Worked examples — Free body diagrams — systematic drawing technique

1.2.4 · D3 · Physics › Newton's Laws & Dynamics › Free body diagrams — systematic drawing technique

Scenario matrix

Kuch bhi work karne se pehle, har case class list karte hain jisme ek FBD problem exist karti hai. Neeche har worked example us cell ke saath tagged hai jo wo fill karta hai.

Hum sirf yeh tools use karenge, har ek tab introduce hoga jab pehli baar zaroorat padegi:

• Newton's 2nd law per axis, $\sum F_x=m{a}_x$, $\sum F_y=m{a}_y$ — dekho Newton's Second Law.
• Vector ko components mein split karna $\sin$ aur $\cos$ se — dekho Resolving Vectors into Components.
• Normal force aur friction ke rules — dekho Normal Force and Friction.
• Tension ideal strings mein — dekho Tension in Strings and Pulleys.
• Incline geometry — dekho Inclined Plane Problems.
• Action–reaction bookkeeping — dekho Newton's Third Law.

Poore notes mein, $g=9.8{\text{m/s}}^2$ lo jab tak kuch aur na bola jaye.

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

Step 1 — FBD draw karo. Yeh step kyun? Har problem yahan se shuru hoti hai; arrows hi equations banti hain. Figure dekho: gravity $mg$ neeche (blue), normal $N$ upar (yellow), tumhara push $F$ right (green), friction $f$ left (red, slide oppose karta hua).

Step 2 — vertical axis. $y:N-mg=0\Rightarrow N=mg=4(9.8)=39.2\text{N}$ $=0$ kyun? Block floor se uthta nahi aur na hi andar jaata hai, toh $a_y=0$.

Step 3 — friction size. $f={\mu }_{k}N=0.25(39.2)=9.8\text{N}$ Abhi kyun? Friction ka magnitude $N$ pe depend karta hai, toh pehle $N$ chahiye tha.

Step 4 — horizontal axis. $x:F-f=ma\Rightarrow a=\frac{20-9.8}{4}=2.55{\text{m/s}}^2$ $F-f$ kyun? $F$ aage drive karta hai, $f$ resist karta hai; wo subtract hote hain.

Ex 2 — Push at an angle (Cell B)

Step 1 — applied force split karo. Split kyun? $F$ kisi bhi axis ke along point nahi karta, aur Newton's law per-axis sums chahta hai. Resolving Vectors into Components use karke: $F_x=F\cos 30°=20(0.8660)=17.32\text{N (forward)}$ $F_y=F\sin 30°=20(0.5)=10\text{N (downward)}$ Horizontal ke liye $\cos$ kyun? Angle horizontal se measure hota hai, toh adjacent side (horizontal) $\cos$ use karti hai.

Step 2 — vertical axis (extra downward push note karo!). $y:N-mg-F_y=0\Rightarrow N=mg+F_y=39.2+10=49.2\text{N}$ $F_y$ add kyun? Tumhara downward push aur gravity dono block ko neeche dabate hain, toh floor ko upar zyada push karna padta hai.

Step 3 — friction. $f={\mu }_{k}N=0.25(49.2)=12.3\text{N}$

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

Ex 3 — Frictionless incline (Cell C)

Step 1 — axes tilt karo. Tilt kyun? $x$ down-slope rakho, $y$ surface ke perpendicular. Tab acceleration purely $x$ along hoga aur $a_y=0$, ek unknown khatam. Dekho Inclined Plane Problems.

Step 2 — gravity split karo. Gravity kyun? Yahi ab single tilted force hai. Figure ke angle geometry se: $\text{along slope}=mg\sin \theta ,\text{into slope}=mg\cos \theta$ Along-slope ke liye $\sin$ kyun? Jab $\theta \to 90°$ slope ek vertical wall ban jaati hai aur saari gravity along it act karti hai; $\sin 90°=1$ ✓.

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

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

Step 1 — "just about to slide" ki condition. Kyun? Threshold pe block abhi bhi static hai lekin friction apne maximum pe pahunch gayi hai, $f_{max}={\mu }_{s}N$, aur sab kuch still equilibrium mein hai ($a=0$).

