PHY131H1/Unit 4

Unit 4 · Introduction to Physics I

Newton's Laws

From describing motion to explaining it. Three laws, one equation, and one drawing habit — the free-body diagram — that does more for your marks than any amount of algebra. Nearly every dynamics error is a force drawn that does not exist, or one left out.

By the end you can

• State all three laws precisely and say what each is for.
• Draw a correct free-body diagram for any body in a system.
• Identify action-reaction pairs and say why they never cancel.
• Solve connected-body problems with tension.

Why it matters & where it comes from

Why it exists

Kinematics tells you what happens given the acceleration. Newton's second law tells you what the acceleration is, from the forces. That link closes the loop and makes mechanics predictive rather than descriptive.

In the real world

Every structure, vehicle and machine is designed against these laws. Seatbelts exist because of the first: in a crash the car stops and the passenger, with no force on them, does not.

Story & open questions

Newton published the laws in 1687. They stood unchallenged for two centuries until relativity showed they fail near light speed, and quantum mechanics showed they fail at atomic scale. Within those limits they remain exact, which is why bridges are still designed with them.

The three laws

Definition

First law. A body at rest stays at rest, and a body in motion continues at constant velocity, unless a net external force acts. This defines what a force does: it changes velocity, it does not maintain it.

Second law. $\sum \vec{F} = m\vec{a}$. The net force equals mass times acceleration, and the acceleration points along the net force.

Third law. If A pushes B, then B pushes A with equal magnitude and opposite direction. The two forces act on different bodies.

Motion needs no force

The deepest misconception in physics is that keeping something moving requires a force. It does not — a hockey puck on ice glides on. What requires force is changing the motion. On Earth we push things continuously only because friction is continuously opposing us.

Example Second law, componentwise

The second law is a vector equation, so it splits into one equation per axis:

$\sum F_x = ma_x$ and $\sum F_y = ma_y$.

If a body slides horizontally, $a_y = 0$, so the vertical forces must balance exactly. That is usually how you find the normal force.

Free-body diagrams

Rule How to draw one

Isolate one body. Draw it as a dot or a box.

Draw only forces acting on that body, as arrows pointing away from it.

Label each with its source: weight $mg$, normal $N$, tension $T$, friction $f$, applied $F$.

Choose axes, tilting them along the motion for an incline.

Write $\sum F = ma$ for each axis.

Forces that do not exist

Two things commonly get drawn that should not:

Centrifugal force. There is no outward force on a body moving in a circle. The net force points inward. The outward feeling is inertia, not a force.

The force of motion. A thrown ball has no forward force on it once it leaves the hand. Only gravity acts. Its forward motion persists by the first law.

Example Block on an incline

A block of mass $m$ rests on a frictionless incline at angle $\theta$.

Forces: weight $mg$ straight down, normal $N$ perpendicular to the surface. That is all — there is no force pushing it down the slope.

Tilt the axes along the incline. Then $mg$ resolves into $mg\sin\theta$ down the slope and $mg\cos\theta$ into it.

Perpendicular: $N - mg\cos\theta = 0$, so $N = mg\cos\theta$ — less than $mg$, which is why a steep slope grips poorly.

Along: $mg\sin\theta = ma$, so $a = g\sin\theta$. Independent of mass, and correctly giving $a = g$ at $90$ degrees and $a = 0$ on the flat.

The third law

Rule Pairs act on different bodies

Action and reaction never cancel, because they are not applied to the same object. Only forces on the same body can cancel in $\sum F = ma$.

Example The horse and cart

If the cart pulls back on the horse as hard as the horse pulls the cart, how does anything move?

Those two forces act on different bodies, so they never appear in the same equation.

For the cart, the relevant forces are the horse's pull forward and friction backward. If the pull wins, the cart accelerates.

The horse accelerates because the ground pushes it forward — its reaction to the horse pushing back on the ground. The cart's pull on the horse is a separate force entirely.

Example Book on a table

The weight $mg$ down and the normal $N$ up are not a third-law pair, even though they are equal and opposite. Both act on the book, and they balance only because the book is not accelerating.

The genuine pairs are: Earth pulls book down / book pulls Earth up; table pushes book up / book pushes table down. Each pair has one force on each of two different bodies.

Connected bodies

Rule One diagram per body

For bodies joined by a rope over a pulley:

Draw a separate free-body diagram for each.

An ideal rope has the same tension throughout, and a massless frictionless pulley only changes its direction.

Connected bodies share the same acceleration magnitude. Choose the positive direction to follow the expected motion so both get the same sign.

Example Atwood machine

Masses $m_1$ and $m_2$ hang either side of a pulley, with $m_1 > m_2$.

For $m_1$ (taking down as positive): $m_1 g - T = m_1 a$.

For $m_2$ (taking up as positive): $T - m_2 g = m_2 a$.

Adding eliminates $T$: $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}$.

