Non-inertial reference frames — pseudo forces: Exercises

1.2.13 · D4 · Physics › Newton's Laws & Dynamics › Non-inertial reference frames — pseudo forces

Shuru karne se pehle, paanch symbols jo tumhe poori tarah yaad hone chahiye:

Definition Har problem ke paanch characters

$m$ — object ki mass (kilograms mein). Zyada $m$ = zyada strong pseudo force.
$a_{0}$ — frame ki acceleration (bus, lift, car, disc ki), jo ek ground observer measure karta hai jo khud accelerate nahi kar raha. Chhota sa arrow matlab yeh ek vector hai: isme size bhi hai aur direction bhi.
$a_{0}$ (bina arrow ke) — us vector ki magnitude (length), ek plain positive number $m/s^{2}$ mein. Toh $a_{0}$ batata hai kitna aur kis taraf; $a_{0}$ batata hai sirf kitna. Jab bhi arrow hata hua dikhe, hum sirf number ki baat kar rahe hain.
$F_{real}$ — real force: har us push aur pull ka sum jiska koi genuine physical source ho (gravity neeche kheench rahi hai, string tug kar rahi hai, floor push kar raha hai, friction pakad raha hai). Ye woh forces hain jo ek ground observer list karega. Ye har frame mein exist karte hain.
$F_{pseudo} = - m a_{0}$ — fake force. Minus sign $a_{0}$ ke arrow ko ulta kar deta hai: frame jis taraf speed up kare, fake force opposite direction mein point karta hai, aur iska size $m a_{0}$ hai (mass times magnitude). $F_{real}$ ke unlike, iska koi physical source nahi — yeh sirf isliye appear karta hai kyunki humne ek accelerating frame choose kiya.

Dono frame ke working law mein combine hote hain: $F_{real} + F_{pseudo} = m a_{in frame}$ , jahan $a_{in frame}$ woh acceleration hai jo accelerating frame ke andar measure hoti hai. Poore note mein, $g = 9.8 m/s^{2}$ real gravity ki strength hai jo sab cheez ko seedha neeche kheeanchti hai.

Intuition Figure 1 — woh do arrows jo tum hamesha draw karte ho

Neeche di gayi figure dekho. Isme ek boxed "frame" dikhaya gaya hai (sochlo ek train car hai) jo do wheels par chal raha hai. Uske upar ek blue arrow hai, jis par $a_{0}$ likha hai, jo right ki taraf point kar raha hai — yani frame jis direction mein speed up kar raha hai. Andar mass $m$ ka ek yellow block baitha hai, aur us block par ek pink arrow hai, jis par $F_{pseudo} = - m a_{0}$ likha hai, jo left ki taraf point karta hai — blue wale ke exactly opposite. Picture ka ek hi lesson hai: fake force (pink) hamesha frame ki acceleration (blue) ka mirror-flip hoti hai. Har problem ke liye yeh pink-opposite-blue image dhyan mein rakhna.

Alt text: a rectangular train car on two wheels; a blue rightward arrow labelled frame acceleration a-zero sits above it; a yellow block labelled m sits inside with a pink leftward arrow labelled pseudo force equals minus m a-zero, pointing opposite the blue arrow.

Figure — Non-inertial reference frames — pseudo forces

Level 1 — Recognition

Problem 1.1

Ek train $a_{0}$ ke saath forward accelerate karti hai jiska magnitude $a_{0} = 3 m/s^{2}$ hai. Mass $m = 8 kg$ ka ek suitcase smooth luggage rack par rakha hai. Train ke frame mein, suitcase par pseudo force ki direction aur size batao.

Recall Solution 1.1

KYA chahiye: fake force $F_{pseudo} = - m a_{0}$ . KYUN: humse motion train ke andar describe karne ko kaha gaya hai, jo ek accelerating (non-inertial) frame hai, isliye hume fake force add karni hogi. Size: $∣ F_{pseudo} ∣ = m a_{0} = 8 \times 3 = 24 N$ (magnitude $a_{0}$ use karke). Direction: minus sign $a_{0}$ ko flip karta hai. Train forward speed up karti hai, toh fake force backward (train ke rear ki taraf) point karti hai. Answer: $24 N$ , backward.

