Torque τ = r × F — definition, physical meaning: Worked examples

1.5.4 · D3 · Physics › Rotational Mechanics › Torque τ = r × F — definition, physical meaning

Yeh page Torque τ = r × F ka "koi surprise nahi bachega" workshop hai. Hum har tarah ki situation cover karenge jo torque mein aa sakti hai: dono directions of spin, pivot ki taraf seedha aim kiya hua force, zero lever arm wala force, maximum-twist case, ek real-world word problem, aur ek exam-style trap. Yahan koi bhi symbol use nahi hoga jo parent note mein pehle se earn nahi kiya — lekin ek one-line refresher ke taur par:

Recall Sirf teen formulas ki zaroorat hai

Magnitude: $τ = r F sin θ$ (poori distance × sideways force). Component (2D, $x y$ -plane mein): $τ_{z} = x F_{y} - y F_{x}$ (positive = anticlockwise = page se bahar aapki taraf). Vector: $τ = r \times F$ . Yahan $r$ hamesha pivot se start hota hai aur wahan point karta hai jahan force touch karti hai.

Scenario matrix

Har torque problem jo aap kabhi bhi miloge woh in cells mein se kisi ek mein aati hai. Table mein cell, uski khaas baat, aur neeche ka worked example diya gaya hai.

Cell	Uski khaas baat	Covered by
A. Perpendicular push	$θ = 9 0^{\circ}$ , maximum twist	Ex 1
B. Angled push	$0^{\circ} < θ < 9 0^{\circ}$ , sirf $F sin θ$ count hota hai	Ex 2
C. Force along the radius	$θ = 0^{\circ}$ ya $18 0^{\circ}$ → zero torque	Ex 3
D. Positive (anticlockwise) sign	$τ_{z} > 0$ , page se bahar	Ex 4
E. Negative (clockwise) sign	$τ_{z} < 0$ , page ke andar	Ex 5
F. Zero lever arm	force line pivot se guzarti hai	Ex 6
G. Real-world word problem	words ko translate karo → $r$ , $F$ , $θ$	Ex 7
H. Exam twist (net torque of two forces)	signed torques add karo, signs dekho	Ex 8
I. Limiting behaviour	$τ$ kaise vary karta hai jab $θ$ sweep karta hai	Ex 9

Example 1 — Perpendicular push (Cell A)

Worked example Statement

Ek mechanic $F = 40 N$ se $0.25 m$ ke spanner ke end par push karta hai, spanner ke perpendicular. Bolt ke baare mein torque find karo.

Forecast: perpendicular matlab "poori force sideways hai" — kya aap expect karte ho ki yeh is $r$ aur $F$ ke liye sabse bada, sabse chhota, ya beech ka twist hoga?

Step 1 — Angle identify karo: $θ = 9 0^{\circ}$ . Yeh step kyun? Torque $r$ aur $F$ ke beech ke angle par depend karta hai; sine touch karne se pehle ise pin down karna zaroori hai.

Step 2 — Sine padho: $sin 9 0^{\circ} = 1$ . Yeh step kyun? $9 0^{\circ}$ par poori force "spinning" component $F sin θ = F$ hai, isliye koi bhi force arm ke saath push karne mein waste nahi hoti.

Step 3 — Multiply karo: $τ = r F sin θ = 0.25 \times 40 \times 1 = 10 N\cdotm$ . Yeh step kyun? $τ = r F sin θ$ mein plug in karo.

Verify: Units $m \times N = N\cdotm$ ✓. Yeh is $r$ aur $F$ ke liye maximum possible twist hai (sine 1 par cap hota hai), forecast se match karta hai.

Figure — Torque τ = r × F — definition, physical meaning

Example 2 — Angled push (Cell B)

Worked example Statement

Same spanner ( $r = 0.25 m$ ), same effort ( $F = 40 N$ ), lekin aap $θ = 3 0^{\circ}$ par spanner se push karte ho. Torque find karo.

Forecast: aap same muscle force use kar rahe ho. Kya twist $10 N\cdotm$ se zyada hoga ya kam? Roughly kitne factor se?

Step 1 — Force split karo: spanner ke along wala part $F cos 3 0^{\circ}$ hai, sideways part $F sin 3 0^{\circ}$ hai. Yeh step kyun? Sirf sideways (perpendicular-to- $r$ ) part hi point ko pivot ke around circulate kar sakta hai; along-part sirf bolt ki taraf/se tug karta hai.

Step 2 — Sideways force $= 40 \times sin 3 0^{\circ} = 40 \times 0.5 = 20 N$ . Yeh step kyun? Yahi wo force hai jo spin produce karti hai.

