Worked examples — Resolution of vectors — into components (any axes)

1.1.9 · D3 · Physics › Measurement, Vectors & Kinematics › Resolution of vectors — into components (any axes)

Scenario matrix

Yeh poora landscape hai un cases ka jo yeh topic tumhare saamne rakh sakta hai. Neeche har worked example us cell ke saath tagged hai jo woh cover karta hai.

(Cell C7 mein symbols $\hat{u}$ aur $\hat{v}$ ka matlab hai do chosen axis directions — har ek ek length $1$ ka arrow hai jo ek oblique axis ke along point karta hai. Inhe Ex 7 mein kisi bhi calculation se pehle properly define kiya gaya hai.)

Ex 1 — Quadrant I, base case (C1)

Figure 1 (neeche): Quadrant I mein length $20$ ka ek blue arrow; uska horizontal shadow (red, adjacent) aur vertical shadow (yellow, opposite) ek right triangle banate hain, origin par $37^{\circ }$ angle marked hai.

1. Right triangle draw karo (Figure 1): hypotenuse $=v$, adjacent side along $x$, opposite side along $y$. Yeh step kyun? Components is triangle ke do perpendicular legs hain — jab tak triangle draw nahi kiya, inhe read nahi kar sakte.
2. $v_x=v\cos \theta =20\cos 37^{\circ }=20(0.7986)=15.97\text{m/s}$. Cos kyun? Horizontal leg angle ko touch karta hai (adjacent hai), aur cosine $=\frac{\text{adjacent}}{\text{hyp}}$ — yeh angle ke hugging direction mein shadow measure karta hai.
3. $v_y=v\sin \theta =20\sin 37^{\circ }=20(0.6018)=12.04\text{m/s}$. Sin kyun? Vertical leg angle ke across hai (opposite), aur sine $=\frac{\text{opposite}}{\text{hyp}}$.

Ex 2 — Quadrant II, cosine negative ho jaata hai (C2)

Figure 2 (neeche): $120^{\circ }$ par ek blue arrow Quadrant II mein land karta hai; uska horizontal shadow (red) ab $-x$ side par backward point karta hai, jabki uska vertical shadow (yellow) abhi bhi upar point karta hai. $+x$ se $120^{\circ }$ ka sweep ek white arc ke roop mein draw kiya gaya hai.

1. Same formulas apply karo — signs khud apna khayal rakh lenge agar $\theta$ ko $+x$ se poori tarah measure kiya gaya ho. $90^{\circ }$ ke baad bhi inpar trust kyun? $\cos$ aur $\sin$ kisi bhi angle ke liye defined hain iss baat se ki arrow unit circle par kahan land karta hai, na ki sirf ek triangle ke andar. $90^{\circ }$ ke baad tip $y$-axis ke baayein hai, toh uska $x$-shadow backward point karta hai → cosine negative hai.
2. $F_x=10\cos 120^{\circ }=10(-0.5)=-5.0\text{N}$. Negative kyun? Shadow $-x$ side par pada hai — minus sign direction hai, automatically encoded.
3. $F_y=10\sin 120^{\circ }=10(0.8660)=8.66\text{N}$ (positive, forecast ke anusaar).

Ex 3 — Quadrant III mein angle recover karna (C3a)

Figure 3 (neeche): true vector (red) Quadrant III mein down-left point karta hai; ek dashed yellow arrow dikhata hai naive ${\tan }^{-1}$ answer Quadrant I mein galat kahan land karta hai. Dono ka slope same hai — exactly yahi wajah hai ki ${\tan }^{-1}$ dhoka khaata hai.

