Normal force — reaction force, not always = mg: Worked examples

1.2.5 · D3 · Physics › Newton's Laws & Dynamics › Normal force — reaction force, not always = mg

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Scenario matrix

Har normal-force problem is table ki ek row hai. Is page ka poora point yahi hai ki har row mein kam se kam ek fully worked example ho taaki tumhe koi unseen case kabhi na mile.

Cell	Situation class	$N$ mein kya badalta hai	Expected result	Example
A	Flat floor, kuch nahi	baseline	$N = m g$	Ex 1
B	Flat floor + neeche push	zyada hold karna hai	$N = m g + F$	Ex 2
C	Flat floor + upar pull (degenerate "lifts off" case ke saath)	kam hold karna hai, $0$ tak aa sakta hai	$N = m g - F$ , floored at $0$	Ex 3
D	Elevator upar accelerate kar raha hai	floor zyada push karta hai	$N = m (g + a)$	Ex 4
E	Elevator neeche accelerate kar raha hai / free fall ( $a = g$ , degenerate)	floor kam push karta hai, $0$ tak aa sakta hai	$N = m (g - a)$	Ex 5
F	Incline, angle $θ$ (general + limits $θ \to 0, θ \to 9 0^{\circ}$ )	sirf gravity ka $⊥$ part	$N = m g cos θ$	Ex 6
G	Vertical wall, horizontal push se hold kiya gaya (word problem)	gravity surface ke saath hai, usmein nahi	$N = F_{push}$	Ex 7
H	Incline horizontally accelerated (exam twist)	$a_{⊥} \neq = 0$	$N = m (g cos θ + a sin θ)$	Ex 8

Neeche use kiye gaye prerequisite tools: Newton's Second Law, Free Body Diagrams, aur rows F/H ke liye Inclined Plane Problems se geometry; row G mein Friction ideas aati hain; rows D/E Apparent Weight & Elevators ka core hain.

Cell A — Flat floor, baseline

Worked example Ex 1 — Table par Book

Ek $2 kg$ ki book flat table par still rakhee hai. $N$ find karo. ( $g = 10$ .)

Forecast: Padhne se pehle guess karo — kya $N$ , $20 N$ weight se bada, chhota, ya barabar hoga?

Forces list karo. Weight $m g = 2 \times 10 = 20 N$ neeche; normal $N$ upar. Yeh step kyun? Free body diagram mein har touching ya pulling agent aana chahiye — yahan sirf Earth (gravity) aur table (normal) hain.
Newton's 2nd law likho, upar positive. $N - m g = m a_{y}$ . Yeh step kyun? Newton's Second Law kehta hai net force $=$ mass $\times$ acceleration, aur hum ise interesting axis (vertical) ke along resolve karte hain.
$a_{y} = 0$ daalo. Book rest par hai aur rest par rehti hai, isliye accelerate nahi karti. $N - 20 = 0 \Rightarrow N = 20 N$ . Yeh step kyun? Koi vertical acceleration nahi — yahi exact special condition hai jo $N = m g$ banati hai.

Verify: $N = m g = 20 N$ . Units: $kg \cdot m/s^{2} = N$ ✔. Weight ke barabar, jaisa baseline row predict karta hai. ✔

Cell B — Extra push neeche

Cell C — Upar pull, lift-off degenerate case ke saath

Worked example Ex 3 — Upar kheenchna, do sub-cases

$2 kg$ ki book table par. (a) Tum upar $F = 8 N$ se pull karte ho. (b) Tum upar $F = 25 N$ se pull karte ho. Dono mein $N$ find karo.

Forecast: Part (b) mein kya $N$ negative ho sakta hai? Socho ki negative "push" ka matlab kya hoga.

Newton's 2nd law, upar positive, $a_{y} = 0$ (jab tak table par hai). $N + F - m g = 0 \Rightarrow N = m g - F$ . Yeh step kyun? Upar ki pull book ko hold karne mein help karti hai, isliye table ko kam karna padta hai.
Part (a): $N = 20 - 8 = 12 N$ . Yeh step kyun? $F < m g$ , isliye surface abhi bhi zaroori hai. Fine.
Part (b): formula deta hai $20 - 25 = - 5 N$ . Yeh step kyun aur catch kya hai: ek surface sirf push kar sakti hai, pull nahi. Negative $N$ impossible hai, isliye iska matlab hai book surface chhod chuki hai: sahi jawab hai $N = 0$ . Kyun: jab $F \geq m g$ ho jata hai toh object lift off kar leta hai aur koi contact nahi rehta.