Step 3 — Torque $= r \times F_{⊥} = 0.25 \times 20 = 5 N\cdotm$ . Yeh step kyun? Torque = poori distance × sideways force.

Verify: $τ = r F sin θ = 0.25 \times 40 \times 0.5 = 5 N\cdotm$ ✓. Yeh exactly Example 1 ke $10 N\cdotm$ ka aadha hai kyunki $sin 3 0^{\circ} = 0.5$ — same effort, aadha twist.

Example 3 — Force along the radius (Cell C)

Example 4 — Positive (anticlockwise) torque (Cell D)

Worked example Statement

Ek force $F = (0, 5, 0) N$ position $r = (3, 0, 0) m$ par act karti hai. $τ$ aur uska spin sense find karo.

Forecast: arm right ( $+ x$ ) ki taraf point karta hai, force up ( $+ y$ ) ki taraf. Apna right hand "right" se "up" ki taraf curl karo — thumb page ke andar point karta hai ya bahar?

Step 1 — 2D component formula $τ_{z} = x F_{y} - y F_{x}$ use karo. Yeh step kyun? Sab kuch $x y$ -plane mein hai, isliye torque $\hat{k}$ ke along ek single number hai — $θ$ nikalne se zyada fast hai.

Step 2 — Substitute karo: $τ_{z} = (3) (5) - (0) (0) = 15$ . Yeh step kyun? $x = 3, y = 0, F_{x} = 0, F_{y} = 5$ ka direct plug-in.

Step 3 — Sign interpret karo: $τ_{z} = + 15 N\cdotm$ , isliye $τ = 15 \hat{k}$ — page se bahar = anticlockwise.

Verify: $r = 3$ , $F = 5$ , aur $r ⊥ F$ isliye $θ = 9 0^{\circ}$ , jo $r F sin θ = 15$ deta hai ✓. Positive sign "right → up" curl se match karta hai (thumb page se bahar). ✓

Example 5 — Negative (clockwise) torque (Cell E)

Example 6 — Zero lever arm (line of action through pivot) (Cell F)

Worked example Statement

Ek force $F = (6, 4, 0) N$ point $r = (3, 2, 0) m$ par apply ki jaati hai. Origin ke baare mein torque find karo.

Forecast: $r = (3, 2)$ aur $F = (6, 4)$ — notice karo ki $(6, 4)$ exactly $2 \times (3, 2)$ hai. Force seedha arm ke along point karti hai. Torque kya hona chahiye?

Step 1 — Geometry check karo: kya $F$ , $r$ ke parallel hai? $(6, 4) = 2 (3, 2)$ , haan. Yeh step kyun? Parallel force matlab uski line of action pivot se guzarti hai → zero lever arm.

Step 2 — Compute karo $τ_{z} = x F_{y} - y F_{x} = (3) (4) - (2) (6) = 12 - 12 = 0$ . Yeh step kyun? Component formula automatically parallel vectors ke liye zero deta hai — hamare geometric claim par ek machine check.

Step 3 — Conclude karo $τ = 0$ .

Verify: Lever arm $d_{⊥} = r sin θ$ ; yahan $θ = 0^{\circ}$ hai isliye $d_{⊥} = 0$ , confirm karta hai ki $F$ kitna bhi bada ho torque zero hai. ✓

Example 7 — Real-world word problem (Cell G)

Worked example Statement

Ek baccha seesaw par central pivot se $1.2 m$ door baithta hai aur uska weight $300 N$ hai. Gravity seedhi neeche pull karti hai jabki seesaw beam horizontal hai. Pivot ke baare mein bacche ka torque find karo.

Forecast: weight vertically act karta hai, arm horizontal hai. Kya yeh perpendicular hain? Isse $sin θ$ kya hota hai?

Step 1 — Setup karo: $r = 1.2 m$ (pivot → baccha), $F = 300 N$ (weight, downward). Yeh step kyun? $r$ pivot se shuru hona chahiye; yahan "force" bacche ka weight hai.

Step 2 — Horizontal arm aur vertical weight ke beech angle: $θ = 9 0^{\circ}$ , isliye $sin θ = 1$ . Yeh step kyun? Horizontal ⟂ vertical, isliye koi force beam ke along waste nahi hoti.

Step 3 — $τ = r F sin θ = 1.2 \times 300 \times 1 = 360 N\cdotm$ .

