Weightlessness — true (free fall) vs apparent: Worked examples

1.2.25 · D3 · Physics › Newton's Laws & Dynamics › Weightlessness — true (free fall) vs apparent

Shuru karne se pehle, un do symbols ko dobara samjhte hain jo hum har line mein use karenge:

$N = m (g + a) (up positive)$

Scenario matrix

Is topic ka har problem exactly inhi cells mein se kisi ek mein aata hai. Neeche ke examples cell ke naam se label kiye gaye hain taaki tum dekh sako ki poora board cover ho gaya hai.

#	Cell (case class)	Sign / value of $a$	Predicted feel	Example
A	At rest / constant velocity	$a = 0$	Normal ( $N = m g$ )	Ex 1
B	Accelerating up	$a > 0$	Heavy ( $N > m g$ )	Ex 2
C	Accelerating down (not free)	$- g < a < 0$	Light ( $N < m g$ )	Ex 3
D	Free fall (degenerate)	$a = - g$	Weightless ( $N = 0$ )	Ex 4
E	"Trap" — moving down, but decelerating	$a > 0$	Heavy	Ex 5
F	Beyond free fall (pushed down harder than $g$ )	$a < - g$	$N = 0$ , body floor chhod deti hai	Ex 6
G	Orbit / circular free fall	$a = g$ inward = " $- g$ " locally	Weightless	Ex 7
H	Different planet / altitude ( $g$ changes)	any $a$ , new $g$	recompute	Ex 8

Do special boundaries dhyan mein rakho: $a = 0$ "normal" ko "heavy/light" se alag karta hai, aur $a = - g$ exactly woh edge hai jahan $N$ zero ho jaata hai. $a = - g$ se neeche sab kuch (cell F) qualitatively naya hai — floor body ke saath nahi reh sakta, toh contact bilkul toot jaata hai.

Figure — Weightlessness — true (free fall) vs apparent

Worked examples

$g = 10 m/s^{2}$ aur $m = 60 kg$ lo jab tak problem kuch aur na kahe.

Worked example

Ex 5 — Cell E: the trap (neeche move karte waqt decelerate karna) Elevator $6 m/s$ se neeche ja raha hai aur $3 m/s^{2}$ se slow ho raha hai (rukne wala hai). Reading kya hogi?

Forecast: "Yeh neeche ja raha hai, toh light feel hoga" — kya yeh sahi hai?

Velocity vs acceleration. Velocity neeche hai, lekin magnitude mein ghatt rahi hai. Downward motion ko slow karne ke liye acceleration upar point karna chahiye: $a = + 3$ . Yeh step kyun? "Neeche jaate waqt slow hona" = acceleration velocity ke opposite = upward. Jaane ki direction ek decoy hai.
Apply karo. $N = 60 (10 + 3) = 780 N$ . Yeh step kyun? Upward acceleration ka matlab hai floor zyada push karta hai → tumhe heavier feel hota hai, chahe tum neeche ja rahe ho.

Verify: $780 > 600$ ✓ — neeche jaane ke bawajood heavier. Yahi woh Non-inertial Frames & Pseudo-forces gotcha hai jiske baare mein parent note ne warning diya tha: velocity direction irrelevant hai.

Worked example

Ex 6 — Cell F: gravity se bhi zyada neeche push kiya gaya Ek rocket floor neeche ki taraf fire hoti hai toh poora cabin $15 m/s^{2}$ se neeche accelerate karta hai (free fall se tez). Scale kya read karta hai?

Forecast: Formula ek negative number deta hai — iska matlab kya hai?

$a$ ka sign. Downward $15 \Rightarrow a = - 15$ , aur $∣ a ∣ = 15 > g = 10$ . Yeh step kyun? Hum free-fall boundary $a = - g$ se neeche cross kar gaye hain.
Naively apply karo. $N = 60 (10 - 15) = 60 (- 5) = - 300 N$ . Yeh step kyun? Taaki dekh sakein ki maths ek problem flag kar raha hai — ek floor tumhare paon ko pull nahi kar sakta, isliye negative $N$ physically impossible hai.
Interpret karo. Ek surface sirf push kar sakti hai ( $N \geq 0$ ). Isliye actual reading $N = 0$ hai: tumhare paon floor chhod dete hain. Tum ceiling ke against float karoge (ya ceiling ko tumhe neeche push karna hoga). Yeh step kyun? Jab demanded $N$ negative ho jaata hai, contact simply toot jaata hai — body us surface se alag ho jaati hai.

Verify: Boundary check: $N = 0$ pehli baar $a = - g = - 10$ par hota hai; kisi bhi $a < - 10$ ke liye floor keep up nahi kar sakta, isliye $N = 0$ aur body alag ho jaati hai. ✓ (Woh $- 300$ N ek signal hai, real force nahi.)