Step 2 — do axes. $y:N=mg\cos \theta$ $x:mg\sin \theta =f_{max}={\mu }_{s}N={\mu }_{s}mg\cos \theta$

Step 3 — solve karo. Divide kyun? $mg$ cancel karo (aur isliye mass — koi farq nahi padta!): $\frac{\sin \theta }{\cos \theta }={\mu }_s\Rightarrow \tan {\theta }^{\ast }={\mu }_s$ $\tan$ kyun? $\tan =\sin /\cos$ exactly "down-slope pull" aur "into-slope press" ka ratio hai, toh yeh measure karta hai ki gravity slope along kitni steeply tug kar rahi hai. ${\theta }^{\ast }=\arctan (0.6)=30.96°$

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

Step 1 — do FBDs. Do kyun? Har body ka apna diagram hota hai; string ka tension $T$ unke beech messenger hai. Ideal string ⇒ same $T$ throughout aur same $|a|$ (inextensible) — dekho Tension in Strings and Pulleys.

$A:T=m_{1}a$ $B:m_{2}g-T=m_{2}a$ Yeh signs kyun? $A$ pe, sirf $T$ horizontally act karta hai (forward). $B$ pe, weight neeche pull karta hai (uski motion ki direction) aur $T$ upar pull karta hai.

Step 2 — add karke $T$ khatam karo. $m_{2}g=(m_1+m_2)a\Rightarrow a=\frac{m_2}{m_1+m_2}g=\frac{2}{5}(9.8)=3.92{\text{m/s}}^2$

Step 3 — $T$ ke liye back-substitute karo. $T=m_{1}a=3(3.92)=11.76\text{N}$

Ex 6 — Degenerate incline angles $\theta =0$ aur $\theta =90°$ (Cell F)

Step 1 — $\theta =0$ plug karo. Yeh check kyun? Ek sahi formula apni boundaries par survive karni chahiye. $a=g\sin 0=0,N=mg\cos 0=mg=39.2\text{N}$ Yeh sirf flat table pe resting block hai ✓.

Step 2 — $\theta =90°$ plug karo. $a=g\sin 90°=9.8{\text{m/s}}^2,N=mg\cos 90°=0\text{N}$ Ek vertical wall block ko touch karti hai lekin zero force se press karti hai — block free-fall mein hai ✓.

Ex 7 — Friction ka sign: do directions (Cell G)

Setup (dono cases). $N=mg\cos 20°=5(9.8)(0.9397)=46.04\text{N},f={\mu }_{k}N=0.30(46.04)=13.81\text{N}$ Gravity down-slope: $mg\sin 20°=5(9.8)(0.3420)=16.76\text{N}$.

Case (i) — neeche slide karna (friction up-slope, yani $-x$). Up-slope kyun? Motion down-slope hai, friction isko oppose karta hai. $mg\sin \theta -f=ma\Rightarrow a=\frac{16.76-13.81}{5}=+0.590{\text{m/s}}^2(\text{down-slope})$

Case (ii) — upar slide karna (friction down-slope, yani $+x$). Down-slope kyun? Motion up-slope hai, friction isko oppose karta hai — toh ab friction aur gravity dono down-slope point karte hain. $mg\sin \theta +f=ma\Rightarrow a=\frac{16.76+13.81}{5}=+6.11{\text{m/s}}^2(\text{down-slope})$ Block fast decelerate karta hai, rukta hai, phir Case (i) ke under neeche slide back kar sakta hai.

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

Step 1 — person ka FBD. Person ka kyun, scale ka nahi? Hum person par force chahte hain; Newton's Third Law se scale equal-and-opposite force read karta hai. Do forces: weight $mg$ neeche, normal $N$ upar.

Step 2 — vertical Newton's law (upar positive). $N-mg=ma\Rightarrow N=m(g+a)$ $+ma$ kyun, $=0$ nahi? Elevator (aur person) genuinely accelerate kar rahe hain; yeh equilibrium nahi hai.

Step 3 — dono cases plug karo. $\text{Up: }N=60(9.8+2)=708\text{N}$ $\text{Down: }N=60(9.8-2)=468\text{N}$

Ex 9 — Exam twist: block lift off kab karta hai? (Cell I)

Step 1 — floor pe rehte waqt FBD. "Floor pe rehte waqt" kyun? Contact abhi bhi exist karta hai, toh $N\ge 0$ valid arrow hai. Forces: $P$ upar, $N$ upar, $mg$ neeche.

Step 2 — vertical equilibrium (still resting, $a=0$). $P+N-mg=0\Rightarrow N=mg-P$ $a=0$ set kyun kar sakte hain? Jab tak lift nahi hota, block motionless hai.

Step 3 — lift-off condition $N=0$ impose karo. $N=0$ kyun? Surface neeche pull nahi kar sakta; minimum possible push zero hai, aur wahi moment hai jab contact break hone wala hota hai. $0=mg-P\Rightarrow P=mg=2(9.8)=19.6\text{N}$

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