Check the limits. Equal masses give $a = 0$. If $m_2 = 0$ then $a = g$, free fall. Both correct, so the algebra is probably right.

Tension is not weight

In the Atwood machine the tension is not $m_1 g$. If it were, the net force on $m_1$ would be zero and nothing would move. Solving gives $T = 2m_1m_2g/(m_1+m_2)$, which lies between the two weights.

Common pitfalls

• Drawing a centrifugal force. Circular motion needs an inward net force.
• Drawing a "force of motion" on a projectile.
• Treating weight and normal force as a third-law pair.
• Assuming $N = mg$ always. On an incline it is $mg\cos\theta$; in an accelerating elevator it is different again.
• Using one diagram for a two-body system. Draw each separately.

Practice

Work each fully before opening the answer. Original exercises written for this guide in the style of the course's problem sets.

Practice 1 · Constant velocity

A crate slides at constant velocity across a floor. What is the net force on it?

Answer

Zero. Constant velocity means zero acceleration, so by the second law the net force vanishes — even though individual forces like push and friction are present.

Practice 2 · Incline

Find the acceleration of a block on a frictionless $30$-degree incline.

Answer

$a = g\sin 30 = 4.9$ m/s$^2$ down the slope, independent of mass.

Practice 3 · Normal force

A $2$ kg book sits on a table. Someone presses down with $10$ N. Find the normal force.

Answer

$N = mg + 10 = 19.6 + 10 = 29.6$ N. The normal force adjusts to whatever keeps the vertical acceleration zero.

Practice 4 · Third law

A truck hits a mosquito. Which experiences the larger force?

Answer

Neither — the forces are equal and opposite. The mosquito's acceleration is enormous because its mass is tiny, which is a second-law consequence, not a third-law one.

Practice 5 · Elevator

You stand on a scale in an elevator accelerating upward at $2$ m/s$^2$. Does the reading rise or fall?

Solution

Rises. The scale reads $N$, and $N - mg = ma$ gives $N = m(g + a)$. For a $70$ kg person that is $826$ N instead of $686$ N. Accelerating downward would lower it, and free fall would give zero — apparent weightlessness.

Practice 6 · Two blocks

Blocks of $3$ kg and $2$ kg touch on a frictionless floor. A $10$ N force pushes the $3$ kg one. Find the contact force between them.

Solution

Whole system: $a = 10/5 = 2$ m/s$^2$. For the $2$ kg block alone the only horizontal force is the contact force, so $F = 2 \times 2 = 4$ N. Analysing the smaller block is the shortcut.

Practice 7 · Atwood check

For $m_1 = 5$ kg and $m_2 = 3$ kg, find $a$ and $T$.

Solution

$a = (2)(9.8)/8 = 2.45$ m/s$^2$; $T = 2(5)(3)(9.8)/8 = 36.75$ N. Note $T$ sits between the two weights, $29.4$ N and $49$ N, as it must.

Practice 8 · Rope with mass

Why is tension the same throughout an ideal rope, and when does that fail?

Solution

Applying the second law to any segment of a massless rope gives net force zero, so the tensions at its two ends match. With a real, heavy rope each segment has weight, so tension is largest at the top where it supports the most rope below it.

Expanded topic checklist

Course coverage. This unit is part of the complete PHY131H1 topic map. It should cover the following items before you move on:

• The three laws. Statements, and what each is used for.
• Free-body diagrams. One body at a time, real forces only, axes chosen to suit the motion.
• Third-law pairs. Why they act on different bodies and never cancel.
• Connected systems. Tension, shared acceleration, and eliminating unknowns.

Textbook/source alignment

The calendar lists "dynamics, force" here. Free-body diagrams are the single most examined skill in the course; check the syllabus on Quercus for assigned problems.

Core idea map

Forces cause acceleration; acceleration feeds Unit 2's kinematics. The free-body diagram is the translation step between the physical situation and the equations, and every later unit assumes you can draw one.

Study routine. For every force you draw, name the object exerting it. If you cannot, the force does not exist and should be erased. This single test removes centrifugal force and the force of motion permanently.

Deep practice extensions

Use these after the main practice set. They push on definitions and edge cases rather than routine calculation.

Extension 1 · First law

What does a net force of zero imply?

Guide

Constant velocity, which includes rest as a special case. Not necessarily stillness.

Extension 2 · Pair identification

Are weight and normal force a third-law pair?

Guide

No. Both act on the same body. A third-law pair always has one force on each of two different bodies.

Extension 3 · Normal force

Is $N$ always equal to $mg$?

Guide

No. On an incline it is $mg\cos\theta$; in an accelerating elevator it is $m(g \pm a)$. Solve for it from the vertical equation each time.

Extension 4 · Massless rope

Why is tension uniform in an ideal rope?

Guide

Because a massless segment has $ma = 0$, so the forces at its ends must balance.

← Unit 03Motion in two dimensions Unit 05 →Friction & circular motion