Non-inertial reference frames — pseudo forces: Worked examples

1.2.13 · D3 · Physics › Newton's Laws & Dynamics › Non-inertial reference frames — pseudo forces

Kuch bhi shuru karne se pehle, chalo har symbol ko plain words mein earn karte hain.

Definition Woh symbols jo hum baar baar use karenge

$m$ — object ki mass, kilograms mein. "Kitna stuff hai."
$a_{0}$ — frame ki acceleration (bus, lift, car) ground ke relative. Ek arrow: iska ek size hai aur ek direction. Hum hamesha pehle ek positive direction choose karte hain, phir $a_{0}$ ek signed number hai: $+$ agar woh us direction mein point kare, $-$ agar doosri taraf.
$g = 9.8 m/s^{2}$ — Earth ki gravity pull ki strength, downward.
$F_{pseudo} = - m a_{0}$ — woh fake force jo hum accelerating frame ke andar add karte hain. Minus sign ka matlab hai bilkul ulti direction $a_{0}$ ke: agar frame forward speed up kare, toh fake force backward point karti hai.
$a_{in frame}$ — woh acceleration jo accelerating frame ke andar observer measure karta hai (parent-note symbol). Har example mein hum ise us frame ke naam se rename karte hain jisme hum hain: $a_{bus}$ , $a_{lift}$ , etc. Yeh woh hai jo passenger actually object ko karte hua dekhta hai, ground wala nahi.
$N$ — normal force: woh push jo ek surface (floor, scale) apne aap ke perpendicular direction mein deta hai. Ek scale $N$ read karta hai (dekho Apparent weight & normal force).
$T$ — tension: woh pull jo ek string ke along carry hoti hai, object se us cheez ki taraf directed hoti hai jis se string latki hoti hai.

Cell	Kya vary karta hai	Example
A. $a_{0}$ forward (+)	horizontal, sign $+$	Ex 1 — bus, ball slides
B. $a_{0}$ backward (−)	horizontal, sign $-$	Ex 2 — braking bus
C. $a_{0}$ up (+)	vertical, sign $+$	Ex 3 — lift going up
D. $a_{0}$ down (−)	vertical, sign $-$	Ex 4 — lift going down
E. Degenerate: free fall	$a_{0} = g$ down	Ex 5 — weightlessness
E′. Beyond free-fall	$a_{0} > g$ down	Ex 6 — driven-down cage
F. Zero input	$a_{0} = 0$	Ex 7 — inertial trap
G. Tilted / 2-D	horizontal + gravity combine	Ex 8 — pendulum tilt-meter
H. Limit behaviour	$a_{0} \to \infty$ , $a_{0} \to 0$	Ex 9 — string kitna tilt ho sakta hai?
I. Word problem	real-world story	Ex 10 — coffee cup on a train
J. Exam twist	block on incline in accelerating truck	Ex 11 — classic trap

Horizontal cases (A & B)

Figure — Non-inertial reference frames — pseudo forces

Figure s01 — Ek forward-accelerating bus (orange arrow, $a_{0}$ ) aur backward pseudo force (red arrow, $- m a_{0}$ ) ek blue ball par jo smooth floor par rakh hai. Ex 1 aur Ex 2 cover karta hai.

Worked example Ex 1 — Smooth bus floor par ball · Cell A (

$a_{0}$ forward) Ek bus forward $a_{0} = 3 m/s^{2}$ par accelerate karti hai. Ek ball jiska mass $m = 0.4 kg$ hai woh smooth (frictionless) floor par rakh hai. Ek passenger kya acceleration dekhega, aur kis direction mein?

Forecast: abhi guess karo — kya ball forward slide karti hai ya backward, aur kya uski acceleration $3 m/s^{2}$ se zyada hai ya kam?

Frame choose karo: bus (non-inertial). Real horizontal forces list karo: koi nahi — floor smooth hai. Yeh step kyun? Ek frame choose karna double-counting rokta hai. Smooth floor ⇒ no friction ⇒ zero real horizontal force.
Pseudo force add karo: $F_{pseudo} = - m a_{0} = - (0.4) (3) = - 1.2 N$ , yaani backward (bus ke forward $a_{0}$ ke opposite). Figure mein red arrow dekho. Yeh step kyun? $- m a_{0}$ mein minus sign ise backward point karwata hai — yahi poora trick hai.
Frame law apply karo: $F_{real} + F_{pseudo} = m a_{bus} \Rightarrow 0 + (- 1.2) = 0.4 a_{bus}$ . Yeh step kyun? Non-inertial bus ke andar, Newton's law tabhi close hota hai jab hum pseudo force ko real forces ke saath include karein; yahan $a_{bus}$ parent note ka $a_{in frame}$ hai — jo passenger actually measure karta hai.
Solve karo: $a_{bus} = - 1.2/0.4 = - 3 m/s^{2}$ — backward, magnitude $3 m/s^{2}$ . Yeh step kyun? Passenger genuinely ball ko exactly isi speed se backward jaate hua dekhta hai.

