1.5.18 · D4 · Physics › Rotational Mechanics › Equilibrium of rigid bodies — translational + rotational

Throughout, use $g = 10 m/s^{2}$ unless told otherwise. Sign rule: counter-clockwise (CCW) torque positive hai, clockwise (CW) negative — socho "ye force body ko kis taraf ghumane ki koshish kar raha hai?"

Wo ek idea jo har problem mein kaam aata hai

Shuru karne se pehle, ek picture bilkul clear kar lete hain kyunki ye almost har solution mein aati hai.

Figure — Equilibrium of rigid bodies — translational + rotational

Figure dekho: moment arm poori length $r$ nahi hai us point tak — balki ye perpendicular drop $d$ hai line of action par. Jab bhi koi force pivot se guzarti hai, to wo perpendicular distance zero hoti hai, isliye uska torque zero ho jaata hai. Yahi ek baat hai jo humein pivot ko unknown reactions ke upar choose karke unhe "khatam" karne deti hai.

L1 · Recognition

Exercise 1.1 — Kya ye equilibrium mein hai?

Ek flat rectangular sign wall par bilkul still latki hai. Uske baare mein teen statements diye gaye hain. Kaun se rigid-body equilibrium describe karte hain? (a) Uska net force zero hai. (b) Net force zero hai aur uske centre ke baare mein net torque bhi zero hai. (c) Ye na accelerate kar raha hai na angularly accelerate kar raha hai.

Recall Solution

Answer: (b) aur (c). Ek rigid body tabhi equilibrium mein hoti hai jab dono conditions poori hon: $\sum F = 0$ aur $\sum τ = 0$ .

Statement (a) akela kaafi nahi hai — zero net force wali body phir bhi spin up ho sakti hai (ek couple ki wajah se).
Statement (b) dono conditions explicitly state karta hai. ✓
Statement (c) in dono conditions ka physical matlab hai: $a_{c m} = 0$ ( $\sum F = 0$ se) aur $α = 0$ ( $\sum τ = 0$ se). ✓

Exercise 1.2 — Kaun sa force koi torque nahi banata?

Ek darwaza apne left edge par hinge hai. Us par equal size ke chaar horizontal pushes lagte hain. Kaun sa hinge line ke baare mein zero torque produce karta hai? (a) Door ke door (right) edge par push, darwaze ke perpendicular. (b) Beech mein push, darwaze ke perpendicular. (c) Push seedha hinge ki taraf (darwaze ke along, hinge ki taraf). (d) Door ke far edge par $4 5^{\circ}$ par push.

Recall Solution

Answer: (c). Hinge ki seedhi taraf ki push ki line of action pivot se guzarti hai, isliye moment arm $d = 0$ aur $τ = F \cdot 0 = 0$ . Isi liye darwaze ko uski hinge ki taraf push karne se wo kabhi nahi khulta. Baaki sabhi pushes ki nonzero perpendicular distance hoti hai aur isliye wo torque create karte hain.

L2 · Application

Exercise 2.1 — Basic seesaw

Ek uniform seesaw apne centre par pivoted hai. $30 kg$ ka ek bachcha pivot ke $1.5 m$ left mein baithta hai. $45 kg$ ka doosra bachcha right mein baithta hai. Doosre bachche ko pivot se kitni doori $x$ par baithna chahiye taaki balance ho jaye?

Recall Solution

Pivot ke baare mein torques lo. Ye axis kyun? Pivot ki upward reaction force pivot se guzarti hai, isliye uska moment arm zero hai aur wo bahar nikal jaati hai — use dhundhne ki zaroorat nahi.

Left bachcha plank ko CCW ghuma raha hai (maan lo $+$ ); right bachcha CW ( $-$ ). Balance ( $\sum τ = 0$ ): $(30) (10) (1.5) - (45) (10) (x) = 0$ $450 = 450 x ⟹ x = 1.0 m .$ Sanity check: bhaari bachcha ( $45$ kg) halke bachche se paas baithta hai ( $1.5$ m vs $1.0$ m). Bhaari weight × chhota arm = halka weight × bada arm. ✓

Exercise 2.2 — Tilted force ka torque

Ek force $F = 20 N$ pivot se $r = 0.40 m$ par ek jagah lagta hai. Force, pivot se application point tak ki line ke saath $3 0^{\circ}$ ka angle banata hai. Pivot ke baare mein torque ka magnitude nikalo.

