Torque τ = r × F — definition, physical meaning

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WHY does torque even exist?

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WHAT is torque (definition)

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HOW to read the formula — deriving the two pictures

Picture 1: "Effective force" (lever arm × force)

We split $\vec F$ into two parts relative to $\vec r$:

• a component along $\vec r$ (radial): $F\cos\theta$
• a component perpendicular to $\vec r$ (tangential): $F\sin\theta$

Picture 2: "Moment arm" (perpendicular distance)

Regroup the same product: $\tau = (r\sin\theta)\,F = d_\perp \cdot F$ where $d_\perp = ==r\sin\theta==$ is the perpendicular distance from the pivot to the line of action of the force (the lever arm / moment arm).

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Deriving the magnitude from scratch (cross-product first principles)

Direction by components (so you trust it)

If $\vec r = (x,y,z)$ and $\vec F = (F_x,F_y,F_z)$:

= (yF_z - zF_y)\,\hat i + (zF_x - xF_z)\,\hat j + (xF_y - yF_x)\,\hat k.$$ For a planar problem in the $xy$-plane ($z=0,\ F_z=0$): $$\vec\tau = (xF_y - yF_x)\,\hat k.$$ This is why 2D torque is just a **signed scalar**: $+$ out of page (anticlockwise), $-$ into page (clockwise). --- ## Worked examples > [!example] 1 — Wrench > You apply $F = 50\ \text{N}$ at the end of a $0.30\ \text{m}$ wrench, pushing perpendicular to it. > **Step 1:** Identify $\theta = 90^\circ$ → *Why?* Perpendicular push = maximum twist. > **Step 2:** $\tau = rF\sin\theta = 0.30 \times 50 \times \sin 90^\circ = 0.30\times50\times1$. > **Result:** $\tau = 15\ \text{N·m}$, directed along the bolt axis. > [!example] 2 — Push at an angle > Same wrench, same force, but now pushing at $30^\circ$ to the wrench. > **Step 1:** $\sin 30^\circ = 0.5$ → *Why?* Only the sideways component $F\sin\theta$ twists. > **Step 2:** $\tau = 0.30 \times 50 \times 0.5 = 7.5\ \text{N·m}$. > **Insight:** Half the twist for the same effort — angle matters a lot. > [!example] 3 — Using the moment arm > A force $\vec F = (4,3,0)\ \text{N}$ acts at $\vec r = (2,0,0)\ \text{m}$. > **Step 1:** Use the component formula: $\tau_z = xF_y - yF_x = (2)(3) - (0)(4) = 6$. *Why component form?* It's faster than finding $\theta$. > **Step 2:** $\vec\tau = 6\,\hat k\ \text{N·m}$ → anticlockwise. > **Check via magnitude:** $r=2$, $F=5$, the angle between $(2,0,0)$ and $(4,3,0)$: $\cos\theta = \frac{8}{2\cdot5}=0.8$, so $\sin\theta=0.6$. Then $rF\sin\theta = 2\cdot5\cdot0.6 = 6$. ✓ Same. --- ## Forecast-then-Verify > [!recall]- Forecast before you read the answer > A force of fixed magnitude $F$ acts at fixed distance $r$. As you rotate the force from $\theta=0^\circ$ to $90^\circ$ to $180^\circ$, sketch how $\tau$ behaves before checking. > > **Verify:** $\tau = rF\sin\theta$ rises from $0$ to a max at $90^\circ$, then falls back to $0$ at $180^\circ$. Maximum twist is always when the force is perpendicular to the radius; pushing toward OR away from the pivot ($0^\circ$ or $180^\circ$) gives zero torque. --- ## Common mistakes (Steel-manned) > [!mistake] "Bigger force always means bigger torque." > **Why it feels right:** In everyday pushing, more force = more effect. **The flaw:** torque also depends on $r$ and $\sin\theta$. A huge force aimed *at* the pivot gives **zero** torque. **Fix:** always ask "what's the lever arm $r\sin\theta$?" before judging. > [!mistake] "Use $\cos\theta$ because the force is at angle $\theta$." > **Why it feels right:** We see $\cos$ for "components along an axis" constantly. **The flaw:** here we need the component **perpendicular** to $\vec r$, which is $F\sin\theta$, not $F\cos\theta$. **Fix:** torque needs the *sideways* part; $\sin$ is zero when vectors are parallel (correctly giving zero torque). > [!mistake] "$\vec r$ is measured from where the force acts." > **Why it feels right:** $\vec r$ is at the force's point of application. **The flaw:** $\vec r$ must start at the **pivot/axis** and point to the application point. Wrong origin → wrong torque. **Fix:** fix your axis first, then draw $\vec r$ from axis → contact point. > [!mistake] "Torque is a scalar (it's in N·m like energy)." > **Why it feels right:** Same units as joules. **The flaw:** torque is a **vector** (it has a rotation axis & sense). **Fix:** report direction (right-hand rule / sign in 2D). N·m for torque, J for energy — never mix. --- ## Mnemonic > [!mnemonic] > **"Sin to spin."** Torque uses $\sin\theta$ (perpendicular force *spins*). > Right-hand rule: fingers curl $\vec r \to \vec F$, thumb = $\vec\tau$. > Door rule: *Far + Sideways = strong twist.