Verify: ground observer ko ball par koi force nahi dikhti, isliye woh still rehti hai; bus floor uske neeche $3 m/s^{2}$ se forward slide karti hai, toh bus ke relative ball $3 m/s^{2}$ se back move karti hai. Same number ✓. Units: $N / kg = m/s^{2}$ ✓.

Worked example Ex 2 — Braking bus · Cell B (

$a_{0}$ backward) Wahi bus ab brake karti hai, $4 m/s^{2}$ par decelerate karti hai. Forward ko positive lo, toh frame ki acceleration signed value $a_{0} = - 4 m/s^{2}$ hai (backward pointing). Same $m = 0.4 kg$ ball smooth floor par. Woh kis taraf slide karti hai?

Forecast: tum ek braking bus mein ho — floor par rakhhi loose cheezein kis taraf jaati hain?

Frame ki signed acceleration: $a_{0} = - 4 m/s^{2}$ (backward).
Pseudo force: $F_{pseudo} = - m a_{0} = - (0.4) (- 4) = + 1.6 N$ — forward (positive). Yeh step kyun? Do minus signs se plus banta hai: "backward" ka ulta "forward" hai. Ball front ki taraf lurch karti hai.
Frame law: $0 + 1.6 = 0.4 a_{bus} \Rightarrow a_{bus} = + 4 m/s^{2}$ forward. Yeh step kyun? Ex 1 jaisi hi closure — jedhi horizontal "force" pseudo force hai, toh wahi $a_{bus}$ set karti hai.

Verify: roz ka experience — jab bus brake maare, unsecured cheezein forward jaati hain. Magnitude $∣ a_{0} ∣$ ke barabar hai smooth floor par ✓. Sign Ex 1 ke versus sahi flip hua ✓.

Vertical cases (C, D, E, E′)

Figure s02 — Ek lift mein ek insaan ka free-body diagram jo upar accelerate kar raha hai (orange). Weight $m g$ down (gray), normal $N$ up (green, scale reading), pseudo force $m a_{0}$ down (red). Ex 3–6 cover karta hai.

Yahan gravity $m g$ (hamesha down) pseudo force ke saath stage share karti hai. Ek insaan jiska mass $m = 60 kg$ hai woh bathroom scale par khada hai; scale normal force $N$ read karta hai (upar define kiya). Up ko positive lo.

Worked example Ex 3 — Lift accelerating up · Cell C

Lift $a_{0} = 2 m/s^{2}$ par upar accelerate karti hai (signed $a_{0} = + 2$ ). Scale kya read karta hai?

Forecast: still khade rehne se zyada bhaari lagega ya halka?

Real forces (up positive): weight $- m g$ (down), normal $+ N$ (up).
Pseudo force: $a_{0}$ up hai, toh $F_{pseudo} = - m a_{0}$ down point karta hai, value $- m a_{0}$ . Yeh step kyun? Frame upar accelerate karta hai ⇒ fake force tumhe scale mein neechay dhakelta hai.
Lift frame mein insaan rest par hai ⇒ $a_{lift} = 0$ ⇒ net $= 0$ : $N - m g - m a_{0} = 0$ . Yeh step kyun? Insaan lift ke relative move nahi karta, toh unka $a_{in frame}$ zero hai aur saari forces (real + pseudo) cancel honi chahiye.
Solve karo: $N = m (g + a_{0}) = 60 (9.8 + 2) = 708 N$ .

Verify: $N > m g = 588 N$ , toh zyada bhaari lagte ho — upar jaane wali lift ke stomach-drop feeling se match karta hai ✓.

Worked example Ex 4 — Lift accelerating down · Cell D (signs dhyaan se!)

Wahi lift ab $2 m/s^{2}$ ki rate se neeche accelerate karti hai. Up positive rakhte hue, signed frame acceleration $a_{0} = - 2 m/s^{2}$ hai. Scale reading?

Forecast: is baar halka lagega ya bhaari?