Recall Solution

Force ka sirf wo component jo $r$ ke perpendicular hai body ko twist karta hai. Us perpendicular part ka size $F sin θ$ hai, isliye: $∣ τ ∣ = F r sin θ = (20) (0.40) sin 3 0^{\circ} = (20) (0.40) (0.5) = 4.0 N\cdotm .$ $sin$ kyun aur $cos$ kyun nahi? $F$ ka wo part jo $r$ ke along hai (wo hai $F cos θ$ ) seedha pivot ki taraf ya pivot se door point karta hai — uski line of action pivot se guzarti hai, isliye wo kuch contribute nahi karta. Sirf sideways part, $F sin θ$ , ka moment arm hota hai.

L3 · Analysis

Exercise 3.1 — Beam on two supports

Ek uniform horizontal beam, length $L = 4.0 m$ aur weight $W = 200 N$ , do supports par rakhti hai: left end par support $A$ aur right end par support $B$ . Left end se $1.0 m$ door $300 N$ ka load latka hai. Upward reaction forces $N_{A}$ aur $N_{B}$ nikalo.

Recall Solution

Beam uniform hai, isliye uska weight $W = 200 N$ centre of mass par act karta hai, yani beech mein, left se $2.0 m$ door. (Dekho Centre of Gravity vs Centre of Mass.)

Step 1 — $A$ (left end) ke baare mein torques. $A$ kyun? Reaction $N_{A}$ , $A$ par act karta hai, isliye uska moment arm zero hai aur wo gayab ho jaata hai — ek unknown turant khatam. Unknowns mein sirf $N_{B}$ bachta hai. CCW positive le kar: $N_{B}$ far end ko uthata hai (CCW, $+$ ); load aur weight neeche kheenchte hain (CW, $-$ ): $N_{B} (4.0) - (300) (1.0) - (200) (2.0) = 0$ $4 N_{B} = 300 + 400 = 700 ⟹ N_{B} = 175 N .$

Step 2 — Translational balance $\sum F_{y} = 0$ : $N_{A} + N_{B} = 300 + 200 = 500 ⟹ N_{A} = 500 - 175 = 325 N .$

$B$ ke baare mein torques se cross-check (wahi $N_{A}$ milna chahiye, kyunki $\sum F = 0$ axis choice ko free banata hai): $N_{A} (4.0) - (300) (3.0) - (200) (2.0) = 0 ⟹ 4 N_{A} = 900 + 400 = 1300 ⟹ N_{A} = 325 N . ✓$ Sanity check: load $A$ ke paas hai, isliye $A$ zyada carry karta hai ( $325 > 175$ ). ✓

L4 · Synthesis

Exercise 4.1 — Cable se thama beam (hinge + tension)

Ek uniform beam, weight $W = 100 N$ aur length $L$ , wall par apne left end par hinge hai. Uska right end ek cable se tika hua hai jo right end se wapas wall tak jaati hai aur beam ke saath $3 0^{\circ}$ ka angle banati hai. Beam horizontal hai. Cable mein tension $T$ nikalo, aur hinge force ke horizontal aur vertical components bhi nikalo.

Recall Solution

Beam par forces: weight $W = 100 N$ centre ( $L /2$ ) par neeche; tension $T$ cable ke along right end par (beam ke $3 0^{\circ}$ upar); hinge force ke unknown components $H_{x}$ (horizontal) aur $H_{y}$ (vertical) left end par.

Step 1 — Hinge ke baare mein torques. Hinge kyun? Dono hinge components hinge par act karte hain, isliye unka moment arm zero hai aur wo gayab ho jaate hain — hum seedha $T$ solve karte hain. Tension ka vertical component $T sin 3 0^{\circ}$ right end ko uthata hai (CCW, $+$ ); weight beech mein neeche kheenchta hai (CW, $-$ ). Tension ka horizontal component beam ke along seedha hinge ki taraf point karta hai, isliye hinge ke baare mein uska moment arm zero hai. $T sin 3 0^{\circ} \cdot L - W \cdot \frac{L}{2} = 0$ $L$ cancel ho jaata hai: $T sin 3 0^{\circ} = \frac{W}{2} = 50 ⟹ T (0.5) = 50 ⟹ T = 100 N .$

