Free body diagrams — systematic drawing technique: Worked examples

The scenario matrix

Before working anything, let's list every case class an FBD problem lives in. Each worked example below is tagged with the cell it fills.

#	Case class	What's special about it	Example
A	Flat surface, horizontal push	Simplest: axes already aligned	Ex 1
B	Flat surface, push at an angle	Applied force must be split; changes $N$	Ex 2
C	Incline, frictionless	Gravity is the tilted vector	Ex 3
D	Incline, with friction (limit: about to slip)	Friction sign + the static/kinetic threshold	Ex 4
E	Connected bodies (string + pulley)	Two FBDs, shared $T$ and $a$	Ex 5
F	Degenerate: $θ = 0$ and $θ = 90°$	Limiting angles — does the formula stay sane?	Ex 6
G	Sign/quadrant: push down-slope vs a block that would slide	Which way does friction point? Both signs	Ex 7
H	Real-world word problem (elevator / scale)	Apparent weight, non-zero vertical $a$	Ex 8
I	Exam twist: normal force going to zero (lift-off)	Degenerate contact: when does $N = 0$ ?	Ex 9

We will use only these tools, each introduced when first needed:

Newton's 2nd law per axis, $\sum F_{x} = m a_{x}$ , $\sum F_{y} = m a_{y}$ — see Newton's Second Law.
Splitting a vector into components with $sin$ and $cos$ — see Resolving Vectors into Components.
Normal force and friction rules — see Normal Force and Friction.
Tension in ideal strings — see Tension in Strings and Pulleys.
Incline geometry — see Inclined Plane Problems.
Action–reaction bookkeeping — see Newton's Third Law.

Throughout, take $g = 9.8 m/s^{2}$ unless told otherwise.

Ex 1 — Flat table, horizontal push (Cell A)

Figure — Free body diagrams — systematic drawing technique

Step 1 — draw the FBD. Why this step? Every problem starts here; the arrows become the equations. Look at the figure: gravity $m g$ down (blue), normal $N$ up (yellow), your push $F$ right (green), friction $f$ left (red, opposing the slide).

Step 2 — vertical axis. $y : N - m g = 0 \Rightarrow N = m g = 4 (9.8) = 39.2 N$ Why $= 0$ ? The block does not leave or sink into the floor, so $a_{y} = 0$ .

Step 3 — friction size. $f = μ_{k} N = 0.25 (39.2) = 9.8 N$ Why now? Friction magnitude depends on $N$ , so we needed $N$ first.

Step 4 — horizontal axis. $x : F - f = ma \Rightarrow a = \frac{20 - 9.8}{4} = 2.55 m/s^{2}$ Why $F - f$ ? $F$ drives forward, $f$ resists; they subtract.

Ex 2 — Push at an angle (Cell B)

Step 1 — split the applied force. Why split? $F$ points along neither axis, and Newton's law wants per-axis sums. Using Resolving Vectors into Components: $F_{x} = F cos 30° = 20 (0.8660) = 17.32 N (forward)$ $F_{y} = F sin 30° = 20 (0.5) = 10 N (downward)$ Why $cos$ for horizontal? The angle is measured from the horizontal, so the adjacent side (horizontal) uses $cos$ .

Step 2 — vertical axis (note the extra downward push!). $y : N - m g - F_{y} = 0 \Rightarrow N = m g + F_{y} = 39.2 + 10 = 49.2 N$ Why add $F_{y}$ ? Your downward push and gravity both press the block down, so the floor must push up harder.

Step 3 — friction. $f = μ_{k} N = 0.25 (49.2) = 12.3 N$

Step 4 — horizontal axis. $x : F_{x} - f = ma \Rightarrow a = \frac{17.32 - 12.3}{4} = 1.255 m/s^{2}$

Ex 3 — Frictionless incline (Cell C)

Step 1 — tilt the axes. Why tilt? Put $x$ down-slope, $y$ perpendicular to the surface. Then acceleration is purely along $x$ and $a_{y} = 0$ , killing one unknown. See Inclined Plane Problems.

Step 2 — split gravity. Why gravity? It's now the only tilted force. From the figure's angle geometry: $along slope = m g sin θ, into slope = m g cos θ$ Why $sin$ along-slope? As $θ \to 90°$ the slope becomes a vertical wall and all gravity acts along it; $sin 90° = 1$ ✓.

Step 3 — the two axes. $y : N - m g cos θ = 0 \Rightarrow N = m g cos 30° = 39.2 (0.8660) = 33.95 N$ $x : m g sin θ = ma \Rightarrow a = g sin 30° = 9.8 (0.5) = 4.9 m/s^{2}$

Ex 4 — Incline WITH friction, at the slip threshold (Cell D)

Step 1 — condition for "just about to slide". Why? At the threshold the block is still static but friction has reached its maximum, $f_{ma x} = μ_{s} N$ , and everything is still in equilibrium ( $a = 0$ ).