Problem 1.2

Ek lift downward $a_{0} = 2 m/s^{2}$ magnitude se speed up karte hue descend karti hai. Mass $m = 60 kg$ ka ek person andar khada hai. Person par pseudo force kis direction mein point karti hai, aur kitni badi hai?

Recall Solution 1.2

$a_{0}$ neeche point karta hai (lift downward speed gain kar rahi hai). Use flip karo: $F_{pseudo} = - m a_{0}$ upar point karta hai. Size $= m a_{0} = 60 \times 2 = 120 N$ . Answer: $120 N$ , upward. (Isliye jab lift neeche jaana shuru karta hai toh tum briefly lighter feel karte ho — upar ki fake force tumhara weight partly cancel kar deti hai.)

Level 2 — Application

Problem 2.1

Problem 1.1 ka smooth-rack suitcase ( $m = 8 kg$ , train magnitude $a_{0} = 3 m/s^{2}$ forward) freely slide kar sakta hai. Train ke andar bathe ek passenger ko yeh kitni acceleration se dikhega?

Recall Solution 2.1

KYA: modified law apply karo $F_{real} + F_{pseudo} = m a_{in frame}$ . KYUN: passenger ka frame non-inertial hai, isliye wahan yahi law kaam karta hai. Horizontally rack smooth hai $\Rightarrow$ real horizontal force $= 0$ . Sirf horizontal "force" pseudo force hai $= 24 N$ backward. $a_{in frame} = \frac{F _{pseudo}}{m} = \frac{24}{8} = 3 m/s^{2} backward.$ Answer: $3 m/s^{2}$ rear ki taraf. (Ground view se sanity check: ground ko suitcase par koi force nahi dikhti, toh woh still rehta hai jabki train $3 m/s^{2}$ par forward kheenchti hai — train ke relative suitcase exactly $3 m/s^{2}$ backward recede karta hai. ✓)

Problem 2.2

Mass $m = 70 kg$ ka ek person ek lift mein scale par khada hai jo upward $a_{0} = 1.5 m/s^{2}$ magnitude se accelerate kar rahi hai. Scale kya read karega (apparent weight $N$ )? $g = 9.8 m/s^{2}$ use karo.

Recall Solution 2.2

KYA: normal force $N$ nikalo (scale ka up-push, jo woh display karta hai). Lift frame mein forces: weight $m g$ neeche, normal $N$ upar, pseudo force $m a_{0}$ neeche (frame upar accelerate karta hai $\Rightarrow$ fake force neeche). Person lift mein rest par hai $\Rightarrow$ net $= 0$ : $N - m g - m a_{0} = 0 \Rightarrow N = m (g + a_{0}) = 70 (9.8 + 1.5) = 70 \times 11.3 = 791 N .$ Answer: $791 N$ (resting $70 \times 9.8 = 686 N$ se zyada). Dekho Apparent weight & normal force.

Level 3 — Analysis

Problem 3.1

Ek pendulum bob ek car ki roof se lata hai. Car horizontally accelerate karti hai aur string vertical se $θ = 1 2^{\circ}$ par settle ho jaati hai. Car ki acceleration ka magnitude $a_{0}$ kya hai? ( $g = 9.8 m/s^{2}$ .)

Recall Solution 3.1

KYA: car frame mein bob equilibrium mein hai tension $T$ (string ke along), weight $m g$ neeche, aur pseudo force $m a_{0}$ horizontal (car ki acceleration ke opposite) ke under. String tension ko horizontal aur vertical pieces mein split karo: $T sin θ = m a_{0} (horizontal balance), T cos θ = m g (vertical balance) .$ KYUN divide karein: divide karne se unknowns $T$ aur $m$ ek saath khatam ho jaate hain, sirf angle aur do accelerations bachte hain: $tan θ = \frac{a _{0}}{g} \Rightarrow a_{0} = g tan θ = 9.8 \times tan 1 2^{\circ} = 9.8 \times 0.2126 = 2.08 m/s^{2} .$ Answer: $a_{0} \approx 2.08 m/s^{2}$ . String ek built-in accelerometer hai.