$F_{pseudo} = - m a_{0}$ mein signed value substitute karo: $F_{pseudo} = - m a_{0} = - (60) (- 2) = + 120 N$ — ek positive (upward) force. Magnitude ke taur par yeh $120 N$ hai, aur direction up hai. Yeh step kyun? Yahi woh signed-vs-magnitude trap hai jo pehle flag ki gayi: kyunki $a_{0}$ already $- 2$ hai, formula $- m a_{0}$ apne aap $+ 120 N$ deta hai — hum manually sign flip nahi karte. Fake force tumhe scale se thoda upar uthata hai.
Lift frame mein equilibrium ( $a_{lift} = 0$ ), up positive: $N - m g + 120 = 0$ . Yeh step kyun? Signed forces ka sum zero hai kyunki insaan lift mein rest par hai. Yahan $+ 120$ step 1 se signed pseudo force hai, woh magnitude nahi jiska sign hum khud chose karein.
Solve karo: $N = m g - 120 = 60 (9.8) - 120 = 588 - 120 = 468 N$ . Equivalently $N = m (g - 2) = 60 (7.8) = 468 N$ .

Verify: $N < 588 N$ — halka lagte ho, yeh downward-accelerating lift ke liye sahi hai ✓. Ex 3 se cross-check: compact formula $N = m (g + a_{0})$ mein signed $a_{0} = - 2$ se $60 (9.8 - 2) = 468$ aata hai ✓ — signed substitution aur by-hand version dono agree karte hain.

Worked example Ex 5 — Free fall · Cell E (degenerate)

Lift ki cable snap karti hai: woh freely girta hai, downward rate $g = 9.8 m/s^{2}$ , toh signed $a_{0} = - 9.8$ . Scale reading?

Forecast: true weightlessness ke liye tum kaunsa number expect karte ho?

$N = m (g + a_{0})$ mein signed $a_{0} = - g$ use karo: $N = m (g - g) = 0 N$ . Yeh step kyun? Yeh degenerate case hai — downward frame acceleration frame ke andar gravity ki pull ko exactly cancel kar deta hai.
Interpretation: pseudo force ( $m g$ up) real weight ( $m g$ down) ko perfectly cancel karta hai. Scale par kuch nahi dabta.

Verify: $N = 0$ ⇒ weightlessness — exactly wahi jo astronauts aur drop-tower riders feel karte hain ✓. Yeh "feel heavy" ( $a_{0} > 0$ , going up) aur "lift off hona" ( $a_{0} < - g$ , driven down faster than gravity) ke beech ki boundary hai.

Worked example Ex 6 — Cage ko gravity se

zyada fast neeche drive karna · Cell E′ (beyond free-fall) Ek motor lift ko gravity se aage $2 m/s^{2}$ ki rate se neeche kheenchti hai, yaani signed frame acceleration $a_{0} = - 12 m/s^{2}$ hai (up positive) — gravity se zyada fast neeche. Insaan ab ek scale par khada hai jo cage ki ceiling par hai (ek spring jo unke sir par neeche push karta hai), jo sirf push kar sakta hai, ek downward normal force $N_{c} \geq 0$ deta hai. Woh ceiling-scale kya read karta hai?

Forecast: floor unhe ab rok nahi sakta — unhe kya push karta hai, aur kitna?

Pehle floor scale test karo $N = m (g + a_{0})$ se, signed $a_{0} = - 12$ : $N = 60 (9.8 - 12) = 60 (- 2.2) = - 132 N$ . Yeh step kyun? Ek floor sirf upar push kar sakta hai ( $N \geq 0$ ). Negative answer matlab floor ko kheenchna padega — jo impossible hai. Toh insaan floor chhod deta hai aur ceiling se press karta hai.
Ceiling ke contact ke saath free body redraw karo. Up positive rakhte hue, har signed force list karo:
- weight: $- m g$
- formula se pseudo force: $- m a_{0} = - (60) (- 12) = + 720$ (up)
- ceiling push: $- N_{c}$ (ceiling sirf sir ko neeche push kar sakti hai, toh uski signed value negative hai) Yeh step kyun? Hum signed $a_{0} = - 12$ seedha $- m a_{0}$ mein substitute karte hain (koi by-hand sign flip nahi), Ex 4 jaisi convention rakhte hue. Yeh $+ 720 N$ ka upward pseudo force deta hai.
Signed sum ko zero karo (insaan cage mein rest par, $a_{cage} = 0$ ): $- m g + (- m a_{0}) - N_{c} = 0 \Rightarrow - 588 + 720 - N_{c} = 0.$ Yeh step kyun? Saari signed forces cancel hoti hain kyunki insaan cage ke relative move nahi karta; yeh wahi equilibrium closure hai jo Ex 3–5 mein use hua, ab ceiling term ke saath.
Reading ke liye solve karo: $N_{c} = 720 - 588 = 132 N$ — ceiling unke sir par $132 N$ se press karti hai.