1. Pythagoras se magnitude: $A=\sqrt{(-3{)}^2+(-4{)}^2}=\sqrt{9+16}=\sqrt{25}=5$. Pythagoras kyun? Components perpendicular legs hain; vector usi right triangle ka hypotenuse hai.
2. Naive angle: ${\tan }^{-1}\left(\frac{A_y}{A_x}\right)={\tan }^{-1}\left(\frac{-4}{-3}\right)={\tan }^{-1}(1.333)=53.13^{\circ }$. Yeh galat kyun hai? $\tan$ har $180^{\circ }$ par repeat karta hai: $\tan (53.13^{\circ })$ aur $\tan (53.13^{\circ }+180^{\circ })$ identical hain, toh ${\tan }^{-1}$ QI aur QIII mein fark nahi kar sakta. Yeh hamesha QI/QIV branch return karta hai.
3. Quadrant picture se fix: hum jaante hain arrow QIII mein hai, toh $180^{\circ }$ add karo: $\theta =53.13^{\circ }+180^{\circ }=233.13^{\circ }.$ $180^{\circ }$ kyun add karo? Half turn rotate karne se dono shadows ke signs flip hote hain, $(+,+)$ ko $(-,-)$ par bhejta hai — exactly hamara quadrant, same tan value ke saath.

Ex 3b — Quadrant IV mein angle recover karna (C3b)

1. Pythagoras se magnitude: $A=\sqrt{3^2+(-4{)}^2}=\sqrt{9+16}=\sqrt{25}=5$. Kyun? Same right triangle; legs ke signs unki squared lengths nahi badlte.
2. Naive angle: ${\tan }^{-1}\left(\frac{-4}{3}\right)={\tan }^{-1}(-1.333)=-53.13^{\circ }$. Yeh QIV branch kyun hai? ${\tan }^{-1}$ hamesha $-90^{\circ }$ aur $+90^{\circ }$ ke beech answer return karta hai. Negative result ka matlab hai "$+x$ axis ke neeche" — jo Quadrant IV hai. Toh yahan naive branch already correct hai.
3. Positive angle ke roop mein express karo (optional): $360^{\circ }$ add karo taaki $-53.13^{\circ }+360^{\circ }=306.87^{\circ }$ mile. Kyun? $-53.13^{\circ }$ aur $306.87^{\circ }$ same direction hain — ek clockwise measure kiya, doosra $+x$ se poora counter-clockwise sweep.

Ex 4 — Degenerate & limiting cases (C4)

1. (a) $A_x=7\cos 0^{\circ }=7(1)=7$, $A_y=7\sin 0^{\circ }=7(0)=0$. Kyun? Arrow $x$ ke along flat pada hai; uska vertical shadow zero length ka hai. $0$ ka cosine $1$ hai (full shadow), $0$ ka sine $0$ hai (koi rise nahi).
2. (b) $B_x=7\cos 90^{\circ }=7(0)=0$, $B_y=7\sin 90^{\circ }=7(1)=7$. Kyun? Ab arrow seedha upar khada hai; kuch bhi $x$ par nahi girta. Yeh (a) ka limiting mirror hai.
3. (c) $|\vec{0}|=0$, aur ${\tan }^{-1}(0/0)$ undefined hai — zero se division. Kyun? Zero-length arrow ki tip kisi jagah point karne ke liye nahi hai, toh "uska angle" ek meaningless question hai. Code mein tumhe ${\tan }^{-1}$ call karne se pehle $A_x=A_y=0$ ke against guard karna hoga.

Ex 5 — Tilted perpendicular axes: inclined plane (C5)

Figure 4 (neeche): $30^{\circ }$ par inclined ek wedge jis par ek block hai. Blue arrow seedha neeche ka weight $W$ hai; yellow arrow uska down-slope part $W\sin 30^{\circ }$ hai; red arrow uska into-surface part $W\cos 30^{\circ }$ hai. $30^{\circ }$ base angle marked hai.

