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

WHAT: just read off the two shadow formulas. WHY cos on x: the x-side touches the angle → 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: $\sqrt{6^2+(6\sqrt{3}{)}^2}=\sqrt{36+108}=\sqrt{144}=12$ ✓ — the shadows rebuild the arrow.

The dot product $A_{\hat{n}}=\vec{A}\cdot \hat{n}=A\cos \varphi$. WHY: a component is a shadow; the dot product measures the shadow of $\vec{A}$ along $\hat{n}$'s direction. Only this one works for any direction, not just x or y.

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

WHAT: run the inverse formulas. $A=\sqrt{3^2+4^2}=\sqrt{25}=5,\theta ={\tan }^{-1}\left(\frac{4}{3}\right)\approx 53.13^{\circ }.$ WHY ${\tan }^{-1}$: $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{A_y}{A_x}$ encodes the steepness; ${\tan }^{-1}$ asks "which angle has this steepness?" Both $A_x,A_y>0$ → first quadrant, so no sign correction needed.

$u_x=20\cos 30^{\circ }=20\left(\frac{\sqrt{3}}{2}\right)\approx 17.32\text{ m/s},u_y=20\sin 30^{\circ }=20(0.5)=10\text{ m/s}.$ WHY split like this: in Projectile motion the horizontal and vertical directions obey independent laws (constant $u_x$; gravity only touches $u_y$). Resolving turns one slanted launch into two clean 1-D problems.

WHY tilt the axes: motion (if any) happens along the slope, so we want one component to be the driving pull and the other the surface-pressing part — see the figure above. $W_{\parallel }=W\sin \theta =100\sin 30^{\circ }=50\text{ N (down the slope)},$ $W_{\perp }=W\cos \theta =100\cos 30^{\circ }=100\left(\frac{\sqrt{3}}{2}\right)\approx 86.60\text{ N (into surface)}.$ WHY $\sin$ for along-slope: the vertical weight and the perpendicular-to-slope axis differ by exactly $\theta$, so the along-slope piece is the "opposite" side → sine. See Inclined plane dynamics.

WHAT: magnitude ignores signs (it squares them): $A=\sqrt{(-3{)}^2+(-4{)}^2}=\sqrt{25}=5.$ The sign subtlety: ${\tan }^{-1}(A_y/A_x)={\tan }^{-1}\left(\frac{-4}{-3}\right)={\tan }^{-1}(1.333)\approx 53.13^{\circ }$ — but that's a first-quadrant answer, and both components are negative, so $\vec{A}$ points into quadrant III (down-left). WHY the calculator lies: $\tan$ repeats every $180^{\circ }$, so it cannot tell quadrant I from quadrant III. Fix: add $180^{\circ }$: $\theta =53.13^{\circ }+180^{\circ }=233.13^{\circ }.$ Look at the plum arrow in the figure below.

$A_x=+2>0$, $A_y=-5<0$ → quadrant IV (right and down). ${\tan }^{-1}\left(\frac{-5}{2}\right)\approx -68.20^{\circ }.$ A negative angle is correct in quadrant IV (it means "below the x-axis"). To state it as a positive angle, add $360^{\circ }$: $291.80^{\circ }$. Both name the same direction.

WHY resolve first: components along a fixed axis add like plain numbers (the whole point — see Vectors — addition (parallelogram & triangle law)). $R_x=6\cos 0^{\circ }+8\cos 90^{\circ }=6+0=6,R_y=6\sin 0^{\circ }+8\sin 90^{\circ }=0+8=8.$ Rebuild: $R=\sqrt{6^2+8^2}=10\text{ N},\theta ={\tan }^{-1}\left(\frac{8}{6}\right)\approx 53.13^{\circ }.$ Both components positive → quadrant I, so no correction needed.

Along $\hat{n}$ (dot product = projection): $A_{\hat{n}}=\vec{A}\cdot \hat{n}=3(0.6)+4(0.8)=1.8+3.2=\mathrm{5.0.}$ Since $|\vec{A}|=5$ and $\hat{n}$ points the same way as $\vec{A}$ (both at $53.13^{\circ }$), the entire length projects — the perpendicular part must be zero. Perpendicular component (check): using $A^2=A_{\hat{n}}^2+A_{\perp }^2$, $A_{\perp }=\sqrt{A^2-A_{\hat{n}}^2}=\sqrt{25-25}=0✓.$ See Dot product & scalar projection.

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

WHY not just dot products: the axes are $60^{\circ }$ apart, not perpendicular, so projection ≠ component (a projection onto $\hat{u}$ would "leak" some of the $\hat{v}$ part). We must demand the pieces rebuild $\vec{A}$. Write the two coordinate equations of $a\hat{u}+b\hat{v}=\vec{A}$: $x:a(1)+b(0.5)=5,y:a(0)+b\left(\frac{\sqrt{3}}{2}\right)=0.$ From the y-equation: $b\cdot \frac{\sqrt{3}}{2}=0\Rightarrow b=0$. Then $a=5$. Interpretation: $\vec{A}$ already lies exactly along $\hat{u}$, so it needs none of $\hat{v}$ — the oblique decomposition correctly returns $a=5,b=0$.

Set up $a\hat{u}+b\hat{v}=\vec{B}$: $x:a+0.5b=4,y:\frac{\sqrt{3}}{2}b=4.$ Solve y first: $b=\frac{4}{\sqrt{3}/2}=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}\approx 4.619$. Back-substitute: $a=4-0.5b=4-0.5(4.619)=4-2.309=1.691$. Contrast with the naive projection $\vec{B}\cdot \hat{u}=4(1)+4(0)=4\ne a=1.691$ — proof that projection ≠ oblique component when axes aren't perpendicular. 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 demands $\sum F_x=0$ and $\sum F_y=0$ separately — resolve everything onto x and y. Assume $|{\vec{F}}_3|=F$. Sum the x-components: $\sum F_x=10\cos 0^{\circ }+10\cos 120^{\circ }+F\cos 240^{\circ }=10+10(-0.5)+F(-0.5)=5-0.5F.$ Set to zero: $5-0.5F=0\Rightarrow F=10\text{ N}$. Check the y-equation with $F=10$: $\sum F_y=10\sin 0^{\circ }+10\sin 120^{\circ }+10\sin 240^{\circ }=0+10\left(\frac{\sqrt{3}}{2}\right)+10\left(-\frac{\sqrt{3}}{2}\right)=0✓.$ Three equal forces at $120^{\circ }$ apart cancel perfectly — a symmetric "Mercedes star" of forces.

Every problem here is the same trick wearing different clothes: cast shadows onto chosen axes, do arithmetic on the shadows, rebuild the arrow. L1–L2 practise the shadows. L3 warns that the calculator's ${\tan }^{-1}$ can't see quadrants — you must read the signs. L4 adds shadows from several arrows. L5 shows what breaks when your axes aren't at right angles, and how equilibrium is just "all shadows cancel."