Problem 3.2

Same pendulum-in-a-car setup. Bob ki mass $m = 0.30 kg$ hai aur car $a_{0} = 4.0 m/s^{2}$ magnitude se accelerate karti hai. (a) tilt angle $θ$ aur (b) string tension $T$ nikalo.

Intuition Figure 2 — pendulum ka force triangle

Neeche di gayi figure dekho. Ek string ceiling bracket se latki hui hai; yellow bob dotted vertical se angle $θ$ par (pale-yellow arc se marked) jhool raha hai. Bob se teen arrows nikalte hain: ek blue straight neeche $m g$ labelled (real weight), ek pink backward pointing $m a_{0}$ labelled (fake force, car ki motion ke opposite), aur ek white string ke along upar $T$ labelled (tension). Neeche ek blue arrow "car $a_{0}$ " labelled hai jo dikhata hai car forward speed up kar rahi hai, isliye bob peeche lag jaata hai — exactly isliye pink arrow backward point karta hai. White tension woh diagonal hai jo blue-neeche aur pink-peeche pair ko balance karti hai.

Alt text: a pendulum hanging from a ceiling, string tilted by angle theta from a dotted vertical line; from the bob a blue arrow points straight down labelled m g, a pink arrow points backward labelled m a-zero, and a white arrow points up along the string labelled T; a separate blue arrow at the bottom labelled car a-zero points forward.

Recall Solution 3.2

(a) $tan θ = \frac{a _{0}}{g} = \frac{4.0}{9.8} = 0.4082 \Rightarrow θ = arctan (0.4082) = 22. 2^{\circ} .$ KYUN arctan: $tan θ$ ratio deta hai; arctan inverse question hai "kis angle ka yeh tangent hai?", jo $θ$ recover karta hai. (b) Bob effectively ek stronger, tilted gravity mein latka hai jiska size $g_{eff} = g^{2} + a_{0}^{2}$ hai (down-weight aur sideways pseudo force ka diagonal). Tension poora utha leti hai: $T = m g^{2} + a_{0}^{2} = 0.30 9. 8^{2} + 4. 0^{2} = 0.30 96.04 + 16 = 0.30 112.04 = 0.30 \times 10.585 = 3.18 N .$ Answer: $θ \approx 22. 2^{\circ}$ , $T \approx 3.18 N$ .

Level 4 — Synthesis

Problem 4.1

Mass $m = 2.0 kg$ ka ek block ek truck ke floor par baitha hai. Block aur floor ke beech static friction ka coefficient $μ_{s} = 0.30$ hai. Truck forward accelerate karta hai. Truck ki maximum forward acceleration magnitude $a_{0}$ kya hai jisse pehle block backward (truck ke relative) slide karna shuru kare? ( $g = 9.8 m/s^{2}$ .)

Recall Solution 4.1

KYA: truck frame mein kaam karo. Block sliding ke verge par hai, toh woh frame mein still at rest hai $\Rightarrow$ net force zero, friction apni maximum par. Friction symbol define karo: $f$ woh static friction force hai — floor block ko sideways exert karta hai taaki woh slide na kare. Static friction khud ko adjust karti hai ek ceiling tak; woh ceiling $f_{m a x} = μ_{s} N$ hai, jahan $μ_{s}$ static friction ka coefficient hai aur $N$ floor ka normal (perpendicular) push hai. Truck frame mein horizontal forces:

Pseudo force $m a_{0}$ backward (frame forward accelerate karta hai).
Static friction $f$ forward, block ko jagah par rokti hai, maximum $f_{m a x} = μ_{s} N$ tak. Vertically kuch accelerate nahi karta aur koi vertical pseudo force nahi hai, toh $N = m g$ . Slipping ke verge par, friction maxed hai aur pseudo force ko balance karti hai: $m a_{0} = f_{m a x} = μ_{s} N = μ_{s} m g \Rightarrow a_{0} = μ_{s} g = 0.30 \times 9.8 = 2.94 m/s^{2} .$ KYUN mass cancel hoti hai: pseudo force ( $m a_{0}$ ) aur friction limit ( $μ_{s} m g$ ) dono $m$ ke saath scale karte hain, toh woh drop ho jaata hai — ek heavy block ko rokna na easier hai na harder. Answer: $a_{0}^{m a x} = 2.94 m/s^{2}$ .

