Newton's laws

Inertia and inertial frames of reference, Newton's three laws: $a = \dfrac{F}{m}$ and $\vec F_{12} = -\vec F_{21}$.

Terms

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Formulas

Newton's second law

a = \dfrac{F}{m}

Quantities
Symbol	Name	Unit
$a$	acceleration of the body	m/s²
$F$	resultant of all the forces on the body	N
$m$	mass of the body	kg

Solved for another quantity:

F = ma

m = \dfrac{F}{a}

Note. The acceleration points the same way as the resultant. $1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2$ : a force of 1 N gives a body of mass 1 kg an acceleration of $1\ \text{m/s}^2$ . At $F = 0$ we get $a = 0$ , the first law.

Newton's third law

\vec F_{12} = -\vec F_{21}

Quantities
Symbol	Name	Unit
$\vec F_{12}$	force with which the first body acts on the second	N
$\vec F_{21}$	force with which the second body acts on the first	N

Solved for another quantity:

F_{12} = F_{21}

Note. For the magnitudes: $F_{12} = F_{21}$ , while the directions are opposite. The forces act on different bodies.

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

A cart under a resultant force

Friction is ignored: F is the resultant of all the forces. The position is drawn to scale: the edges of the track correspond to ±50 m.

Time0 s

Graphs

a(t)

v(t)

x(t)

a

2 m/s² (right →)

v

0 m/s

x

0 m

Formulas with your numbers

a = \dfrac{F}{m} = \dfrac{4}{2} = 2\ \text{m/s²}

v = v_0 + at = 0 + 2\cdot 0 = 0\ \text{m/s}

x = v_0 t + \dfrac{at^2}{2} = 0\ \text{m}

Newton's third law: two bodies push each other

The arrows are equal in length and point in opposite directions, but act on different bodies: F₁₂ on the second, F₂₁ on the first. The accelerations differ because the masses differ.

|F_{12}| = |F_{21}| = 6\ \text{N},\quad \dfrac{|a_1|}{|a_2|} = \dfrac{m_2}{m_1} = 2

Interaction force,

F

6 N

Mass of the first body,

m_1

2 kg

Mass of the second body,

m_2

4 kg

How to read the graphs

At $F = 0$ the $a(t)$ graph lies on the time axis, $v(t)$ is a horizontal line and $x(t)$ is a straight line: the body moves uniformly (first law).
For a constant resultant $a(t)$ is a horizontal line at the height $a = \dfrac{F}{m}$ : the greater the force, the higher the line; the greater the mass, the lower it is (second law).
$v(t)$ is a straight line whose slope equals the acceleration $a$ ; it starts at the point $v_0$ . If the force points against the velocity, the speed decreases.
For $a \neq 0$ the $x(t)$ graph is a parabola; the greater the acceleration, the more sharply it bends.
In the third-law mini-scene the force arrows are always equal in length and point in opposite directions, while the accelerations of the bodies are inversely proportional to their masses.

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Terms

Studied 0 of 5

Formulas

Newton's second law

a = \dfrac{F}{m}

Quantities
Symbol	Name	Unit
$a$	acceleration of the body	m/s²
$F$	resultant of all the forces on the body	N
$m$	mass of the body	kg

Solved for another quantity:

F = ma

m = \dfrac{F}{a}

Newton's third law

\vec F_{12} = -\vec F_{21}

Quantities
Symbol	Name	Unit
$\vec F_{12}$	force with which the first body acts on the second	N
$\vec F_{21}$	force with which the second body acts on the first	N

Solved for another quantity:

F_{12} = F_{21}

Note. For the magnitudes: $F_{12} = F_{21}$ , while the directions are opposite. The forces act on different bodies.

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

A cart under a resultant force

Friction is ignored: F is the resultant of all the forces. The position is drawn to scale: the edges of the track correspond to ±50 m.

Time0 s

Graphs

a(t)

v(t)

x(t)

a

2 m/s² (right →)

v

0 m/s

x

0 m

Formulas with your numbers

a = \dfrac{F}{m} = \dfrac{4}{2} = 2\ \text{m/s²}

v = v_0 + at = 0 + 2\cdot 0 = 0\ \text{m/s}

x = v_0 t + \dfrac{at^2}{2} = 0\ \text{m}

Newton's third law: two bodies push each other

The arrows are equal in length and point in opposite directions, but act on different bodies: F₁₂ on the second, F₂₁ on the first. The accelerations differ because the masses differ.

|F_{12}| = |F_{21}| = 6\ \text{N},\quad \dfrac{|a_1|}{|a_2|} = \dfrac{m_2}{m_1} = 2

Interaction force,

F

6 N

Mass of the first body,

m_1

2 kg

Mass of the second body,

m_2

4 kg

How to read the graphs

F = 0

the

a(t)

graph lies on the time axis,

v(t)

is a horizontal line and

x(t)

is a straight line: the body moves uniformly (first law).

For a constant resultant

a(t)

is a horizontal line at the height

a = \dfrac{F}{m}

: the greater the force, the higher the line; the greater the mass, the lower it is (second law).

v(t)

is a straight line whose slope equals the acceleration

a

; it starts at the point

v_0

. If the force points against the velocity, the speed decreases.

For

a \neq 0

the

x(t)

graph is a parabola; the greater the acceleration, the more sharply it bends.

In the third-law mini-scene the force arrows are always equal in length and point in opposite directions, while the accelerations of the bodies are inversely proportional to their masses.