Newton's Third Law Practice: Identify the Force Pair

A 2 kg box rests on a shelf. Its weight is 20 N downward, and the shelf pushes it upward with 20 N. Those forces balance. They still aren't a Newton's third law pair: both act on the box.

Equal numbers and opposite arrows can hide the mistake. This Newton's third law practice worksheet uses a force ledger and changed conditions to help you identify each interaction. Try the nine questions on paper before reading the answers. Use g = 10 m/s² throughout.

Name the source, receiver, and interaction

Work within introductory Newtonian mechanics. If A exerts a force on B, B exerts a force of equal magnitude and opposite direction on A. The two forces act on different objects simultaneously; “reaction” doesn't mean a later response. See OpenStax's explanation of Newton's third law.

Write each force as a three-part entry:

source → receiver | interaction
Earth → box       | gravity
box → Earth       | gravity

To find the partner, swap source and receiver while keeping the interaction. The Physics Classroom's interaction-pair lesson uses the same object-swapping check.

A free-body diagram for the box includes only forces whose receiver is the box. The box's pull on Earth belongs on Earth's diagram. The net force is the vector sum of the forces acting on the chosen object or system. Don't add it to the ledger as another interaction.

The resting box involves two interactions

Assume the box is stationary on a horizontal shelf, with no other significant forces on it. Its weight is (2 kg)(10 m/s²) = 20 N.

Source → receiver	Interaction	Force direction and magnitude
Earth → box	Gravity	Downward, 20 N
Box → Earth	Gravity	Upward toward the box, 20 N
Shelf → box	Normal contact	Upward, 20 N
Box → shelf	Normal contact	Downward, 20 N

The first two rows form one third-law pair. The last two form another. For the box's vertical force balance, select the two rows with box as receiver. Call the normal-force magnitude F_normal and take upward as positive:

F_normal − mg = ma = 0
F_normal = 20 N

A normal force acts perpendicular to the contact surface. Its magnitude depends on the setup; it doesn't always equal weight. See OpenStax's common-forces lesson.

Now press downward on the same box with an additional 8 N. Keep it stationary, and assume the shelf can support the load:

F_normal − 20 N − 8 N = 0
F_normal = 28 N

The gravity pair remains 20 N in each direction, neglecting any change in height. The shelf–box contact pair becomes 28 N in each direction. There's also a hand–box pair: hand on box, 8 N downward; box on hand, 8 N upward.

Weight and normal force no longer match, even though the box still has zero net force. Each third-law pair retains equal magnitudes. The extra push exposes why matching numbers alone can't identify a pair.

Follow the forces through two accelerating boxes

Put box A, mass 2 kg, immediately left of box B, mass 3 kg, on a horizontal frictionless surface. A hand pushes A rightward with 20 N. Assume the boxes stay in contact and move together, with no other horizontal forces. Vertical forces balance. Choose right as positive.

Let F_contact be the magnitude of the force between the boxes. A pushes B rightward; B pushes A leftward. The separate horizontal equations are:

A: 20 N − F_contact = (2 kg)a
B:        F_contact = (3 kg)a

Add them. Both receivers are now inside the combined system A + B, so the internal contact terms cancel:

20 N = (2 kg + 3 kg)a
a = 4 m/s²

Substitute into B's equation:

F_contact = (3 kg)(4 m/s²) = 12 N

A pushes B with 12 N rightward, and B pushes A with 12 N leftward. Check A separately: 20 N − 12 N = 8 N, which equals (2 kg)(4 m/s²).

The contact forces form a 12 N pair. The net horizontal forces are 8 N on A and 12 N on B. Those net forces describe different sums; they aren't third-law partners.

The boxes share an acceleration because they stay together. Third-law partners can act on objects with different accelerations, as the collision problem below shows. The contact forces cancel in the combined sum, while each remains on its receiver's individual diagram. OpenStax explains this internal-force cancellation.

Nine practice problems

For a claimed pair, name the sources, receivers, and interaction. For calculations, state the system and positive direction. These original questions have answers below.