Problem 4.2

Ek person ( $m = 65 kg$ ) ek lift ke andar scale par khada hai. Scale $520 N$ read karta hai. Kya lift upar ya neeche accelerate kar rahi hai, aur $a_{0}$ ki magnitude kya hai? ( $g = 9.8 m/s^{2}$ .)

Recall Solution 4.2

KYA: general lift result hai $N = m (g \pm a_{0})$ , jahan $+$ upward acceleration ke liye aur $-$ downward ke liye. Resting weight $m g = 65 \times 9.8 = 637 N$ hai. Scale $520 N < 637 N$ read karta hai, matlab person lighter hai, toh pseudo force upar point karti hogi, jo tab hota hai jab $a_{0}$ neeche point kare. Minus form use karo: $N = m (g - a_{0}) \Rightarrow 520 = 65 (9.8 - a_{0}) \Rightarrow 9.8 - a_{0} = \frac{520}{65} = 8.0 \Rightarrow a_{0} = 1.8 m/s^{2} .$ Answer: lift magnitude $a_{0} = 1.8 m/s^{2}$ se neeche ki taraf accelerate kar rahi hai.

Level 5 — Mastery

Problem 5.1

Mass $m$ ka ek block ek frictionless wedge par rakha hai jiska incline angle $α$ hai. Wedge ko horizontally push kiya jaata hai taaki block wedge ke relative fixed rahe (koi sliding nahi). Wedge ki required horizontal acceleration magnitude $a_{0}$ , $g$ aur $α$ ke terms mein nikalo. Phir $α = 3 0^{\circ}$ ke liye evaluate karo ( $g = 9.8 m/s^{2}$ ). Aakhir mein, discuss karo kya hoga agar wedge is value se faster ya slower accelerate kare.

Recall Solution 5.1

KYA: wedge ke frame mein jao. Block wahan rest par hai, toh real forces + pseudo force = 0. Block par forces:

Weight $m g$ straight neeche.
Normal force $N$ incline surface ke perpendicular.
Pseudo force $m a_{0}$ horizontal, wedge ki acceleration ke opposite. KYUN axes along/across incline use karein: unknown $N$ surface ke perpendicular hai, toh incline ke along aur perpendicular axes choose karne se ek equation $N$ -free ho jaati hai.

$cos α$ aur $sin α$ kahan se aate hain. Incline surface horizontal se $α$ tilt karti hai. Har horizontal ya vertical vector ko us par project karo — yeh exactly wahi "arrow ko do perpendicular legs mein todne" wala move hai jo opening definition mein tha:

Pseudo force horizontal hai. Ek horizontal arrow aur up-slope direction ke beech angle $α$ hai, toh uska along-slope component $m a_{0} cos α$ hai (adjacent-side projection). Yeh slope ke upar point karta hai.
Weight vertical hai. Ek vertical arrow aur up-slope direction ke beech angle $9 0^{\circ} - α$ hai, toh uska along-slope component $m g cos (9 0^{\circ} - α) = m g sin α$ hai, slope ke neeche point karta hai.

Incline ke along (up-slope positive), balance: $m a_{0} cos α = m g sin α \Rightarrow a_{0} = g tan α .$ Incline ke perpendicular (poornata ke liye — yeh $N$ fix karta hai): perpendicular components bhi zero sum karne chahiye. Weight $m g cos α$ surface mein contribute karta hai; pseudo force $m a_{0} sin α$ surface mein contribute karta hai; normal $N$ bahar push karta hai. Toh $N = m g cos α + m a_{0} sin α .$ Yeh confirm karta hai ki surface sach mein needed push supply kar sakta hai, aur dikhata hai ki $N$ wedge ke harder accelerate karne par badhta hai. $α = 3 0^{\circ}$ ke liye: $a_{0} = 9.8 \times tan 3 0^{\circ} = 9.8 \times 0.5774 = 5.66 m/s^{2} .$

Edge cases — kya ho agar $a_{0} \neq = g tan α$ ?

