Exercises — Resolution of vectors — into components (any axes)

WHAT: bas do shadow formulas padh lo. WHY cos on x: x-side angle ko touch karti hai → adjacent → cosine. $A_x=12\cos 60^{\circ }=12(0.5)=6,A_y=12\sin 60^{\circ }=12\left(\frac{\sqrt{3}}{2}\right)=6\sqrt{3}\approx \mathrm{10.39.}$ Sanity: $\sqrt{6^2+(6\sqrt{3}{)}^2}=\sqrt{36+108}=\sqrt{144}=12$ ✓ — shadows arrow ko rebuild kar deti hain.

Dot product $A_{\hat{n}}=\vec{A}\cdot \hat{n}=A\cos \varphi$. WHY: ek component ek shadow hota hai; dot product $\vec{A}$ ki shadow ko $\hat{n}$ ki direction mein measure karta hai. Sirf yahi kisi bhi direction ke liye kaam karta hai, sirf x ya y ke liye nahi.

$F_x=50\cos 37^{\circ }=50(0.8)=40\text{ N},F_y=50\sin 37^{\circ }=50(0.6)=30\text{ N}.$ Check: $\sqrt{40^2+30^2}=\sqrt{1600+900}=\sqrt{2500}=50$ ✓.

WHAT: inverse formulas chalao. $A=\sqrt{3^2+4^2}=\sqrt{25}=5,\theta ={\tan }^{-1}\left(\frac{4}{3}\right)\approx 53.13^{\circ }.$ WHY ${\tan }^{-1}$: $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{A_y}{A_x}$ steepness encode karta hai; ${\tan }^{-1}$ puchta hai "kaunsa angle is steepness rakhta hai?" Dono $A_x,A_y>0$ → first quadrant, toh koi sign correction nahi chahiye.

$u_x=20\cos 30^{\circ }=20\left(\frac{\sqrt{3}}{2}\right)\approx 17.32\text{ m/s},u_y=20\sin 30^{\circ }=20(0.5)=10\text{ m/s}.$ WHY aise split karte hain: Projectile motion mein horizontal aur vertical directions independent laws follow karti hain (constant $u_x$; gravity sirf $u_y$ ko touch karti hai). Resolve karne se ek tedha launch do saaf 1-D problems mein badal jaata hai.

WHY axes tilt karo: motion (agar ho) slope ke along hoti hai, toh hum chahte hain ek component driving pull ho aur doosra surface-pressing part — upar ki figure dekho. $W_{\parallel }=W\sin \theta =100\sin 30^{\circ }=50\text{ N (slope ke neeche)},$ $W_{\perp }=W\cos \theta =100\cos 30^{\circ }=100\left(\frac{\sqrt{3}}{2}\right)\approx 86.60\text{ N (surface mein)}.$ WHY along-slope ke liye $\sin$: vertical weight aur slope ke perpendicular axis mein exactly $\theta$ ka difference hota hai, toh along-slope piece "opposite" side hai → sine. Inclined plane dynamics dekho.

WHAT: magnitude signs ignore karta hai (unhe square karta hai): $A=\sqrt{(-3{)}^2+(-4{)}^2}=\sqrt{25}=5.$ Sign subtlety: ${\tan }^{-1}(A_y/A_x)={\tan }^{-1}\left(\frac{-4}{-3}\right)={\tan }^{-1}(1.333)\approx 53.13^{\circ }$ — lekin yeh first-quadrant answer hai, aur dono components negative hain, toh $\vec{A}$ quadrant III (neeche-left) mein point karta hai. WHY calculator jhooth bolta hai: $\tan$ har $180^{\circ }$ par repeat hota hai, toh woh quadrant I aur quadrant III mein fark nahi kar sakta. Fix: $180^{\circ }$ add karo: $\theta =53.13^{\circ }+180^{\circ }=233.13^{\circ }.$ Neeche ki figure mein plum arrow dekho.

$A_x=+2>0$, $A_y=-5<0$ → quadrant IV (right aur neeche). ${\tan }^{-1}\left(\frac{-5}{2}\right)\approx -68.20^{\circ }.$ Quadrant IV mein negative angle sahi hai (matlab hai "x-axis ke neeche"). Ise positive angle mein batane ke liye, $360^{\circ }$ add karo: $291.80^{\circ }$. Dono same direction name karte hain.