Verify: (a) $N = 12 N$ , $[0, m g]$ mein ✔. (b) formula $= - 5$ ⇒ clamp karke $N = 0$ ✔ (degenerate lift-off cell handle ho gaya).

Cell D — Elevator upar accelerate kar raha hai

Worked example Ex 4 — Upar jaate lift mein heavy feeling

Ek $60 kg$ banda ek elevator mein khada hai jo upar $a = 2 m/s^{2}$ se accelerate kar raha hai. Floor ki push $N$ find karo (unka apparent weight).

Forecast: Unke true weight $600 N$ se heavier ya lighter?

Free body: sirf $N$ upar, $m g$ neeche. Yeh step kyun? Banda sirf floor se touch kar raha hai; elevator ki motion acceleration ke through enter hoti hai, kisi naye force ke through nahi.
Newton's 2nd law, upar positive, $a_{y} = + a$ . $N - m g = ma$ . Yeh step kyun? Banda sach mein cabin ke saath upar accelerate kar raha hai, isliye right side zero nahi hai — yahi core Apparent Weight & Elevators idea hai.
Solve karo. $N = m (g + a) = 60 \times (10 + 2) = 720 N$ . Kyun: floor ko weight hold karna bhi hai aur banda speed up ho sake isliye extra push bhi supply karni hai.

Verify: $N = 720 N > 600 N$ ✔ — upar accelerate karte hue heavier feel hota hai, row D se match karta hai.

Cell E — Neeche accelerate karna, aur free-fall degenerate case

Worked example Ex 5 — Neeche jaata lift aur toota hua cable

Wahi $60 kg$ banda. (a) Elevator neeche $a = 4 m/s^{2}$ se accelerate karta hai. (b) Cable toot jaata hai aur cabin free fall mein hai ( $a = g = 10$ ). Dono mein $N$ find karo.

Forecast: Part (b) mein, jab sab kuch saath girta hai toh floor kya push karta hai?

Newton's 2nd law, upar positive, neeche acceleration matlab $a_{y} = - a$ . $N - m g = m (- a) \Rightarrow N = m (g - a)$ . Yeh step kyun? Neeche ki taraf acceleration zaroori net upward force kam kar deti hai, isliye floor relax karta hai.
Part (a): $N = 60 \times (10 - 4) = 360 N$ . Kyun: $600 N$ se kam — neeche jaate hue lighter feel hota hai. ✔
Part (b), free fall: $N = 60 \times (10 - 10) = 0 N$ . Kyun: banda aur floor ek hi rate se girte hain, isliye koi cheez kisi pe press nahi karti — weightlessness, is row ki degenerate limit.

Verify: (a) $N = 360 N$ ( $[0, m g]$ mein) ✔; (b) $N = 0 N$ ✔ — free-fall weightlessness confirm hua.

Cell F — Incline, general angle aur dono limits

Yahan surface tilted hai, isliye gravity seedhi usmein point nahi karti. Hum weight ko do arrows mein split karte hain: ek slope mein aur ek slope ke along. Sirf "slope mein" wala part $N$ se balance hona chahiye.

Figure — Normal force — reaction force, not always = mg

Worked example Ex 6 —

$3 7^{\circ}$ frictionless ramp par block, aur limits bhi Ek $4 kg$ ka block frictionless incline par $θ = 3 7^{\circ}$ par rakha hai. $N$ find karo. Phir do extreme angles $θ = 0^{\circ}$ aur $θ = 9 0^{\circ}$ check karo. ( $cos 3 7^{\circ} = 0.8$ use karo.)

Forecast: $m g = 40 N$ se zyada ya kam? Aur vertical cliff ( $9 0^{\circ}$ ) par $N$ kya hona chahiye?

Axes surface ke along/perpendicular choose karo (figure dekho). Yeh step kyun? Block sirf ramp ke along slide kar sakta hai, isliye motion wahan hoti hai, aur perpendicular wahan constraint hota hai. Axes tilting se constraint ek clean equation ban jaati hai.
Gravity resolve karo. Perpendicular part $= m g cos θ$ (ramp mein press karta hai); along-slope part $= m g sin θ$ (ramp neeche kheenchta hai). Perpendicular part ke liye $cos θ$ kyun? Jab ramp $θ$ se tilt hota hai, weight arrow surface normal se $θ$ door tilt hota hai, aur us chhote right triangle ka adjacent (perpendicular) side $cos θ$ se scale hota hai — figure mein red/mint split dekho.
Newton's 2nd law perpendicular, $a_{⊥} = 0$ (block surface par rehta hai). $N - m g cos θ = 0 \Rightarrow N = m g cos θ$ . $a_{⊥} = 0$ kyun? Block kabhi ramp mein dhoosta ya ramp se uchalta nahi, isliye surface mein uski acceleration zero hai.
Number: $N = 4 \times 10 \times 0.8 = 32 N$ .
Limit $θ = 0^{\circ}$ : $cos 0 = 1 \Rightarrow N = 40 N = m g$ — flat ground, wapas Cell A. ✔
Limit $θ = 9 0^{\circ}$ : $cos 9 0^{\circ} = 0 \Rightarrow N = 0$ — ek vertical wall jis par block seedha neeche slide karta hai, surface mein kuch press nahi karta. Kyun: gravity ab poori tarah surface ke along hai, usmein nahi.