Agar $a_{0} = g tan α$ exactly ho, along-slope forces cancel hote hain aur block still rehta hai. ✓
Agar $a_{0} > g tan α$ ho, pseudo force ka up-slope push jeet jaata hai, toh wedge frame mein block ramp ke upar slide karta hai.
Agar $a_{0} < g tan α$ ho (including $a_{0} = 0$ , ek unaccelerated wedge), gravity ka down-slope pull jeet jaata hai aur block neeche slide karta hai — ordinary "block frictionless ramp par slip karta hai" wala case. Answer: $a_{0} = g tan α$ ; $3 0^{\circ}$ ke liye, $a_{0} \approx 5.66 m/s^{2}$ . Notice karo yeh pendulum wala same law hai ( $tan θ = a_{0} / g$ ) — incline surface tilted "effective vertical" ka role play karta hai.

Problem 5.2

Ek stone lift ki ceiling se string ke through lata hai. Lift upward $a_{0} = 3.0 m/s^{2}$ magnitude se accelerate karti hai jabki stone (mass $m = 0.50 kg$ ) ko simultaneously hold kiya jaata hai taaki string vertical se $θ = 2 0^{\circ}$ banaye aur stone lift frame mein fixed rahe. Isse ek effective gravity ke under motion maanke, (a) lift ke andar feel hone wali effective gravity $g_{eff}$ aur (b) stone ko $2 0^{\circ}$ par rokne ke liye required horizontal hold force $F_{h}$ nikalo. ( $g = 9.8 m/s^{2}$ .)

Recall Solution 5.2

Symbol define karo: $F_{h}$ horizontal holding force hai — ek real sideways push (imagine ek light spring ya lift wall se finger) jo tilted stone ko vertical par wapas swing karne se rokti hai. Yeh horizontal hai, isliye subscript $h$ hai. (a) Effective gravity. Upar accelerate karte lift mein downward pull boost ho jaata hai: pseudo force $m a_{0}$ neeche add karti hai, toh per unit mass effective downward gravity hai $g_{eff} = g + a_{0} = 9.8 + 3.0 = 12.8 m/s^{2} .$ KYUN add karein: frame upar accelerate karta hai $\Rightarrow$ pseudo force neeche $\Rightarrow$ real gravity par stack ho jaati hai. (b) Sideways hold. Lift frame mein stone equilibrium mein hai: tension $T$ (string ke along), effective weight $m g_{eff}$ neeche, aur horizontal hold $F_{h}$ . Tension ko pieces mein todo: Vertical: $T cos θ = m g_{eff}$ . Horizontal: $T sin θ = F_{h}$ . $T$ hatane ke liye divide karo: $F_{h} = m g_{eff} tan θ = 0.50 \times 12.8 \times tan 2 0^{\circ} = 0.50 \times 12.8 \times 0.36397 = 2.33 N .$ Answer: $g_{eff} = 12.8 m/s^{2}$ , $F_{h} \approx 2.33 N$ .

Active recall

Recall Quick self-check

Problem 1.1 mein fake-force ki size ::: $24 N$ , backward. Block ke slide karne se pehle truck ki max acceleration (Problem 4.1) ::: $a_{0} = μ_{s} g = 2.94 m/s^{2}$ . Block ko freeze karne ke liye wedge acceleration (Problem 5.1) ::: $a_{0} = g tan α$ . Wedge $g tan α$ se faster accelerate kare — block kis taraf slide karta hai? ::: Ramp ke upar (pseudo force ka up-slope push jeet jaata hai). Lift Problem 4.2 mein kis taraf accelerate karti hai? ::: Neeche, $1.8 m/s^{2}$ (scale light read karta hai).

Connections

Newton's Second Law — har solution $F = m a$ hai jisme fake term add hai.
Apparent weight & normal force — Problems 2.2 aur 4.2 apparent-weight questions hain.
Centrifugal force · Coriolis force — in fake forces ke rotating-frame cousins.
Galilean relativity — yaad dilata hai ki kon se frames ko pseudo force ki zarurat nahi.