Worked example

Ex 7 — Cell G: orbit (curved free fall) $70 kg$ mass ki ek astronaut us altitude par orbit kar rahi hai jahan local gravitational acceleration $g^{'} = 8.7 m/s^{2}$ hai. Station ek circular orbit mein hai jiska centripetal acceleration $g^{'}$ ke barabar hai (iska matlab hi "orbit" hai). Apparent weight kya hai?

Forecast: $g^{'} = 8.7$ ke saath, kya uska weight $70 \times 8.7 = 609 N$ hoga, ya zero?

System ka acceleration kya hai? Circular orbit ke liye, gravity hi centripetal force hai, isliye $a = g^{'}$ Earth ki taraf directed — same direction jis mein gravity pull karti hai. Yeh step kyun? Circular Motion & Centripetal Acceleration: net inward acceleration poori tarah gravity supply karta hai, kuch aur nahi.
Falling frame mein jao. Station ke relative, astronaut exactly usi direction mein $g^{'}$ se accelerate karti hai jis mein gravity pull karti hai — yeh ek curve ke saath free fall hai. Locally "down" Earth ki taraf point karta hai, isliye up-positive local frame mein $a = - g^{'}$ . Yeh step kyun? Same setup jaise tooti cable (Ex 4), bas straight ki jagah curved.
Apply karo. $N = m (g^{'} + a) = 70 (8.7 - 8.7) = 0 N$ . Yeh step kyun? Station aur astronaut same acceleration share karte hain, isliye koi bhi dono mein se kisi par press nahi karta.

Verify: $N = 0$ ✓ despite $g^{'} = 8.7$ (surface gravity ka lagbhag 89%, Gravitation — Variation of g with altitude se). Float karna shared acceleration ki wajah se hota hai, zero gravity ki wajah se nahi.

Worked example

Ex 8 — Cell H: different planet, elevator accelerating Moon par ( $g_{M} = 1.6 m/s^{2}$ ) $80 kg$ mass ka ek astronaut ek lander mein baith kar upar ki taraf $2.0 m/s^{2}$ se accelerate karta hai. Scale reading kya hogi? Phir: kaunsa upward-or-downward $a$ use weightless karega?

Forecast: Kya reading uske Earth weight ke kareeb hogi ya bahut chhoti?

Local $g$ use karo. $g$ ki jagah $g_{M} = 1.6$ rakho. Up positive hai, $a = + 2.0$ . Yeh step kyun? Formula ka $g$ hamesha local gravitational acceleration hota hai — isme Earth-specific kuch bhi nahi.
Apply karo. $N = 80 (1.6 + 2.0) = 80 (3.6) = 288 N$ . Yeh step kyun? Uske standing Moon weight $80 \times 1.6 = 128 N$ ke relative heavy, upward acceleration ki wajah se.
Weightless condition. $N = 0$ chahiye $\Rightarrow g_{M} + a = 0 \Rightarrow a = - g_{M} = - 1.6 m/s^{2}$ (lander Moon ki taraf free fall mein). Yeh step kyun? Weightless edge hamesha $a = - g$ hoti hai local $g$ ke saath — yahan $- 1.6$ , $- 10$ nahi.

Verify: Standing Moon weight $80 \times 1.6 = 128 N$ ; upar accelerate karne par $288 > 128$ ✓. Free-fall $a = - 1.6$ deta hai $N = 80 (1.6 - 1.6) = 0$ ✓.

Recall

Recall Which cell am I in?

Constant speed se upar move karna — heavy, light, ya normal? ::: Normal ( $a = 0$ , isliye $N = m g$ ). Neeche jaate waqt slow ho raha hai — heavy ya light? ::: Heavy (acceleration upar point karta hai, $a > 0$ ). Formula $N = - 200$ N deta hai — physical reading kya hai? ::: $0$ N; body surface chhod deti hai (ek floor pull nahi kar sakta). Moon par, kaunsa $a$ weightlessness deta hai? ::: $a = - g_{M} = - 1.6 m/s^{2}$ (Moon ki taraf free fall). Orbit mein $g^{'} = 8.7$ ke saath, apparent weight kya hai? ::: $0$ N — shared acceleration, zero gravity nahi.

Connections

Parent: Weightlessness — yahan build ki gayi master formula.
Newton's Second Law — har example $F_{n e t} = ma$ vertical hai.
Normal Force — kyun $N \geq 0$ aur " $N = 0$ " ka physically kya matlab hai.
Newton's Third Law — kyun scale tumhara push read karta hai.
Non-inertial Frames & Pseudo-forces — "neeche jaana light feel karta hai" wala trap.
Circular Motion & Centripetal Acceleration — orbit as curved free fall (Ex 7).
Gravitation — Variation of g with altitude — $g^{'} = 8.7$ kahan se aata hai.
Free Fall Kinematics — motion mein $a = - g$ case.