Classical-Mechanics · Unit 6 · Video 1 · Interactive Practice

Defining Force: Mass, Acceleration, and Force as a Vector

IKey Formulas

Formula	Name	What it says
$\vec{F} \equiv m\,\vec{a}$	Newton's definition of force	Force is inertial mass times acceleration
$\lvert\vec{F}\rvert = m\,\lvert\vec{a}\rvert$	Magnitude & direction	Direction of $\vec{F}$ = direction of $\vec{a}$
$\vec{F}^{\,T} = \vec{F}_1 + \vec{F}_2$	Forces add as vectors	Head-to-tail, because accelerations do
$m_{\text{in}} = m_s\,\dfrac{a_s}{a_{\text{in}}}$	Mass calibration	Weigh any body against a standard

Key Insight: $\vec{F}^{\,T} = \vec{F}_1 + \vec{F}_2$ is not a definition — it is forced on us by experiment. Applying both forces together gives $\vec{a} = \vec{a}_1 + \vec{a}_2$ ; multiplying through by $m_s$ carries the same head-to-tail rule over to the forces.

IIVisualization 1 — Forces Add Head-to-Tail

A force points where its acceleration points, so two forces combine tip-to-tail into one resultant — the reason force is a vector.

💡 Notice: drag the tips so the resultant shrinks to the zero vector — the pushes and pulls cancel and the body has zero acceleration. That balanced case is exactly the post in Example 7.1 below.

IIIVisualization 2 — Example 7.1: The Ground Force on a Post

A post in equilibrium: add the two rope pulls and gravity component by component, then read off the ground's reaction — its $1\ \text{kg}$ mass never enters.

Step 1

IVVisualization 3 — Calibrating an Unknown Mass

Push two bodies with the same force: the one that accelerates less must carry more mass, in exact inverse proportion.

Unknown acceleration aᵢₙ (m/s²) 4

💡 Notice: repeat with a different force and $a_s,\ a_{\text{in}}$ both change — but their ratio, and therefore $m_{\text{in}}$ , does not. Inertial mass is a property of the body, not of the force used to weigh it.

VQuiz Questions

Problem 1 · Newton's Definition

Given: a force gives a body of mass $m = 2\ \text{kg}$ an acceleration of magnitude $\lvert\vec{a}\rvert = 6\ \text{m/s}^2$ . Find the magnitude of the force $\lvert\vec{F}\rvert$ .

✅ Correct!

\lvert\vec{F}\rvert = m\,\lvert\vec{a}\rvert = 2 \times 6 = 12\ \text{N}

❌ Close — check the operation. That is

a/m = 3

. Force is

m\,\lvert\vec{a}\rvert

: multiply, giving

2 \times 6 = 12\ \text{N}

❌ Not quite. Force is proportional to acceleration through the mass:

\lvert\vec{F}\rvert = m\,\lvert\vec{a}\rvert

, not

m + a

a - m

Show solution

Newton defines force as inertial mass times acceleration:

\lvert\vec{F}\rvert = m\,\lvert\vec{a}\rvert = (2\ \text{kg})(6\ \text{m/s}^2) = 12\ \text{N}

Force depends on acceleration — how the motion is changing — not on speed or position.

Problem 2 · Vectors, Not Magnitudes

Given: two forces act on a body, $\vec{F}_1 = 30\,\hat{\imath}\ \text{N}$ and $\vec{F}_2 = 40\,\hat{\jmath}\ \text{N}$ . Find the magnitude of the net force $\lvert\vec{F}_1 + \vec{F}_2\rvert$ .

✅ Correct! The net is

30\,\hat{\imath} + 40\,\hat{\jmath}

, so

\lvert\vec{F}\rvert = \sqrt{30^2 + 40^2} = 50\ \text{N}

❌ Close, but forces add as vectors. You added the magnitudes (

30 + 40

). Because these forces are perpendicular, use Pythagoras:

\sqrt{30^2 + 40^2} = 50\ \text{N}

❌ Not quite. Add the components to get

30\,\hat{\imath} + 40\,\hat{\jmath}

, then take

\sqrt{30^2 + 40^2}

Show solution

Step 1 — add component by component:

\vec{F}_1 + \vec{F}_2 = (30 + 0)\,\hat{\imath} + (0 + 40)\,\hat{\jmath} = 30\,\hat{\imath} + 40\,\hat{\jmath}\ \text{N}

Step 2 — magnitude:

\lvert\vec{F}\rvert = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \text{N}

Only when forces are parallel do magnitudes simply add.