Verify: magnitude step 1 ke impossible floor reading se match karta hai ( $∣ - 132∣ = 132$ ) ✓ — physics ne sirf contact floor se ceiling par shift kar diya. Yeh free-fall ke bilkul aage ka natural case hai (Ex 5 ka $N = 0$ crossover hai $a_{0} = - g$ par).

Zero-input trap (F)

Worked example Ex 7 — Koi acceleration nahi · Cell F

Ek train constant $20 m/s$ par move karti hai (toh $a_{0} = 0$ ). Ek student "helpfully" ek pseudo force add karta hai yeh explain karne ke liye ki unki coffee kyun tiki rahti hai. Pseudo force kya hai?

Forecast: kya koi pseudo force honi chahiye?

$F_{pseudo} = - m a_{0} = - m \cdot 0 = 0 N$ . Yeh step kyun? Constant velocity ⇒ $a_{0} = 0$ ⇒ frame inertial hai (dekho Galilean relativity). Koi correction ki zaroorat nahi.
Agar tum yahan ek nonzero pseudo force add karo, toh tum Newton's law ko todoge jo already theek tha — ek double count. Yeh step kyun? Ek inertial frame already $F = m a$ satisfy karta hai; ek extra fake force over-count karega aur galat motion dega.

Verify: coffee still rehti hai kyunki koi real force act nahi karti aur koi fake force bhi zaroorat nahi; $F = m a$ as-is kaam karta hai ✓. Yeh exactly woh third [!mistake] hai jiske baare mein parent note ne warn kiya tha.

Tilted / two-dimensional case (G)

Figure s03 — Pendulum bob (blue) ek forward-accelerating car mein (orange). String tension string ke along, weight $m g$ down (gray), pseudo force $m a_{0}$ backward (red); vertical se tilt angle $θ$ (red arc) $tan θ = a_{0} / g$ satisfy karta hai.

Worked example Ex 8 — Pendulum accelerometer · Cell G

Ek pendulum bob car ki roof se latka hua hai. Car $a_{0} = 4.9 m/s^{2}$ par horizontally accelerate karti hai. Bob vertical se $θ$ angle par settle ho jaata hai. $θ$ find karo.

Forecast: $θ$ guess karo — $4 5^{\circ}$ se bada hoga ya chhota? (Hint: $a_{0}$ ko $g$ se compare karo.)

Car frame mein bob equilibrium mein hai. Teen forces (figure dekho): tension $T$ string ke along, weight $m g$ down, pseudo force $m a_{0}$ backward (car ke forward $a_{0}$ ke opposite). Yeh step kyun? Steady angle ⇒ car frame mein zero acceleration ( $a_{car} = 0$ ) ⇒ forces sum to zero.
Tension ko pieces mein split karo. Tan kyun aur sin ya cos kyun nahi? Hum angle chahte hain do known perpendicular forces se — horizontal pull ( $m a_{0}$ ) aur vertical pull ( $m g$ ). $θ$ ke opposite side ka adjacent side se ratio $tan θ$ hai; woh ratio exactly "kitna tilted hai" bataata hai, toh tan sahi tool hai.
Horizontal balance: $T sin θ = m a_{0}$ . Vertical balance: $T cos θ = m g$ . Yeh step kyun? String vertical se $θ$ tilt hai, toh uski tension ka horizontal part $T sin θ$ hai (pseudo force ko rok ke rakha) aur vertical part $T cos θ$ hai (weight ko hold kiya). Equilibrium mein har direction alag se balance hoti hai.
Unknown $T$ eliminate karne ke liye dono equations divide karo: $\frac{T sin θ}{T cos θ} = \frac{m a _{0}}{m g} \Rightarrow tan θ = \frac{a _{0}}{g} = \frac{4.9}{9.8} = 0.5$ . Yeh step kyun? Hum $T$ nahi jaante aur uski zaroorat bhi nahi, toh divide karne se woh cancel ho jaata hai aur $m$ bhi — sirf angle aur do accelerations ke beech ek pure relation reh jaata hai.
arctan se tan undo karo (sawaal "kaunse angle ka tan yeh hai?"): $θ = arctan (0.5) \approx 26.5 7^{\circ}$ . Yeh step kyun? $tan$ ne angle ko ratio mein badal diya; $arctan$ uska inverse hai, ratio ko wapas angle mein badalta hai jo hum actually chahte hain.