1. Axes ko slope ke saath align karne ke liye tilt karo (Figure 4): ek axis slope ke neeche, ek perpendicular (surface ke andar). Tilt kyun? Motion slope ke along hoti hai. Tilted axes ek component ko driving force aur doosre ko balanced force banate hain — plain $x$–$y$ se kaafi simple. Inclined plane dynamics dekho.
2. Geometry (Figure 4): vertical weight aur slope-ke-perpendicular axis ke beech angle $\theta$ ke barabar hai. Toh along-slope leg $\theta$ ke opposite hai: $W_{\parallel }=W\sin \theta =50\sin 30^{\circ }=50(0.5)=25\text{N}.$ Sin kyun? Opposite side → sine.
3. Perpendicular leg $\theta$ ke adjacent hai: $W_{\perp }=W\cos \theta =50\cos 30^{\circ }=50(0.8660)=43.30\text{N}.$ Cos kyun? Adjacent side → cosine. Ise normal force $N$ balance karta hai.

Ex 6 — Kisi arbitrary direction par projection (C6)

Figure 5 (neeche): blue arrow $\vec{A}=(6,8)$, $\hat{n}$ direction ke liye $30^{\circ }$ par ek dashed yellow ray, aur ek green arrow $\vec{A}$ ka shadow us ray par perpendicularly drop karte hue dikhata hai — uski length projection $9.196$ hai.

1. Unit vector banao: $\hat{n}=(\cos 30^{\circ },\sin 30^{\circ })=(0.8660,0.5)$. Unit vector kyun? Ek component ek length hai; dot product $\vec{A}\cdot \hat{n}$ tabhi pure length read karta hai jab $\hat{n}$ ki length $1$ ho.
2. Dot product = projection: $A_{\hat{n}}=\vec{A}\cdot \hat{n}=6(0.8660)+8(0.5)=5.196+4.0=\mathrm{9.196.}$ Dot product kyun? Jaise parent note mein prove kiya gaya, $\vec{A}\cdot \hat{n}=A\cos \varphi$ exactly $\vec{A}$ ka $\hat{n}$ ke along shadow hai, jahan $\varphi$ abhi define kiya gaya vector-to-vector angle hai. Dot product & scalar projection dekho.
3. Us angle $\varphi$ se cross-check: yahan $\varphi =53.13^{\circ }-30^{\circ }=23.13^{\circ }$ ($\vec{A}$ ke $53.13^{\circ }$ aur $\hat{n}$ ke $30^{\circ }$ ke beech ka gap), toh $A\cos \varphi =10\cos 23.13^{\circ }=10(0.9196)=9.196$. ✓ Dono routes kyun? Inhe agree karna chahiye — "dot product IS projection" ka poora content yahi hai.

Ex 7 — Oblique axes: projection ≠ component (C7)

Figure 6 (neeche): blue diagonal $\vec{A}=(4,3)$; horizontal axis ke along yellow side $a\hat{u}$; $60^{\circ }$ axis ke along green side $b\hat{v}$, $a\hat{u}$ ki tip se $\vec{A}$ ki tip tak draw ki gayi taaki do sides parallelogram ko wapas $\vec{A}$ par close kare.

1. Sirf project kyun nahi karte? $60^{\circ }$ axes ke liye, $\hat{v}$ piece ka $\hat{u}$ par apna shadow hai — ek "leak." Projection us leak ko double-count karega. Hume instead demand karni hogi ki pieces add hokar $\vec{A}$ bane. Yeh parallelogram idea hai (upar definition dekho): $\vec{A}$ diagonal hai; $a\hat{u}$ aur $b\hat{v}$ do sides hain.
2. Har coordinate ke liye ek equation likho ($\vec{A}=a\hat{u}+b\hat{v}$): $x:4=a(1)+b(0.5),y:3=a(0)+b(0.8660).$ Do equations kyun, aur yeh "system solve" kyun karta hai? Single vector statement $\vec{A}=a\hat{u}+b\hat{v}$ really do statements hain — $x$-parts match hone chahiye AUR $y$-parts match hone chahiye. Do matching conditions do unknowns $a,b$ exactly pin down karti hain. Inhe solve karna forces karta hai ki reassembled pieces $\vec{A}$ ke barabar hon dono directions mein ek saath, jo exactly component ki definition hai.
3. $y$-equation se: $b=\frac{3}{0.8660}=3.4641$. $x$-equation mein substitute karo: $4=a+0.5(3.4641)=a+1.7320$, toh $a=4-1.7320=2.2680$.
4. Naive projection se compare karo: $\vec{A}\cdot \hat{u}=4$. Lekin true component $a=2.268$ hai — kaafi chhota. Chhota kyun? $\hat{v}$-piece pehle se $1.732$ horizontal reach supply kar chuka hai; parallelogram ko use recount nahi karna chahiye. Projection $=4$; component $=2.268$; yeh differ karte hain exactly kyunki axes orthogonal nahi hain.

