foundation25 min studyLesson 3 in Physics

Newton's laws and choosing the system

Use free-body diagrams to connect interactions to acceleration without confusing forces on different objects.

Before you begin

On this page

The question for today

What you’ll understand

Draw forces acting on one chosen system.
Apply Newton's second law component by component.
Identify action–reaction pairs and distinguish them from balanced forces.

Kinematics describes what a cart does. Dynamics asks how interactions change its motion. Newton's laws provide a remarkably effective model for ordinary speeds and macroscopic objects. Their practical power comes from disciplined bookkeeping: choose a system, identify interactions crossing its boundary, and connect the net force to that system's momentum change.

An inertial frame is part of the law

The first law says that a body without net external force maintains constant velocity in an inertial frame. Constant velocity includes rest, and does not require zero position. A cart slowing on a bench is not evidence against the law: contact friction and air resistance are interactions that must be included. Motion does not need a continuing forward force; acceleration does.

An accelerating bus is an inconvenient frame for applying the simplest form of Newton's law. A loose bag can appear to accelerate backward although no backward interaction is acting on it. You can either analyze in an approximately inertial ground frame or introduce an appropriate inertial force in the bus frame. State which choice you make rather than treating the two descriptions as contradictory.

\sum\mathbf F_{\mathrm{ext}}=\frac{d\mathbf p}{dt}=m\mathbf a

The final equality assumes constant mass and Newtonian momentum p = mv in an inertial frame. Force is measured in newtons, with 1 N = 1 kg·m/s². Variable-mass systems need careful momentum-flux accounting.↔ Scroll wide equations horizontally.

Draw interactions, not motion arrows

Isolate one object and draw each force acting on it: gravity from Earth, a normal force from a surface, tension from a taut rope, or friction from contact. A velocity arrow can be useful beside the diagram, but it is not another force. Labels should name both the interaction and, when ambiguity matters, the object exerting it.

The normal force adjusts to the contact constraint; it is not always equal to weight. A horizontal table supporting a motionless block gives that equality, but an inclined surface or vertical acceleration changes it. Likewise, static friction adjusts up to a limit rather than always reaching its maximum. Decide whether the surfaces actually slide before substituting a kinetic-friction model.

|f_s|\leq\mu_sN,\qquad |f_k|=\mu_kN

This approximate dry-contact model uses nonnegative dimensionless friction coefficients. Static friction takes the needed value within its bound; kinetic friction opposes relative sliding. Real friction can depend on speed, material state, and temperature.↔ Scroll wide equations horizontally.

The third law connects systems

If a hand pushes a cart, the cart pushes the hand with an equal and opposite force. Those forces act on different objects, so they do not cancel in the cart's free-body diagram. In contrast, weight and the table's normal force can balance on one block, but they are not a third-law pair: their partners act on Earth and on the table respectively.

Choosing a larger system can eliminate internal interaction pairs from its total momentum equation. For two carts joined by a rope, the rope forces cancel when both carts belong to the system, leaving external forces to determine their total acceleration. This separation of internal exchange and external influence prepares the language of conservation laws, including the momentum carried by fields in later physics.

\mathbf F_{A\to B}=-\mathbf F_{B\to A}

For Newtonian interaction pairs this relates forces on different objects. Electromagnetic systems require including field momentum for a complete local momentum account; instantaneous particle-force pairs alone are not the most general formulation.↔ Scroll wide equations horizontally.

Work it through

Pulling a rough-surface cart

A $4.0\,\mathrm{kg}$ cart is pulled horizontally with $14\,\mathrm{N}$ while sliding in the pulling direction on a level surface with $\mu_k=0.20$ . Use $g=10\,\mathrm{m\,s^{-2}}$ . Find its acceleration.

Choose the cart as the system. Vertically it has no acceleration, so $N-mg=0$ and $N=40\,\mathrm{N}$ . This conclusion follows from the vertical equation, not from the definition of normal force.
Because sliding is specified, use kinetic friction: $f_k=\mu_kN=8\,\mathrm{N}$ . It points opposite the sliding direction; assume the cart is sliding in the pull direction for this calculation.
Choose that direction positive. The horizontal net force is $14-8=6\,\mathrm{N}$ , so $a=6/4.0=1.5\,\mathrm{m\,s^{-2}}$ . A positive result means the existing forward motion speeds up.

The result

The acceleration is $1.5\,\mathrm{m\,s^{-2}}$ in the pulling direction. If the cart were initially at rest, static-friction information would be needed to decide whether it starts moving.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Forces are vectors before they are numbers

Change one normalized vector's angle and magnitude, then inspect its signed x/y components. Sketch an opposing force separately before summing. The lab illustrates force decomposition; it does not calculate a cart's contact or friction forces.

Vectorx / y projectionsTrace marker

x component: 1.393
y component: 0.975
Magnitude: 1.7

Angle35°Magnitude1.7

t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forWrite the signed component sum before dividing by mass.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The partner reverses the roles of the interacting bodies: book on Earth. The normal force balances weight in this situation but comes from a different interaction and has its own partner on the table.

Your turn

Try it yourself

Two carts of masses $2\,\mathrm{kg}$ and $3\,\mathrm{kg}$ are joined by a light rope on a frictionless track. A $10\,\mathrm{N}$ force pulls the $3\,\mathrm{kg}$ cart. Find the common acceleration and rope tension.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Analyze both carts together for acceleration, then isolate the cart that is pulled only by the rope.

Compare with the solution

The total mass is $5\,\mathrm{kg}$ and the only horizontal external force is $10\,\mathrm{N}$ , giving $a=2\,\mathrm{m\,s^{-2}}$ . The $2\,\mathrm{kg}$ cart is accelerated by tension alone, so $T=ma=4\,\mathrm{N}$ . For the other cart, $10-T=3a$ confirms the same answer. The rope force is internal only when both carts are included.

Keep the idea

Take it with you

Choose the system and inertial frame before applying a force law.
A free-body diagram contains interactions acting on that system.
Balanced forces act on one object; third-law partners act on different objects.

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