Verify: Units N·m ✓. Agar doosri taraf ka ek doosra baccha ise balance karna chahta hai, toh use same $360 N\cdotm$ opposite-sense torque supply karna hoga — roz ka balancing rule (dekho Equilibrium of Rigid Bodies). ✓

Example 8 — Exam twist: net torque of two forces (Cell H)

Worked example Statement

Centre par pivoted ek horizontal rod par, ek force $x = + 2 m$ par $F_{1} = 10 N$ se neeche push karti hai, aur doosri $x = - 3 m$ par $F_{2} = 10 N$ se neeche push karti hai. Dono forces vertical (down) hain. Pivot ke baare mein net torque aur resulting sense of spin find karo.

Forecast: dono forces neeche push karti hain, lekin opposite sides par. Kya unke torques add up hote hain ya ek doosre se ladte hain?

Step 1 — Force 1 ka torque: $r_{1} = (2, 0)$ , $F_{1} = (0, - 10)$ , isliye $τ_{1} = x F_{y} - y F_{x} = (2) (- 10) - 0 = - 20 N\cdotm$ (clockwise). Yeh step kyun? Right end par downward push rod ko clockwise rotate karta hai.

Step 2 — Force 2 ka torque: $r_{2} = (- 3, 0)$ , $F_{2} = (0, - 10)$ , isliye $τ_{2} = (- 3) (- 10) - 0 = + 30 N\cdotm$ (anticlockwise). Yeh step kyun? Left end par downward push rod ko doosri taraf rotate karta hai — sign flip hota hai kyunki $x$ negative hai.

Step 3 — Net torque: $τ_{net} = τ_{1} + τ_{2} = - 20 + 30 = + 10 N\cdotm$ . Yeh step kyun? Same axis ke baare mein torques 2D mein signed scalars hain, isliye hum unhe bas add kar dete hain.

Verify: $τ_{net} = + 10 N\cdotm$ → net anticlockwise. Trap yeh sochna hai ki "dono forces same taraf point karti hain isliye torques magnitude mein add hote hain" — lekin opposite sides opposite signs deti hain. Dono torques ladte hain, aur longer arm ( $3 > 2$ ) jeet jaata hai. ✓

Example 9 — Limiting behaviour as the angle sweeps (Cell I)

Worked example Statement

$r = 2 m$ aur $F = 5 N$ fixed rakho, $θ = 0^{\circ}, 3 0^{\circ}, 9 0^{\circ}, 15 0^{\circ}, 18 0^{\circ}$ par torque magnitude tabulate karo aur pattern describe karo.

Forecast: jaise $θ$ , $0 \to 90 \to 180$ jaata hai, aage padhne se pehle $τ$ ka curve sketch karo.

Step 1 — $τ = r F sin θ = 10 sin θ$ use karo. Yeh step kyun? Sirf $sin θ$ vary karta hai; $r F = 10$ ek constant multiplier hai.

Step 2 — Evaluate karo:

$θ = 0^{\circ}$ : $10 sin 0^{\circ} = 0$
$θ = 3 0^{\circ}$ : $10 \times 0.5 = 5$
$θ = 9 0^{\circ}$ : $10 \times 1 = 10$ (maximum)
$θ = 15 0^{\circ}$ : $10 \times 0.5 = 5$
$θ = 18 0^{\circ}$ : $10 \times 0 = 0$

Yeh step kyun? Yeh points rise-then-fall shape dikhate hain.

Step 3 — Describe karo: torque $0$ se $9 0^{\circ}$ par $10 N\cdotm$ ke peak tak chadha, phir symmetrically $18 0^{\circ}$ par $0$ par wapas aata hai.

Verify: $sin θ = sin (18 0^{\circ} - θ)$ , isliye $3 0^{\circ}$ aur $15 0^{\circ}$ identical $5 N\cdotm$ dete hain — curve ek symmetric hump hai. ✓

Wrap-up

Recall Yeh problem kaunse cell mein hai?

Errors se bachne ka sabse fast tarika: compute karne se pehle cell ka naam lo. Force arm ke perpendicular hai? ::: Cell A — poora $r F$ . Force kisi angle par hai? ::: Cell B — $r F sin θ$ use karo. Force arm ke along hai (ya line pivot se guzarti hai)? ::: Cells C/F — torque zero hai. 2D mein kaun sa sign? ::: $τ_{z} = x F_{y} - y F_{x}$ ; $+$ anticlockwise, $-$ clockwise (Cells D/E). Kai forces hain? ::: Cell H — same axis ke baare mein signed torques add karo.

Related deeper dives: in torques ko Newton's Second Law for Rotation mein use karo, cross product itself ko Cross Product (Vector Algebra) mein connect karo, aur inhe Equilibrium of Rigid Bodies mein balance karo.