Part 1 · Chapter 05

Force and Motion — I

What makes things accelerate — Newton's three laws and the forces behind them

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 35 min
i What you'll learn
  • That a force is what causes a body to accelerate, and that forces add as vectors.
  • The meaning of mass as a body's resistance to acceleration.
  • Newton's three laws of motion — and what an inertial reference frame is.
  • How to use Fnet = ma component by component, and how to build a free-body diagram.
  • The everyday forces: gravity, weight, the normal force, friction, and tension.
Section 5-1

What Is Physics?

Chapters 2–4 described how things move — position, velocity, and acceleration. This chapter asks why. What can cause a body to accelerate? The answer is a force: loosely, a push or a pull. When a dragster's tires grip the track, when a guard knocks down a quarterback, when a car slams into a pole — in each case a force acts on the body to change its velocity. The study of the relation between force and acceleration is the heart of mechanics.

Section 5-2

Newtonian Mechanics

The link between a force and the acceleration it produces was first set out by Isaac Newton (1642–1727), and the framework he built is called Newtonian mechanics. It rests on three laws of motion. It is not the whole story — at speeds near that of light we need Einstein's special relativity, and at the atomic scale we need quantum mechanics — but it is a remarkably good approximation, accurate for objects from a speck of dust up to a galaxy.

Section 5-3

Newton's First Law

For centuries it was thought that a steady push was needed to keep a body moving — a puck slid across a wooden floor slows and stops, after all. But slide it across ice, and it travels much farther; across a near-frictionless air table, it barely slows at all. Extrapolating to a perfectly frictionless surface, Newton saw that no force is needed to maintain motion at all:

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Newton's First Law (law of inertia)
If no net force acts on a body, its velocity cannot change; the body cannot accelerate.

A body at rest stays at rest; a body in motion keeps the same speed and direction. Several forces may act, but if their net force is zero, the motion is unchanged. This tendency to resist a change in motion is called inertia.

The first law only holds in certain frames. An inertial reference frame is one in which Newton's laws hold — for most problems, the ground works fine. A frame that is itself accelerating (a braking car, a spinning Earth viewed over long distances) is noninertial, and objects in it can appear to accelerate with no force in sight.

Section 5-4

Force

We define the unit of force by the acceleration it gives a standard body. Place the standard 1 kg mass on a frictionless table; the pull that gives it an acceleration of \(1 m/s^{2}\) is defined to be 1 newton (N). A pull producing \(2 m/s^{2}\) is 2 N, and so on.

Forces are vectors
When several forces act on a body, the net force is their vector sum.

This is the principle of superposition: a single force equal to the vector sum has exactly the same effect as all the individual forces together. Forces add by the rules of Chapter 3 — component by component, not by simple arithmetic.

Section 5-5

Mass

Give a baseball and a bowling ball the same sharp kick and the baseball flies off faster — the same force produces a smaller acceleration on the more massive body. Mass is the property that sets this ratio. Apply one fixed force to the standard body (mass \(m_{0}\)) and to an unknown body X, and compare accelerations:

Mass from acceleration ratio
\[ m_{X} / m_{0} = a_{0} / a_{X} \]
The same force gives a body half the acceleration when its mass is doubled. Mass is a scalar and an intrinsic property of the body.
Mass is not weight, size, or density. It is simply the characteristic that relates the force on a body to the resulting acceleration — a measure of the body's inertia.
Section 5-6

Newton's Second Law

Everything so far folds into one compact statement — the workhorse equation of mechanics:

Newton's Second Law
The net force on a body equals the product of its mass and its acceleration.

In vector form, \(F_{\text{net}} = m a\). The acceleration points the same way as the net force, and its size is the net force divided by the mass.

Second law — vector and components
\[\begin{gathered} F_{\text{net}} = m a \\ F_{net,x} = m a_{x}, \qquad F_{net,y} = m a_{y}, \qquad F_{net,z} = m a_{z} \end{gathered}\]
A vector equation is three scalar equations in one. Each axis is independent: the force along x produces only the x-acceleration.
The newton
\[ 1 N = (1 \mathrm{kg})(1 m/s^{2}) = 1 \mathrm{kg}\cdot m/s^{2} \]
! Two cautions

First, always state which body you are applying the law to, and include only the forces acting on that body — not forces it exerts on something else. Second, \(F_{\text{net}}\) is the net force; if it is zero the body is in equilibrium — at rest or moving at constant velocity — even though real forces still act and merely balance.

Section 5-7

Some Particular Forces

Five forces appear again and again in mechanics problems. Knowing each on sight is half the battle.

Gravitational force & weight
\[ F_{g} = m g \qquad \text{(downward)} \qquad W = m g \]
In vector form Fg = −mg ĵ. The weight W is the magnitude of the gravitational pull, measured where the body is not accelerating vertically.
! Weight is not mass

Mass is intrinsic and unchanging; weight depends on the local \(g\). A 7.2 kg bowling ball weighs about 71 N on Earth but only 12 N on the Moon — same mass, different weight, because the Moon's \(g\) is about 1.6 m/s².

