Free body diagrams — systematic drawing technique

WHAT is a free body diagram?

The three words that matter:

• Free — the body is freed (isolated) from its surroundings.
• Body — one chosen object (the "system").
• Diagram — pure force picture, nothing else.

HOW: the systematic recipe (the 80/20 core)

Deriving the equations FROM the diagram

The FBD is the bridge to algebra. Once arrows are drawn, Newton's 2nd law splits per axis:

$\sum F_x = ma_x \qquad \sum F_y = ma_y$

Worked Example 1 — Block on a table, you push horizontally

A block (mass $m$) sits on a rough table. You push horizontally with force $F$. Draw the FBD and find acceleration.

Step 1–2: System = block, shrink to dot. Why this step? Fixes what "external" means — the table and your hand are environment.

Step 3: Gravity $mg$ down. Why? Long-range, always there.

Step 4 (contacts): Block touches the table (→ normal $N$ up, friction $f$ backward opposing motion) and your hand (→ applied $F$ forward). Why backward friction? Friction opposes the tendency to slide; block tends to slide forward, so $f$ points back.

Step 6 — resolve: $y:\quad N - mg = 0 \Rightarrow N = mg$ Why = 0? No vertical acceleration (block stays on table). $x:\quad F - f = ma \Rightarrow a = \frac{F-f}{m}$ Why? Net horizontal force drives horizontal acceleration.

Worked Example 2 — Block on a frictionless incline (angle $\theta$)

Steps 1–4: System = block. Gravity $mg$ down. Surface contact → normal $N$ ⊥ to incline. No friction.

Step 6 — tilt axes so $x$ points down-slope, $y$ ⊥ slope. Why tilt? So acceleration is purely along $x$; $a_y=0$.

Resolve gravity (the only tilted force):

• Component along incline (down-slope): $mg\sin\theta$
• Component perpendicular (into surface): $mg\cos\theta$

Why $\sin$ for along-slope? As $\theta\to90°$ (vertical wall) the slope direction becomes vertical, and all of gravity acts along it — $\sin90°=1$. ✓ This sanity check tells you which trig function goes where.

$y:\quad N - mg\cos\theta = 0 \Rightarrow N = mg\cos\theta$ $x:\quad mg\sin\theta = ma \Rightarrow a = g\sin\theta$

Why no $m$ in $a$? Mass cancels — heavier blocks aren't faster, just like free-fall.

Worked Example 3 — Two blocks joined by a string over a pulley (Atwood-ish, on a table)

Block $A$ ($m_1$) on a frictionless table, string over a pulley to a hanging block $B$ ($m_2$).

Two FBDs — one per body! Why two? Each body needs its own diagram; tension links them.

FBD of $A$ (horizontal): tension $T$ forward, $N$ up, $m_1g$ down. $T = m_1 a$ FBD of $B$ (hanging): weight $m_2g$ down, tension $T$ up. $m_2g - T = m_2 a$ Why same $T$, same $a$? One ideal string (massless, inextensible) → equal tension throughout and equal speed.

Add the two equations (eliminate $T$): $m_2 g = (m_1+m_2)a \Rightarrow a = \frac{m_2}{m_1+m_2}\,g$ Why add? It cancels the internal tension, leaving only the driving weight vs total inertia.

Common Mistakes (Steel-manned)

Active Recall

Connections

• Newton's Second Law — the FBD feeds directly into $\sum F = m\vec a$.
• Newton's Third Law — explains why reaction forces live on other bodies' diagrams.
• Normal Force and Friction — the contact forces you'll draw most often.
• Tension in Strings and Pulleys — multi-body FBD coupling.
• Inclined Plane Problems — the canonical axis-tilting application.
• Resolving Vectors into Components — the math step after drawing arrows.

Concept Map

Hinglish (regional understanding)