* --- ## Feynman (explain to a 12-year-old) > [!recall]- Explain it simply > Imagine opening a heavy door. If you push at the far edge, it swings easily. If you push right next to the hinges, it's super hard. And if you try to push the door *into* its hinges (sideways into the wall), it doesn't open at all — you're just squishing it. **Torque** is just a number that tells you how good your push is at *spinning* something. Push **far** from the spinning point and push **sideways** (not toward it), and you get the most spin. --- ## #flashcards/physics What is the definition of torque as a vector? ::: $\vec\tau = \vec r \times \vec F$, with $\vec r$ from the pivot to the point of force application. What is the magnitude of torque? ::: $\tau = rF\sin\theta$, where $\theta$ is the angle between $\vec r$ and $\vec F$. Why does torque use $\sin\theta$ and not $\cos\theta$? ::: Only the component of force perpendicular to $\vec r$ ($F\sin\theta$) causes rotation; the radial part can't spin anything. What is the moment (lever) arm? ::: $d_\perp = r\sin\theta$, the perpendicular distance from the pivot to the line of action of the force. When is torque zero for a nonzero force? ::: When $\vec F$ is parallel or antiparallel to $\vec r$ ($\theta=0$ or $180^\circ$), i.e. force aimed along the line to the pivot. SI unit of torque, and how is it distinct from energy? ::: N·m; same dimensions as joule but torque is a vector quantity (energy is scalar). 2D torque from components in the xy-plane? ::: $\tau_z = xF_y - yF_x$; positive = anticlockwise (out of page). Direction of torque vector? ::: Perpendicular to the plane of $\vec r$ and $\vec F$, by the right-hand rule (curl $\vec r$ into $\vec F$). Geometric meaning of $|\vec r \times \vec F|$? ::: The area of the parallelogram spanned by $\vec r$ and $\vec F$. Where must $\vec r$ start? ::: At the chosen pivot/axis, pointing to the point where the force is applied. --- ## Connections - [[Cross Product (Vector Algebra)]] — torque is its prototypical physical example. - [[Moment of Inertia]] and [[Newton's Second Law for Rotation]] — $\vec\tau = I\vec\alpha$. - [[Angular Momentum]] — $\vec\tau = \dfrac{d\vec L}{dt}$. - [[Equilibrium of Rigid Bodies]] — $\sum\vec\tau = 0$ condition. - [[Work-Energy Theorem (Rotational)]] — rotational work $= \tau\,\theta$. - [[Center of Mass & Gravity]] — line of action and lever arms. ## 🖼️ Concept Map ```mermaid flowchart TD F[Force F] R[Position vector r from pivot] ANGLE[Angle theta between r and F] CROSS[Cross product r x F] TAU[Torque vector tau] MAG[Magnitude rF sin theta] FPERP[Tangential component F sin theta] LEVER[Moment arm d_perp = r sin theta] DIR[Direction by right-hand rule] AREA[Parallelogram area] DOOR[Door example] F -->|acts at| R R -->|separated by| ANGLE R -->|combined via| CROSS F -->|combined via| CROSS CROSS -->|defines| TAU CROSS -->|gives| MAG MAG -->|read as| FPERP MAG -->|read as| LEVER ANGLE -->|only sideways part twists| FPERP ANGLE -->|sets perpendicular distance| LEVER CROSS -->|equals| AREA AREA -->|explains| MAG TAU -->|orientation from| DIR DOOR -->|motivates| TAU ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, torque ka matlab hai "ghumane ki taqat" — yaani koi force kisi cheez ko spin karne mein kitna effective hai. Sirf force ka magnitude kaafi nahi hota. Heavy door socho: agar tum handle pe (hinge se door) push karo to door aaram se khulta hai, lekin agar hinge ke paas push karo to mushkil se hilta hai. Aur agar tum door ko seedha hinge ki line mein dabaao, to wo bilkul nahi ghoomta. Isiliye torque distance $r$ aur angle dono pe depend karta hai. > > Formula simple hai: $\tau = rF\sin\theta$. Yahan $\theta$ hai $\vec r$ aur $\vec F$ ke beech ka angle. $\sin\theta$ kyun? Kyunki sirf force ka woh part jo $\vec r$ ke perpendicular (sideways) hai, wahi spin karwata hai. Jo part $\vec r$ ke along hai (radial), woh sirf push-pull karega, ghuma nahi sakta. Jab force aur radius parallel ho ($\theta=0$), torque zero — yahi door wala intuition maths mein. > > Do tarike se padho same cheez: $\tau = r \times (F\sin\theta)$ matlab "poora distance × sideways force", ya $\tau = (r\sin\theta) \times F$ matlab "perpendicular distance (lever arm) × poora force". Dono ka answer same aata hai. Aur yaad rakho — torque ek **vector** hai, sirf number nahi. Direction right-hand rule se: fingers ko $\vec r$ se $\vec F$ ki taraf curl karo, thumb torque ki direction batata hai (anticlockwise = bahar, clockwise = andar). > > Common galti: log $\cos\theta$ laga dete hain components ke chakkar mein — galat! Torque mein $\sin\theta$ aata hai. Aur $\vec r$ hamesha **pivot se** start hota hai, force ke point tak. Exam mein pehle axis fix karo, phir $\vec r$ banao. "Sin to spin" yaad rakho, kaam ho jayega. ![[audio/1.5.04-Torque-τ-=-r-×-F-—-definition,-physical-meaning.mp3]]