Verify: $a_{0} < g$ ⇒ $tan θ < 1$ ⇒ $θ < 4 5^{\circ}$ ✓ (forecast se match). Bob backward jhukta hai, car ki motion ke opposite, jaise ek real accelerometer karta hai ✓. Effective gravity ki magnitude $g^{2} + a_{0}^{2} = 9. 8^{2} + 4. 9^{2} \approx 10.96 m/s^{2}$ hai, string ke along point karti hai ✓.

Limit behaviour (H)

Worked example Ex 9 — String kitna tilt ho sakta hai? · Cell H (limits)

Ex 8 se $tan θ = a_{0} / g$ use karke, do extremes examine karo: (a) $a_{0} \to 0$ , (b) $a_{0} \to \infty$ .

Forecast: bahut badi acceleration par, kya string $9 0^{\circ}$ tak pahunch jaati hai?

Small limit: jaise $a_{0} \to 0$ , $tan θ \to 0 \Rightarrow θ \to 0^{\circ}$ . Yeh step kyun? Koi horizontal fake force nahi ⇒ string seedha neeche latkti hai. Range ke bottom ka sanity anchor.
Large limit: jaise $a_{0} \to \infty$ , $tan θ \to \infty \Rightarrow θ \to 9 0^{\circ}$ lekin kabhi reach nahi karta. Yeh step kyun? $tan (9 0^{\circ})$ undefined hai (vertical line ki "infinite steepness" hoti hai); string horizontal ke paas aati hai lekin koi bhi finite $a_{0}$ $θ$ ko $9 0^{\circ}$ se thoda neeche rakhta hai.
Concrete midpoint: $a_{0} = g = 9.8 \Rightarrow tan θ = 1 \Rightarrow θ = arctan (1) = 4 5^{\circ}$ exactly. Yeh step kyun? Equal horizontal aur vertical pulls se $4 5^{\circ}$ tilt aata hai — do limits ke beech ek yaadgar landmark.

Verify: $0^{\circ}$ (rest par) se monotone rise $4 5^{\circ}$ ( $a_{0} = g$ par) se $9 0^{\circ}$ (infinite $a_{0}$ ) ki taraf. $a_{0} = g$ par: $arctan (1) = 4 5^{\circ}$ ✓; $tan (0) = 0$ low end ko anchor karta hai ✓.

Word problem (I)

Worked example Ex 10 — Starting train par coffee cup · Cell I

Ek train start karti hai aur $5 s$ mein $15 m/s$ reach karti hai (uniform acceleration). Ek coffee cup ( $m = 0.3 kg$ ) ek table par rakhha hai jisme cup aur table ke beech friction coefficient $μ = 0.15$ hai. Kya cup backward slide karega? (Lo $g = 9.8$ .)

Forecast: kya fake force friction ko beat kar deti hai?

Frame acceleration: $a_{0} = Δ v /Δ t = 15/5 = 3 m/s^{2}$ forward. Yeh step kyun? Kisi bhi pseudo force se pehle hum $a_{0}$ chahte hain; yeh seedha "speed mein change over time" se aata hai.
Cup par pseudo force: $F_{pseudo} = m a_{0} = 0.3 \times 3 = 0.9 N$ backward. Yeh step kyun? Train frame ke andar yeh backward ghost-shove woh hai jo cup ko slide karwaane ki koshish karta hai.
Available maximum static friction: $f_{m a x} = μ m g = 0.15 \times 0.3 \times 9.8 = 0.441 N$ . Yeh step kyun? Friction sirf iss amount tak grip supply kar sakta hai; agar pseudo force ise exceed kare, toh cup slip karta hai.
Compare karo: $0.9 N > 0.441 N$ ⇒ cup backward slide karta hai. Yeh step kyun? Demand ( $0.9$ N) supply ( $0.441$ N) se zyada hai, toh static friction overwhelm ho jaati hai.

Verify: ghost-shove ( $0.9$ N) grip ( $0.441$ N) ko haraata hai, toh coffee slide karti hai — exactly isliye spill hoti hai jab train forward jerk karti hai ✓. Units sab newtons ✓.