WHY pehle resolve karo: ek fixed axis ke along components plain numbers ki tarah add hote hain (poora point yahi hai — Vectors — addition (parallelogram & triangle law) dekho). $R_x=6\cos 0^{\circ }+8\cos 90^{\circ }=6+0=6,R_y=6\sin 0^{\circ }+8\sin 90^{\circ }=0+8=8.$ Rebuild: $R=\sqrt{6^2+8^2}=10\text{ N},\theta ={\tan }^{-1}\left(\frac{8}{6}\right)\approx 53.13^{\circ }.$ Dono components positive → quadrant I, toh koi correction nahi chahiye.

$\hat{n}$ ke along (dot product = projection): $A_{\hat{n}}=\vec{A}\cdot \hat{n}=3(0.6)+4(0.8)=1.8+3.2=\mathrm{5.0.}$ Kyunki $|\vec{A}|=5$ hai aur $\hat{n}$ same direction mein point karta hai jis mein $\vec{A}$ (dono $53.13^{\circ }$ par), poori length project hoti hai — perpendicular part zero hona chahiye. Perpendicular component (check): $A^2=A_{\hat{n}}^2+A_{\perp }^2$ use karke, $A_{\perp }=\sqrt{A^2-A_{\hat{n}}^2}=\sqrt{25-25}=0✓.$ Dot product & scalar projection dekho.

Componentwise add karo: ${\vec{v}}_g=(4+2)\hat{x}+(3+0)\hat{y}=6\hat{x}+3\hat{y}.$ $|{\vec{v}}_g|=\sqrt{6^2+3^2}=\sqrt{45}=3\sqrt{5}\approx 6.71\text{ m/s},\theta ={\tan }^{-1}\left(\frac{3}{6}\right)\approx 26.57^{\circ }\text{ N of E}.$

WHY sirf dot products nahi: axes $60^{\circ }$ apart hain, perpendicular nahi, toh projection ≠ component ($\hat{u}$ par projection $\hat{v}$ wale part mein "leak" karega). Hume demand karni hai ki pieces $\vec{A}$ ko rebuild karein. $a\hat{u}+b\hat{v}=\vec{A}$ ki do coordinate equations likho: $x:a(1)+b(0.5)=5,y:a(0)+b\left(\frac{\sqrt{3}}{2}\right)=0.$ y-equation se: $b\cdot \frac{\sqrt{3}}{2}=0\Rightarrow b=0$. Phir $a=5$. Interpretation: $\vec{A}$ already exactly $\hat{u}$ ke along hai, toh use $\hat{v}$ ki kuch bhi zaroorat nahi — oblique decomposition sahi se $a=5,b=0$ return karta hai.

$a\hat{u}+b\hat{v}=\vec{B}$ set up karo: $x:a+0.5b=4,y:\frac{\sqrt{3}}{2}b=4.$ Pehle y solve karo: $b=\frac{4}{\sqrt{3}/2}=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}\approx 4.619$. Back-substitute: $a=4-0.5b=4-0.5(4.619)=4-2.309=1.691$. Naive projection se contrast: $\vec{B}\cdot \hat{u}=4(1)+4(0)=4\ne a=1.691$ — proof ki projection ≠ oblique component jab axes perpendicular nahi hain. Check reassembly: $a\hat{u}+b\hat{v}=(1.691+0.5\cdot 4.619,\frac{\sqrt{3}}{2}\cdot 4.619)=(4.000,4.000)$ ✓.

WHY components: Equilibrium of concurrent forces demand karta hai $\sum F_x=0$ aur $\sum F_y=0$ alag alag — sab kuch x aur y par resolve karo. Maano $|{\vec{F}}_3|=F$. x-components ka sum: $\sum F_x=10\cos 0^{\circ }+10\cos 120^{\circ }+F\cos 240^{\circ }=10+10(-0.5)+F(-0.5)=5-0.5F.$ Zero set karo: $5-0.5F=0\Rightarrow F=10\text{ N}$. $F=10$ ke saath y-equation check karo: $\sum F_y=10\sin 0^{\circ }+10\sin 120^{\circ }+10\sin 240^{\circ }=0+10\left(\frac{\sqrt{3}}{2}\right)+10\left(-\frac{\sqrt{3}}{2}\right)=0✓.$ $120^{\circ }$ apart teen equal forces perfectly cancel ho jaati hain — forces ka ek symmetric "Mercedes star."

Yahan har problem same trick hai alag kapde mein: chosen axes par shadows daalo, shadows par arithmetic karo, arrow rebuild karo. L1–L2 shadows practice karaate hain. L3 warn karta hai ki calculator ka ${\tan }^{-1}$ quadrants nahi dekh sakta — tumhe signs padhne honge. L4 kai arrows ke shadows add karta hai. L5 dikhata hai kya toot ta hai jab axes right angles par nahi hain, aur equilibrium bas "sab shadows cancel" hai.