Step 2 — Horizontal balance $\sum F_{x} = 0$ . Cable beam ke right end ko wall ki taraf horizontal size $T cos 3 0^{\circ}$ se kheenchta hai, isliye hinge ko wapas push karna padta hai: $H_{x} = T cos 3 0^{\circ} = 100 \times \frac{3}{2} \approx 86.6 N .$

Step 3 — Vertical balance $\sum F_{y} = 0$ : $H_{y} + T sin 3 0^{\circ} - W = 0 ⟹ H_{y} = 100 - 100 (0.5) = 50 N (upward) .$

Sanity check: cable aur hinge dono milkar poore $100 N$ ko vertically support karte hain: $T sin 3 0^{\circ} + H_{y} = 50 + 50 = 100$ . ✓

L5 · Mastery

Exercise 5.1 — Fisalne wali ladder (numbers + limiting case)

Ek uniform ladder, weight $W = 200 N$ , horizontal ke saath $θ = 6 0^{\circ}$ par frictionless wall ke against tiki hai, rough floor par. (a) Friction force $f$ aur floor normal $N_{f}$ nikalo. (b) Minimum coefficient of static friction $μ_{m i n}$ nikalo. (c) Ab $600 N$ weight ka ek aadmi foot se ek-chauthai upar khada ho jaata hai. $μ_{m i n}$ recalculate karo.

Recall Solution

Ladder ki length $L$ maan lo. Forces: weight $W$ beech mein; wall normal $N_{w}$ (horizontal, kyunki wall frictionless hai); floor normal $N_{f}$ (upar); floor friction $f$ (horizontal, wall ki taraf point karta hai taaki foot bahar na phisale).

(a) Vertical balance: sirf $N_{f}$ aur $W$ vertical hain: $N_{f} = W = 200 N .$ Foot ke baare mein torques ( $N_{f}$ aur $f$ dono foot par hain, wo khatam ho jaate hain). Wall push height $L sin θ$ par act karta hai aur ladder ko foot ke upar CCW ghuma deta hai ( $+$ ); weight horizontal distance $\frac{L}{2} cos θ$ par act karta hai aur use CW ghuma deta hai ( $-$ ): $N_{w} (L sin θ) - W (\frac{L}{2} cos θ) = 0 ⟹ N_{w} = \frac{W}{2} cot θ .$ $θ = 6 0^{\circ}$ ke saath, $cot 6 0^{\circ} = 1/ 3 \approx 0.5774$ : $N_{w} = \frac{200}{2} (0.5774) = 57.74 N .$ Horizontal balance: $f = N_{w} = 57.74 N .$

(b) No-slip ke liye chahiye $f \leq μ N_{f}$ , isliye minimum equality par hai: $μ_{m i n} = \frac{f}{N _{f}} = \frac{57.74}{200} = 0.2887 = \frac{1}{2} cot 6 0^{\circ} .$ (Dekho Static Friction $f \leq μ N$ ke liye.)

(c) Ab ek aadmi $P = 600 N$ ladder ke $\frac{L}{4}$ upar hai, yani foot se horizontal distance $\frac{L}{4} cos θ$ par. Vertical: $N_{f} = W + P = 200 + 600 = 800 N .$ Foot ke baare mein torques: $N_{w} (L sin θ) = W (\frac{L}{2} cos θ) + P (\frac{L}{4} cos θ)$ $N_{w} = cot θ (\frac{W}{2} + \frac{P}{4}) = 0.5774 (100 + 150) = 0.5774 \times 250 = 144.34 N .$ $f = N_{w} = 144.34 N, μ_{m i n} = \frac{144.34}{800} = 0.1804.$

Surprising result: ladder par neeche ek aadmi khadd karne se $μ_{m i n}$ kam ho gaya ( $0.289$ se $0.180$ tak). Unka weight $N_{f}$ (denominator) ko bahut badhata hai, jabki foot ke paas rehne se tipping torque mein thoda sa hi add hota hai. Upar khada aadmi iska ulta karega.

Limiting-case check: jaise $θ \to 9 0^{\circ}$ (seedhi khadi ladder), $cot θ \to 0$ , isliye $μ_{m i n} \to 0$ — seedhi ladder ko koi friction nahi chahiye. Jaise $θ \to 0^{\circ}$ (almost flat), $cot θ \to \infty$ , isliye $μ_{m i n} \to \infty$ — koi real floor almost flat leaning ladder nahi rok sakta. ✓