1. Return to the stationary 2 kg box with no hand pressing down. What is the partner of Earth's gravitational force on the box? Does that partner belong on the box's free-body diagram?
2. A student pairs “shelf on box, upward” with “Earth on box, downward” because both are 20 N. Identify the failed ledger check and name each force's actual partner.
3. Restore the downward 8 N hand push. Find the shelf's force on the stationary box and the box's force on the hand. Are those two forces a pair?
4. A 0.3 kg ornament has been released and is falling freely near Earth. Ignore air resistance. Find Earth's force on the ornament and its third-law partner. Does the absence of a support remove the partner?
5. A suitcase moves in a straight line at constant velocity. A horizontal rope pulls it rightward with 18 N; kinetic friction from the floor acts leftward with 18 N. Are these forces a pair? Name the partner of each.
6. During a collision, a 1.5 kg cart pushes a 4.5 kg cart rightward with 9 N. Neglect other horizontal forces at the instant considered. Find the force on the lighter cart and each cart's horizontal acceleration. Does the larger acceleration require a larger interaction force?
7. Replace the original 20 N push in the two-box example with a 10 N leftward push on B, driving B against A. Keep the same masses, positions, frictionless surface, and shared motion. Find the common acceleration and both box-to-box contact forces.
8. A straight, taut, ideal massless rope connects a hand and a sled. Its tension is 15 N. A student pairs the rope's force on the hand with the rope's force on the sled. These forces have equal magnitudes and opposite directions. Are they third-law partners?
9. In the original 20 N two-box example, a student says, “For A + B, the 12 N forces cancel, so the system's net force is zero.” Repair the argument without changing the system boundary.

Answers and the step that matters

1. Box → Earth, gravity: 20 N upward toward the box. Swap the objects in Earth → box. The partner's receiver is Earth, so it doesn't belong on the box's diagram.
2. Both proposed forces have the same receiver. Shelf → box pairs with box → shelf, normal contact. Earth → box pairs with box → Earth, gravity. The shared 20 N magnitude doesn't join those two interactions.
3. Shelf on box: 28 N upward. Box on hand: 8 N upward. For the box, 28 N − 20 N − 8 N = 0. The 28 N force pairs with box on shelf, 28 N downward. The 8 N force pairs with hand on box, 8 N downward. The requested forces belong to different interactions.
4. Earth on ornament: 3 N downward; ornament on Earth: 3 N upward toward the ornament. mg = (0.3 kg)(10 m/s²) = 3 N. Gravity needs no contact. The ornament accelerates downward at 3 N / 0.3 kg = 10 m/s²; its gravitational partner acts on Earth.
5. They balance on the suitcase. Their sum is zero, consistent with constant velocity. Rope → suitcase pairs with suitcase → rope, 18 N leftward. Floor → suitcase pairs with suitcase → floor, a friction force of 18 N rightward. Neither partner acts on the suitcase.
6. The lighter cart receives 9 N leftward. With right positive, its acceleration is −9 N / 1.5 kg = −6 m/s²; the heavier cart's is 9 N / 4.5 kg = +2 m/s². Equal interaction forces produce different accelerations here because the masses differ. Each net horizontal force equals its contact force under the stated assumption of no other horizontal forces.
7. Both accelerate at 2 m/s² leftward. For A + B, with right positive, a = −10 N / (2 kg + 3 kg) = −2 m/s². A receives only B's horizontal push, so B → A is 4 N leftward: (2 kg)(2 m/s²) = 4 N. A → B is 4 N rightward. Check B: −10 N + 4 N = −6 N = (3 kg)(−2 m/s²). The contact magnitude depends on which box receives the external push as well as on that push's magnitude.
8. They aren't partners: both have the rope as source. Rope → hand pairs with hand → rope. Rope → sled pairs with sled → rope. Equal tension at the ends of this ideal rope doesn't turn its two separate contacts into one interaction pair.
9. The external 20 N push remains. Only the internal box-to-box forces cancel: 20 N − 12 N + 12 N = 20 N. The 5 kg system accelerates at 20 N / 5 kg = 4 m/s² rightward. Internal cancellation doesn't imply zero net external force.

Turn a missed distinction into one small card

Keep the full worked problem on paper. Make a recall card for the step you missed:

Front	Back
Two equal, opposite forces both act on one box. Can they be third-law partners?	No. Partners swap source and receiver; these share a receiver. They may balance in the box's force sum.
A stationary 2 kg box has an extra 8 N downward push. With g = 10 m/s², what is the shelf's normal force?	28 N upward: F_normal − 20 N − 8 N = 0. Its partner is the box's 28 N downward contact force on the shelf.
When A and B become one system, what happens to their mutual forces?	They cancel in the combined force sum as internal forces. Count the remaining external forces to find the system's acceleration.

These prompts can go into Nibomo, another flashcard app, or a paper deck. The guide to making better flashcards helps keep each prompt answerable; the algebra-based physics study guide explains where cards fit alongside full problems.

Then change one condition and solve again. Swap which box is pushed, change the downward hand force, or choose a different system. Rebuilding the ledger tests whether you can identify the forces in a new setup.

AI tools assisted substantially with research, writing, editing, and original exercise construction, including checking force balances and arithmetic against the stated Newtonian models. The illustration is AI-generated.