Step 2 — the two axes. $y : N = m g cos θ$ $x : m g sin θ = f_{ma x} = μ_{s} N = μ_{s} m g cos θ$

Step 3 — solve. Why divide? Cancel $m g$ (and hence mass — it doesn't matter!): $\frac{s i n θ}{c o s θ} = μ_{s} \Rightarrow tan θ^{*} = μ_{s}$ Why $tan$ ? $tan = sin / cos$ is exactly the ratio of "down-slope pull" to "into-slope press", so it measures how steeply gravity is tugging along the slope. $θ^{*} = arctan (0.6) = 30.96°$

Ex 5 — Connected blocks over a pulley (Cell E)

Step 1 — two FBDs. Why two? Each body gets its own diagram; the string's tension $T$ is the messenger between them. Ideal string ⇒ same $T$ throughout and same $∣ a ∣$ (inextensible) — see Tension in Strings and Pulleys.

$A : T = m_{1} a$ $B : m_{2} g - T = m_{2} a$ Why these signs? On $A$ , only $T$ acts horizontally (forward). On $B$ , weight pulls down (its direction of motion) and $T$ pulls up.

Step 2 — add to kill $T$ . $m_{2} g = (m_{1} + m_{2}) a \Rightarrow a = \frac{m _{2}}{m _{1} + m _{2}} g = \frac{2}{5} (9.8) = 3.92 m/s^{2}$

Step 3 — back-substitute for $T$ . $T = m_{1} a = 3 (3.92) = 11.76 N$

Ex 6 — Degenerate incline angles $θ = 0$ and $θ = 90°$ (Cell F)

Step 1 — plug $θ = 0$ . Why check this? A correct formula must survive its boundaries. $a = g sin 0 = 0, N = m g cos 0 = m g = 39.2 N$ This is just a block resting on a flat table ✓.

Step 2 — plug $θ = 90°$ . $a = g sin 90° = 9.8 m/s^{2}, N = m g cos 90° = 0 N$ A vertical wall touches the block but presses with zero force — the block is in free-fall ✓.

Ex 7 — Sign of friction: two directions (Cell G)

Setup (both cases). $N = m g cos 20° = 5 (9.8) (0.9397) = 46.04 N, f = μ_{k} N = 0.30 (46.04) = 13.81 N$ Gravity down-slope: $m g sin 20° = 5 (9.8) (0.3420) = 16.76 N$ .

Case (i) — sliding down (friction up-slope, i.e. $- x$ ). Why up-slope? Motion is down-slope, friction opposes it. $m g sin θ - f = ma \Rightarrow a = \frac{16.76 - 13.81}{5} = + 0.590 m/s^{2} (down-slope)$

Case (ii) — sliding up (friction down-slope, i.e. $+ x$ ). Why down-slope? Motion is up-slope, friction opposes it — so now friction and gravity both point down-slope. $m g sin θ + f = ma \Rightarrow a = \frac{16.76 + 13.81}{5} = + 6.11 m/s^{2} (down-slope)$ The block decelerates fast, stops, then may slide back down under Case (i).

Ex 8 — Real-world word problem: elevator scale (Cell H)

Step 1 — FBD of the person. Why the person, not the scale? We want the force on the person; by Newton's Third Law the scale reads the equal-and-opposite force. Two forces: weight $m g$ down, normal $N$ up.

Step 2 — vertical Newton's law (up positive). $N - m g = ma \Rightarrow N = m (g + a)$ Why $+ ma$ , not $= 0$ ? The elevator (and person) genuinely accelerate; this is not equilibrium.

Step 3 — plug both cases. $Up: N = 60 (9.8 + 2) = 708 N$ $Down: N = 60 (9.8 - 2) = 468 N$

Ex 9 — Exam twist: when does the block lift off? (Cell I)

Step 1 — FBD while still on the floor. Why "while on floor"? Contact still exists, so $N \geq 0$ is a valid arrow. Forces: $P$ up, $N$ up, $m g$ down.

Step 2 — vertical equilibrium (still resting, $a = 0$ ). $P + N - m g = 0 \Rightarrow N = m g - P$ Why can we set $a = 0$ ? Until it lifts, the block is motionless.

Step 3 — impose the lift-off condition $N = 0$ . Why $N = 0$ ? A surface cannot pull down; the smallest possible push is zero, and that's the moment contact is about to break. $0 = m g - P \Rightarrow P = m g = 2 (9.8) = 19.6 N$

Active Recall

Why does pushing a block at an angle below horizontal reduce its acceleration?

The downward component adds to

N

, increasing friction

f = μ_{k} N

, which opposes motion more.

At the slip threshold on an incline, what equation gives the angle?

tan θ = μ_{s}

(mass and

g

cancel).

In Ex 7, why is the up-slope block's deceleration larger than the down-slope block's acceleration?

Moving up, friction and gravity both point down-slope and add; moving down they oppose and subtract.

What condition marks a block lifting off a surface?

N = 0

; beyond that the equilibrium formula gives impossible negative

N

Scale reading in an elevator with upward acceleration

a

N = m (g + a)

, larger than the true weight

m g