Classical-Mechanics · Unit 8 · Video 3 · Interactive Practice

The Free-Body Diagram: Deciding Which Forces Matter

IKey Formulas

Formula	Name	What it says
$\vec{F} = \vec{F}_1 + \vec{F}_2 + \cdots$	Total force	Vector sum of every force on the system
$\vec{F} = F_x\,\hat{\imath} + F_y\,\hat{\jmath} + F_z\,\hat{k}$	Cartesian components	Resolve the total force onto the axes
$F_x = F_{1,x} + F_{2,x} + \cdots$	Component-wise sum	Each component adds independently

Key Insight: A free-body diagram counts only the forces acting on the system — never the forces the system exerts on everything else.

IINet Force as a Vector Sum

Two forces on a system combine tip-to-tail into one net force — and the magnitudes generally don't add.

💡 Challenge: arrange the two forces so the net force is zero — the system is then in equilibrium.

IIIResolving a Force into Components

Every force casts perpendicular shadows on the axes: $\vec{F} = F_x\,\hat{\imath} + F_y\,\hat{\jmath}$ , with $F_x = |\vec{F}|\cos\theta$ and $F_y = |\vec{F}|\sin\theta$ .

Magnitude |F| 5.0 Angle θ 35°

💡 Each component is itself a sum: $F_x = F_{1,x} + F_{2,x} + \cdots$ — the diagram does the bookkeeping, one axis at a time.

IVWhich Forces Belong in the Model

The model is a choice — include the forces that matter, drop those both small and hard to compute.

💡 Whenever objects move, some friction is always present; the only question is whether the model can afford to ignore it.

VQuiz Questions

Problem 1 · Add the Forces

Given: two forces act on a system, $\vec{F}_1 = (3,\,-2)$ N and $\vec{F}_2 = (-1,\,5)$ N — find the net force $\vec{F}$ .

✅ Correct! Add components with their signs:

3 + (-1) = 2

and

-2 + 5 = 3

❌ Watch the signs. You added magnitudes; the components are signed, so

3 + (-1) = 2

, not

4

❌ Not quite. Add the two vectors component by component, keeping every sign.

Show solution

Vectors add component by component:

\vec{F} = \vec{F}_1 + \vec{F}_2 = (3 + (-1),\ -2 + 5) = (2,\,3)\text{ N}

The net force is $(2,\,3)$ N — the $x$ -components combine to $2$ and the $y$ -components to $3$ .

Problem 2 · Magnitudes Don't Add

Given: $\vec{F}_1$ points east with magnitude $3$ N and $\vec{F}_2$ points north with magnitude $4$ N — find the magnitude of the net force $|\vec{F}|$ . ⚠️ Direction matters!

✅ Correct! Perpendicular forces combine by the Pythagorean theorem:

\sqrt{3^2 + 4^2} = 5

❌ Close, but that's $3 + 4$ . Magnitudes only add when the forces are parallel; here they are perpendicular.

❌ Not quite. The forces are at right angles — use

|\vec{F}| = \sqrt{F_x^2 + F_y^2}

Show solution

East and north are perpendicular, so the components are $F_x = 3$ N and $F_y = 4$ N:

|\vec{F}| = \sqrt{F_x^2 + F_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ N}

Adding magnitudes ( $3 + 4 = 7$ ) would only be right if both forces pointed the same way.

Problem 3 · Resolve into Components

Given: a force of magnitude $|\vec{F}| = 10$ N acts at $\theta = 30\degree$ above the $+x$ axis — find $F_x$ and $F_y$ .

What is $F_x$ ?

What is $F_y$ ?

✅ Correct!

F_x = 10\cos 30\degree \approx 8.66

N and

F_y = 10\sin 30\degree = 5

❌ Check the horizontal leg. The

x

-component uses cosine:

F_x = |\vec{F}|\cos\theta

❌ Check the vertical leg. The

y

-component uses sine:

F_y = |\vec{F}|\sin\theta

Show solution

Drop perpendiculars from the tip of $\vec{F}$ onto each axis:

F_x = |\vec{F}|\cos\theta = 10\cos 30\degree = 10 \cdot \tfrac{\sqrt{3}}{2} \approx 8.66\text{ N}

F_y = |\vec{F}|\sin\theta = 10\sin 30\degree = 10 \cdot \tfrac{1}{2} = 5\text{ N}

Swapping sine and cosine is the classic slip — cosine goes with the axis the angle is measured from.

Problem 4 · Build the Model

Given: a metal block is pushed by hand across a rough table. Which set of forces gives the best working model?

✅ Correct! The essentials plus surface friction — simple enough to keep, small hard effects left out.

❌ Too few. Surface friction is usually significant here and its formula is simple, so dropping it makes the prediction wrong.

❌ Too many. Nothing is magnetic, and air resistance is small yet messy — including them makes the math intractable for no gain.

❌ Not quite. Gravity and the normal force are essential — leaving either out breaks the model.

Show solution

Keep the essentials: gravity, the normal force from the table, and the applied push all clearly act and all matter.

Keep surface friction: on a rough table it is significant, and its formula is simple — so include it.

Drop the small-and-hard: air resistance is often negligible here and its mathematics is messy; the magnetic force isn't acting at all.

Best model: gravity, normal, applied push, and surface friction. Include too few and the prediction is wrong; include too many and the math becomes impossible.

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