Problem 3 · Equilibrium of a Post

Given: a post in equilibrium is pulled by ropes $\vec{F}_1 = 50\,\hat{\imath} + 20\,\hat{\jmath}\ \text{N}$ and $\vec{F}_2 = -20\,\hat{\imath} + 30\,\hat{\jmath}\ \text{N}$ , plus gravity $\vec{F}_g = -10\,\hat{\jmath}\ \text{N}$ . Find the ground force $\vec{F}_{\text{ground}}$ and its magnitude.

The ground force $\vec{F}_{\text{ground}}$ is:

Its magnitude $\lvert\vec{F}_{\text{ground}}\rvert$ is:

✅ Correct! The net pull is

30\,\hat{\imath} + 40\,\hat{\jmath}

; the ground opposes it with

-30\,\hat{\imath} - 40\,\hat{\jmath}\ \text{N}

, magnitude

50\ \text{N}

❌ Check the direction. That is the net applied force. For equilibrium the ground must oppose it:

\vec{F}_{\text{ground}} = -(30\,\hat{\imath} + 40\,\hat{\jmath})

❌ Recompute the net. Sum

x

50 - 20 = 30

; sum

y

20 + 30 - 10 = 40

. Then negate for the reaction.

❌ Take the square root.

\lvert\vec{F}\rvert = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50\ \text{N}

, not

2500

Show solution

Step 1 — net applied force (post is not accelerating, so all forces sum to the reaction's negative):

\vec{F}_{\text{net}} = (50 - 20)\,\hat{\imath} + (20 + 30 - 10)\,\hat{\jmath} = 30\,\hat{\imath} + 40\,\hat{\jmath}\ \text{N}

Step 2 — the ground opposes it exactly:

\vec{F}_{\text{ground}} = -\vec{F}_{\text{net}} = -30\,\hat{\imath} - 40\,\hat{\jmath}\ \text{N}

Step 3 — magnitude:

\lvert\vec{F}_{\text{ground}}\rvert = \sqrt{(-30)^2 + (-40)^2} = \sqrt{2500} = 50\ \text{N}

The post's mass never entered — vector addition of the pushes and pulls is the whole computation.

Problem 4 · Weighing Against a Standard

Given: the same force gives a standard body ( $m_s = 1\ \text{kg}$ ) an acceleration $a_s = 6\ \text{m/s}^2$ , and gives an unknown body an acceleration $a_{\text{in}} = 2\ \text{m/s}^2$ . Find the unknown inertial mass $m_{\text{in}}$ .

✅ Correct!

m_{\text{in}} = m_s\,\dfrac{a_s}{a_{\text{in}}} = 1 \cdot \dfrac{6}{2} = 3\ \text{kg}

— the slower body is heavier.

❌ The ratio is inverted. You used

a_{\text{in}}/a_s

. The formula is

m_{\text{in}} = m_s\,a_s/a_{\text{in}} = 6/2 = 3\ \text{kg}

: less acceleration means more mass.

❌ Not quite. Set the shared force equal for both bodies:

m_{\text{in}}\,a_{\text{in}} = m_s\,a_s

, then solve for

m_{\text{in}}

Show solution

The same force acts on both bodies, so:

m_{\text{in}}\,a_{\text{in}} = m_s\,a_s

Solve for the unknown mass:

m_{\text{in}} = m_s\,\frac{a_s}{a_{\text{in}}} = (1\ \text{kg}) \cdot \frac{6}{2} = 3\ \text{kg}

The unknown body accelerates one-third as fast ( $2$ vs $6$ ), so it has three times the mass. A bigger acceleration always means a smaller mass.

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