3.03MechanicsCore

Newton’s Laws of Motion

Kinematics describes how things move; $\textbf{dynamics}$ explains why. Newton’s three laws connect force, mass and acceleration, and with a clear force diagram they turn almost any mechanics problem — a box dragged across a floor, a sledge on a hill, a person in a lift — into an equation you can solve.

50 min Forces and Newton’s Laws

The big picture

Every dynamics question follows the same routine: draw a force diagram, choose directions, resolve forces, and apply $F = ma$ in each direction. The first law says an object’s velocity only changes when there is a resultant force; the second quantifies that change; the third says forces come in equal and opposite pairs acting on different objects. Real problems add three complications: forces at angles, which must be resolved into components; slopes, where it pays to resolve parallel and perpendicular to the surface; and friction, which opposes motion but can only grow up to a limit of $\mu R$ . Master the diagram and the resolving, and connected particles and equilibrium in the next lessons use exactly the same steps.

What you'll be able to do

State and apply Newton’s three laws of motion
Draw force diagrams including weight, normal reaction, tension, thrust and friction
Resolve forces and apply $F = ma$ in two perpendicular directions, including on inclined planes
Use the friction model $F \le \mu R$ , with $F = \mu R$ when sliding or on the point of sliding

The three laws

$\textbf{First law}$ : an object remains at rest or moves with constant velocity unless acted on by a resultant force.

$\textbf{Second law}$ : the resultant force equals mass times acceleration, $F = ma$ , in the direction of the acceleration. Force is measured in newtons, where $1$ N accelerates $1$ kg at $1$ m s⁻².

$\textbf{Third law}$ : if $A$ exerts a force on $B$ , then $B$ exerts an equal and opposite force on $A$ . The two forces act on $\textit{different}$ objects, so they never cancel in a single force diagram.

The $\textbf{weight}$ of a mass $m$ is $mg$ , acting vertically downwards.

F = ma, \qquad W = mg

Newton’s second law and weight.

Tip — Label forces with standard letters — $R$ for normal reaction, $T$ for tension, $F$ for friction, $mg$ for weight — and draw the acceleration as a separate double-headed arrow.

Resolving forces

A force $P$ at angle $\theta$ to the horizontal has horizontal component $P\cos\theta$ and vertical component $P\sin\theta$ .

Apply $F = ma$ separately in each direction. If there is no acceleration in a direction, the forces in that direction balance.

Example 1A 5 kg box on rough horizontal ground (

\mu = 0.3

) is pulled by a rope with tension 30 N at

25^{\circ}

above the horizontal. Find its acceleration.

1Vertically (no acceleration):

R + 30\sin 25^{\circ} = 5 \times 9.8

, so

R = 49 - 12.68 = 36.32

2Sliding, so friction

F = \mu R = 0.3 \times 36.32 = 10.90

3Horizontally:

30\cos 25^{\circ} - 10.90 = 5a

, so

27.19 - 10.90 = 5a

Answer

a = 3.26

m s⁻²

Why this works

Pulling upwards at an angle reduces the normal reaction, which reduces friction. That is why pulling a sledge with a raised rope is easier than pushing it down into the snow.

Inclined planes

On a slope at angle $\alpha$ , resolve parallel and perpendicular to the surface. Weight splits into $mg\sin\alpha$ down the slope and $mg\cos\alpha$ into the slope.

Perpendicular to the slope there is no acceleration, so $R = mg\cos\alpha$ (if no other forces act at an angle).

Along the slope, apply $F = ma$ with friction opposing the direction of motion.

Example 1A 4 kg block slides down a slope inclined at

30^{\circ}

with coefficient of friction

0.2

. Find its acceleration.

1Perpendicular:

R = 4g\cos 30^{\circ} = 33.95

2Friction up the slope:

0.2 \times 33.95 = 6.79

3Along the slope:

4g\sin 30^{\circ} - 6.79 = 4a

, so

19.6 - 6.79 = 4a

Answer

a = 3.20

m s⁻² down the slope

Friction

Friction opposes motion (or the tendency to move) along the surface. Its size adjusts to what is needed to prevent slipping, up to a maximum: $F \le \mu R$ , where $\mu$ is the $\textbf{coefficient of friction}$ .

When an object is sliding, or on the point of sliding ( $\textbf{limiting equilibrium}$ ), friction takes its maximum value $F = \mu R$ .

A surface described as $\textbf{smooth}$ has $\mu = 0$ ; a $\textbf{rough}$ surface has friction.

F \le \mu R

With equality when sliding or about to slide.

Example 1A 70 kg person stands in a lift accelerating upwards at 1.5 m s⁻². Find the normal reaction from the floor.

1Up positive:

R - 70g = 70 \times 1.5

R = 686 + 105

Answer

R = 791

N — the person feels heavier than their weight of 686 N.

Tip — Do not assume $F = \mu R$ for a stationary object. It only applies if you are told it is on the point of moving; otherwise find $F$ from equilibrium and check $F \le \mu R$ .

Think like an examiner

Draw a clear force diagram with all forces and the acceleration shown.
Resolve in two stated directions and write $F = ma$ for each.
Justify using $F = \mu R$ by saying the object is moving or in limiting equilibrium.
Use $g = 9.8$ m s⁻² and give answers to 3 significant figures.

Common misconceptions

Taking

R = mg

when a pulling force acts at an angle.

Resolve vertically: the vertical component of the pull changes

R

Cancelling third-law pairs in one diagram.

Third-law pairs act on different bodies.

Using

mg\cos\alpha

down the slope.

Down the slope is

mg\sin\alpha

; into the slope is

mg\cos\alpha

Assuming friction always equals

\mu R

Only when sliding or on the point of sliding.

Newton’s laws

F = ma

Second law.

W = mg

Weight.

F \le \mu R

Friction.

mg\sin\alpha,\ mg\cos\alpha

Weight on a slope.

Stretch yourself

A 2 kg box rests on a rough plane inclined at $20^{\circ}$ . The coefficient of friction is $0.5$ . Show that the box does not slide, and find the least force applied parallel to the slope, directed up the slope, that will make the box move up it.

Hint — Compare the component of weight down the slope with the maximum friction. For the second part, friction acts down the slope once the box is about to move up.

Reveal solution

SolutionPerpendicular:

R = 2g\cos 20^{\circ} = 18.42

N, so maximum friction is

0.5 \times 18.42 = 9.21

N. The component of weight down the slope is

2g\sin 20^{\circ} = 6.70

N. Since

6.70 < 9.21

, friction can hold the box: it does not slide (friction is

6.70

N, below its limit). To make it move up, friction acts down the slope at its maximum. On the point of moving:

P = 6.70 + 9.21 = 15.9

N. So the least force is

15.9

\blacksquare

Questions students ask

What does "resultant force" mean?

The single force equivalent to all the forces acting, found by adding them as vectors. It is what goes on the left of

F = ma

Why do I feel heavier in a lift accelerating upwards?

The floor must push with more than your weight to give you an upward acceleration, and it is that reaction you feel.

\mu

always less than 1?

Usually, but not necessarily — very grippy surfaces can have

\mu > 1

Which directions should I resolve in?

Any two perpendicular directions work. Choose them so that as many forces as possible lie along an axis — usually along and perpendicular to a slope.

Key takeaways

Newton’s first law: no resultant force means constant velocity.
Second law: $F = ma$ along the direction of acceleration.
Third law: equal and opposite forces on different bodies.
Resolve forces into perpendicular components; on slopes, along and perpendicular to the surface.
Friction $F \le \mu R$ , equal to $\mu R$ when sliding or about to slide.