Ex 8 — Word problem + equilibrium (C8)

1. Components mein translate karo. Rope 1: $(+T_1\cos 50^{\circ },+T_1\sin 50^{\circ })$. Rope 2: $(-T_2\cos 40^{\circ },+T_2\sin 40^{\circ })$. Weight: $(0,-100)$. Resolve kyun? Equilibrium ka matlab vector sum zero hai; lekin parent note ka rule kehta hai ek fixed axis ke along components sirf numbers ki tarah add hote hain. Toh "sum = 0" do scalar equations ban jaata hai. Equilibrium of concurrent forces dekho.
2. $x$-balance: $T_1\cos 50^{\circ }-T_2\cos 40^{\circ }=0\Rightarrow T_1(0.6428)=T_2(0.7660)$, toh $T_1=1.1918T_2$. Kyun? Koi horizontal motion nahi → horizontal shadows cancel hote hain.
3. $y$-balance: $T_1\sin 50^{\circ }+T_2\sin 40^{\circ }=100\Rightarrow T_1(0.7660)+T_2(0.6428)=100$. Kyun? Vertical shadows ko milkar $100\text{N}$ weight uthana hoga.
4. $T_1=1.1918T_2$ substitute karo: $1.1918T_2(0.7660)+0.6428T_2=100\Rightarrow (0.9129+0.6428)T_2=100\Rightarrow T_2=\frac{100}{1.5557}=64.28\text{N}.$ Phir $T_1=1.1918(64.28)=76.60\text{N}$.

Ex 9 — Exam twist: angle y-axis se measure kiya gaya (C9)

Figure 7 (neeche): ek blue launch-velocity arrow seedhe-upar se sirf $20^{\circ }$ jhuka hua. Vertical shadow (yellow) lambi, adjacent leg hai; horizontal shadow (red) chhoti, opposite leg hai. $20^{\circ }$ angle vertical axis ko hug karte hua draw kiya gaya hai.

1. Triangle ko sahi corner par angle ke saath redraw karo (Figure 7). $20^{\circ }$ velocity aur vertical axis ke beech baitha hai. Redraw kyun? "Adjacent → cos" ka rule kin leg angle ko touch karta hai ke baare mein hai, na ki kaun horizontal hai. Yahan vertical leg $20^{\circ }$ ke adjacent hai.
2. Vertical component (angle ke adjacent): $v_y=v\cos 20^{\circ }=25\cos 20^{\circ }=25(0.9397)=23.49\text{m/s}$. Cos kyun? Vertical leg $20^{\circ }$ angle ko hug karta hai → yeh adjacent side hai → cosine.
3. Horizontal component (angle ke opposite): $v_x=v\sin 20^{\circ }=25\sin 20^{\circ }=25(0.3420)=8.55\text{m/s}$. Sin kyun? Horizontal leg angle ke across hai → opposite side → sine. Yeh value seedha Projectile motion mein constant horizontal speed ke roop mein jaati hai.

Self-test

Connections

• Resolution of vectors — into components (any axes) (parent)
• Vectors — addition (parallelogram & triangle law)
• Dot product & scalar projection
• Unit vectors and Cartesian coordinates
• Inclined plane dynamics
• Projectile motion
• Equilibrium of concurrent forces