Exam twist (J)

Figure s04 — Ek block (blue) ek frictionless $3 0^{\circ}$ wedge par jo ek forward-accelerating truck (orange) ke andar hai. Weight $m g$ down (gray) aur pseudo force $m a_{0}$ backward (red) slope ke along balance karte hain jab $a_{0} = g tan α$ .

Worked example Ex 11 — Frictionless wedge par block ek accelerating truck ke andar · Cell J

Ek wedge jiska incline angle $α = 3 0^{\circ}$ hai truck bed par rakhhi hai. Ek block frictionless incline par rakhha hai. Truck forward (wedge ke high side ki taraf) $a_{0}$ par accelerate karta hai. Woh $a_{0}$ find karo jo block ko wedge ke relative stationary rakhta hai (koi sliding nahi).

Forecast: kya $a_{0}$ bada hona chahiye ya chhota $3 0^{\circ}$ slope par block ko pin karne ke liye?

Frame: truck. Block par forces: weight $m g$ down, normal $N$ incline ke perpendicular, pseudo force $m a_{0}$ backward. Yeh step kyun? Frictionless ⇒ sirf gravity, normal, aur fake force act karte hain. Madad karne ke liye koi friction nahi.
"Wedge ke relative stationary" ⇒ incline surface ke along net force zero hai. Yeh step kyun? Sirf along-slope direction hi ise slide karwa sakti hai; ise balance karna condition hai.
Along incline resolve karo (up-slope positive lo). Gravity ka along-slope pull: $- m g sin α$ . Pseudo force ka along-slope component: backward force, slope ke upar project kiya, $+ m a_{0} cos α$ deta hai. Yeh step kyun? Hume sirf surface ke parallel components chahiye; normal force ka koi along-slope component nahi hota.
Balance: $m a_{0} cos α - m g sin α = 0 \Rightarrow a_{0} = g tan α$ .
Number: $a_{0} = 9.8 \times tan 3 0^{\circ} = 9.8 \times 0.5774 \approx 5.66 m/s^{2}$ . Yeh step kyun? Yeh wahi acceleration hai jo block ko pin karta hai; isse slow aur woh neeche slide karega, isse fast aur woh upar slide karega.

Verify: pendulum se form mein identical, $tan α = a_{0} / g$ — "effective gravity" $g^{2} + a_{0}^{2}$ incline ke perpendicular point karni chahiye taaki block slide na kare. Centrifugal force-style effective-gravity reasoning ke saath consistent ✓. Number check: $9.8 tan 3 0^{\circ} \approx 5.66$ ✓.

Active recall

Recall Quick self-test

Smooth bus floor par ball, bus $3 m/s^{2}$ forward accelerate karti hai — bus frame mein ball ki acceleration? ::: $3 m/s^{2}$ backward. $60 kg$ ke liye scale reading lift mein $2 m/s^{2}$ upar accelerate karte hue? ::: $N = 60 (9.8 + 2) = 708 N$ . Wahi lift $2 m/s^{2}$ neeche accelerate karti hai (signed $a_{0} = - 2$ )? ::: $N = 60 (9.8 - 2) = 468 N$ . Free-fall mein scale reading? ::: $0 N$ (weightless). Cage $12 m/s^{2}$ neeche driven — floor "reading" aur actually kya hota hai? ::: Floor formula $- 132 N$ deta hai (impossible); insaan ceiling ko $132 N$ se press karta hai. Pendulum tilt $a_{0} = 4.9$ , $g = 9.8$ ke liye? ::: $θ = arctan 0.5 \approx 26.5 7^{\circ}$ . $a_{0}$ ek frictionless $3 0^{\circ}$ wedge par block pin karne ke liye? ::: $a_{0} = g tan 3 0^{\circ} \approx 5.66 m/s^{2}$ . Pseudo force jab frame constant velocity par move kare? ::: Zero — frame inertial hai.

Connections

Parent topic — yeh page har case class fill karta hai.
Newton's Second Law — har cell force-balance reduce ho jaata hai jab ghost force add ho jaaye.
Apparent weight & normal force — Ex 3–6 lift family hain.
Newton's Third Law — yaad rakho pseudo force ka kabhi koi reaction partner nahi hota.
Galilean relativity — Ex 7 ka constant-velocity frame inertial hai.
Centrifugal force · Coriolis force — rotating-frame pseudo forces usi $- m a_{0}$ idea par build karte hain.

Worked examples — Non-inertial reference frames — pseudo forces

Scenario matrix