Normal force

When a body presses on a surface, the surface pushes back perpendicular to itself. On a level table with no vertical acceleration, \(F_{N} = mg\); in general \(F_{N} = m(g + a_{y})\).

Friction

A surface resists sliding with a force \(f\) directed along the surface, opposite the intended motion. "Frictionless" means we take this force to be zero.

Tension

A taut cord pulls on a body with force \(T\) directed along the cord, away from the body. An ideal cord is massless and unstretchable, so it transmits the same T at both ends — even over a massless, frictionless pulley.

Section 5-8

Newton's Third Law

Two bodies interact when each pushes or pulls on the other. Lean a book B against a crate C: the crate pushes the book, and the book pushes the crate. These two forces are always paired:

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Newton's Third Law
When two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction.

As a vector relation, \(F_{\text{BC}} = -F_{\text{CB}}\). The two forces of a third-law pair always act on different bodies — which is exactly why they never cancel each other.

The classic trap. A book resting on a table feels gravity down and the normal force up, and these balance — but they are not a third-law pair (both act on the book). The partner of the table's push on the book is the book's push on the table; the partner of Earth's pull on the book is the book's pull on Earth.
Section 5-9

Applying Newton's Laws

Almost every force problem yields to the same recipe. The single most important skill is translating a picture of a situation into a clean free-body diagram.

1 Choose the body

Pick one body (or a rigidly connected system) and isolate it. Decide what you are applying \(F_{\text{net}} = ma\) to before drawing anything.

2 Draw the forces

Represent the body as a dot and draw every force acting on it as an arrow from the dot — gravity, normal, tension, friction, applied pushes. Nothing the body exerts on others.

3 Choose axes

Align an axis with the acceleration where you can (e.g. along an incline). Resolve every force into components.

4 Solve per axis

Write \(F_{net,x} = ma_{x}\) and \(F_{net,y} = ma_{y}\). For linked bodies, write the law for each and combine the equations.

Worked Examples

Putting It to Work

1 Forces on a puck — one and two dimensions

Problem. A puck of mass \(m = 0.20 \mathrm{kg}\) slides on frictionless ice along the x axis. Find its acceleration when (A) only \(F_{1} = 4.0 N\) acts along +x; (B) \(F_{1} = 4.0 N\) (+x) and \(F_{2} = 2.0 N\) (−x) act; (C) \(F_{2} = 2.0 N\) (−x) acts together with \(F_{3} = 1.0 N\) at 30° above +x.

Apply Fnet,x = m ax
\[\begin{gathered} A: \qquad a_{x} = F_{1}/m = 4.0/0.20 = 20 m/s^{2} \\ B: \qquad a_{x} = (F_{1} - F_{2})/m = (4.0 - 2.0)/0.20 = 10 m/s^{2} \\ C: \qquad a_{x} = (F_{3} \cos 30^{\circ} - F_{2})/m = (0.866 - 2.0)/0.20 \approx -5.7 m/s^{2} \end{gathered}\]

Only the x-components matter for x-motion: the vertical part of \(F_{3}\) is cancelled by ice and gravity and never enters. In case C the net force points in −x, so the puck accelerates backward.

2 Cookie tin — finding a missing force

Problem. A 2.0 kg cookie tin accelerates at \(3.0 m/s^{2}\) at 50° above the x axis on a frictionless surface. Two of the three forces on it are \(F_{1} = 10 N\) (at −150°) and \(F_{2} = 20 N\) (at +90°). Find the third force \(F_{3}\).

Solution. From \(F_{1} + F_{2} + F_{3} = m a\), solve \(F_{3} = m a - F_{1} - F_{2}\) component by component:

Components, then magnitude–angle
\[\begin{gathered} F_{3,x} = (2.0)(3.0)\cos 50^{\circ} - (10)\cos (-150^{\circ}) - (20)\cos 90^{\circ} \approx 12.5 N \\ F_{3,y} = (2.0)(3.0)\sin 50^{\circ} - (10)\sin (-150^{\circ}) - (20)\sin 90^{\circ} \approx -10.4 N \\ F_{3} = (13 \hat{\imath} - 10 \hat{\jmath}) N \approx 16 N at -40^{\circ} \end{gathered}\]

Two dimensions force a true vector subtraction — you cannot just subtract magnitudes. The missing force has magnitude ≈ 16 N, pointing 40° below the x axis.

3 Block on a table pulling a hanging block

Problem. A sliding block \(S\) of mass \(M = 3.3 \mathrm{kg}\) on a frictionless table is joined by a cord over a massless, frictionless pulley to a hanging block \(H\) of mass \(m = 2.1 \mathrm{kg}\). Find the acceleration and the cord tension.

