kg and the car’s mass is 1200 kg. Find the magnitude of each vehicle’s acceleration during the interaction (assume the force is the dominant horizontal force during the short collision).

• N3: the truck exerts 12,000 N on the car in the opposite direction.
• Truck acceleration: a_truck = F/m = 12000 / 6000 = 2.0 m/s^2
• Car acceleration: a_car = 12000 / 1200 = 10 m/s^2
• Interpretation: same interaction force, different accelerations because masses differ.