Solution. The cord is inextensible, so both blocks share the same acceleration magnitude \(a\). Apply the second law to each: for S (horizontal), \(T = Ma\); for H (vertical), \(T - mg = -ma\). Subtract to eliminate \(T\):

Acceleration and tension
\[\begin{gathered} a = m g / (M + m) = (2.1)(9.8)/(3.3 + 2.1) \approx 3.8 m/s^{2} \\ T = Mm g / (M + m) = (3.3)(2.1)(9.8)/(5.4) \approx 13 N \end{gathered}\]

Two sanity checks: \(a \lt g\) always (the hanging block is not in free fall — the cord pulls up on it), and \(T \lt mg\) always (if it weren't, H would accelerate upward).

4 A cord pulling a box up a ramp

Problem. A cord pulls a 5.00 kg box of sea biscuits up a frictionless plane inclined at \(\theta = 30^{\circ}\), with tension \(T = 25.0 N\) directed up the slope. Find the box's acceleration along the plane.

Solution. Tilt the axes so x runs up the plane. Only the along-plane forces set the along-plane acceleration: the tension pulls up the slope (+T), and gravity's component down the slope is \(mg \sin \theta\). The normal force is perpendicular and drops out of the x-equation.

Newton's second law along the incline
\[\begin{gathered} T - mg \sin \theta = m a \\ a = [25.0 - (5.00)(9.8)\sin 30^{\circ}] / 5.00 = (25.0 - 24.5)/5.00 = 0.100 m/s^{2} \end{gathered}\]

The positive sign confirms the box accelerates up the plane — but only just, since the tension barely beats the gravitational pull along the slope.

Review

Chapter Summary

Force & the first law

A force causes acceleration. With zero net force a body keeps constant velocity (inertia). Newton's laws hold in inertial frames.

Mass

An intrinsic scalar measuring resistance to acceleration: \(m_{X}/m_{0} = a_{0}/a_{X}\) for the same force.

Second law

\(F_{\text{net}} = m a\), i.e. one equation per axis. \(1 N = 1 \mathrm{kg}\cdot m/s^{2}\). Zero net force ⇒ equilibrium.

Particular forces

Gravity \(F_{g} = mg\); weight \(W = mg\); normal force ⊥ surface; friction opposes sliding; tension along the cord.

Third law

\(F_{\text{BC}} = -F_{\text{CB}}\): equal, opposite, and acting on two different bodies — so they never cancel.

Free-body method

Isolate the body, draw every force on it, choose smart axes, then solve \(F_{\text{net}} = ma\) axis by axis.

Practice

Problems

For every problem, name the body first and draw its free-body diagram before writing a single equation. Take \(g = 9.8 m/s^{2}\) unless told otherwise.

  1. Two horizontal forces, 9.0 N east and 8.0 N at 62° north of east, act on a 2.0 kg sled on frictionless ice. Find the magnitude and direction of its acceleration.
  2. A 0.340 kg puck is pushed by a single horizontal force and accelerates at 12.0 m/s². What is the magnitude of the force?
  3. A constant net force gives a 4.0 kg body an acceleration of 6.0 m/s². What acceleration would the same force give a 12.0 kg body?
  4. An astronaut weighs 730 N at sea level on Earth. (a) What is her mass? (b) What would she weigh on the Moon, where \(g = 1.6 m/s^{2}\)?
  5. A 5.0 kg block sits on a horizontal floor. Find the normal force on it when the floor is (a) at rest, (b) in an elevator rising at constant speed, and (c) in an elevator accelerating upward at 2.0 m/s².
  6. A 65 kg person stands in an elevator on a scale. What does the scale read (in newtons) when the elevator accelerates (a) upward at 1.5 m/s² and (b) downward at 1.5 m/s²?
  7. A 6.0 kg and a 4.0 kg block sit in contact on a frictionless surface. A 20 N horizontal push is applied to the 6.0 kg block, toward the 4.0 kg block. Find (a) the acceleration of the pair and (b) the contact force between them.
  8. In an Atwood machine, two blocks of mass 3.0 kg and 5.0 kg hang from a cord over a massless, frictionless pulley. Find (a) the acceleration of the blocks and (b) the cord tension.
  9. A 12 kg crate is pulled up a frictionless 25° incline by a cord parallel to the slope. What tension gives the crate an acceleration of 1.2 m/s² up the incline?
  10. A 2.0 kg block hangs from a spring scale that is attached to the ceiling of an elevator. State whether the scale reads more than, less than, or equal to 19.6 N when the elevator is (a) accelerating upward, (b) moving up at constant speed, and (c) accelerating downward — and give the reading for case (a) if the acceleration is 3.0 m/s².
Tip: a third-law pair never appears together in the same free-body diagram — the two forces act on different bodies. If two forces you've drawn act on the same body, they are not partners, even when they happen to balance.