Verify: $N = 32 N < 40 N$ ✔; flat limit $= 40$ ✔; vertical limit $= 0$ ✔.

Cell G — Vertical wall (word problem)

Ab surface ka normal horizontal hai, isliye gravity $N$ mein koi role nahi khelti. Yahi woh case hai jo log bhool jaate hain.

Worked example Ex 7 — Poster wall par pin karna

Tum ek $0.5 kg$ ka poster vertical wall ke against horizontal push $F = 6 N$ seedha wall mein daba ke press karte ho. Wall ki normal force $N$ poster par find karo.

Forecast: Kya $N$ poster ke weight par depend karta hai yahan?

Normal direction identify karo. Wall vertical hai, isliye uska normal horizontally outward point karta hai (figure dekho). Yeh step kyun? Normal force hamesha surface ke $⊥$ hoti hai — wall ke liye matlab sideways, upar nahi.
Newton's 2nd law horizontal (into-wall) axis ke along, $a_{x} = 0$ (poster wall mein/se accelerate nahi kar raha). $F - N = 0 \Rightarrow N = F$ . Yeh step kyun? Tumhari push aur wall ki push sirf horizontal forces hain; gravity vertical hai aur ek alag equation mein rehti hai.
Number: $N = 6 N$ . Gravity ( $m g = 5 N$ ) horizontal balance mein kabhi appear nahi karti. Kyun: yeh wall ke normal ke perpendicular hai, isliye $N$ affect nahi kar sakti.

Verify: $N = F = 6 N$ , $m g$ se independent ✔ — "normal $\neq = m g$ " ka sabse extreme lesson: yahan $N$ ka weight se koi lena-dena nahi.

Cell H — Incline jo khud accelerate karta hai (exam twist)

Worked example Ex 8 — Ramp par block jo sideways push ho raha hai

Ek $2 kg$ ka block angle $θ = 3 0^{\circ}$ ke frictionless wedge par rakha hai. Poora wedge horizontally push kiya jata hai toh block wedge ke saath $a = 5 m/s^{2}$ se accelerate karta hai, slope se pressed rehta hai. $N$ find karo. ( $cos 3 0^{\circ} = 0.866$ , $sin 3 0^{\circ} = 0.5$ use karo.)

Forecast: Static-ramp answer $m g cos θ = 17.3 N$ se zyada ya kam?

Along/perpendicular axes rakho, lekin $a_{⊥}$ ab zero nahi hai. Horizontal acceleration $a$ ka ek component surface mein hai. Yeh step kyun? Block wedge ke saath horizontally move karta hai; us horizontal motion ka kuch part slope mein directed hai, isliye $a_{⊥} \neq = 0$ — parent note ne yahi trap bataya tha.
Perpendicular components. Into-surface gravity part $= m g cos θ$ . Horizontal acceleration, inward normal par project hoke, surface ki taraf $m a sin θ$ contribute karta hai. $sin θ$ kyun? Normal vertical se $θ$ angle par point karta hai, isliye horizontal push usse $sin θ$ factor ke saath lagta hai.
Newton's 2nd law perpendicular. "Surface se bahar" positive leke: $N - m g cos θ - ma sin θ = 0$ , isliye $N = m (g cos θ + a sin θ)$ . Yeh step kyun: $N$ ko gravity balance aur required acceleration ka perpendicular part dono supply karna hai.
Number: $N = 2 \times (10 \times 0.866 + 5 \times 0.5) = 2 \times (8.66 + 2.5) = 2 \times 11.16 = 22.32 N$ .

Verify: $N = 22.32 N > 17.3 N$ ✔ — ramp ko block mein accelerate karna zyada press karta hai, aur $a = 0$ set karne se $m g cos θ = 17.32 N$ wapas milta hai ✔.

Recall Kya har cell cover hua?

Row A → Ex 1; B → Ex 2; C (lift-off ke saath) → Ex 3; D → Ex 4; E (free fall ke saath) → Ex 5; F ( $0^{\circ}$ / $9 0^{\circ}$ limits ke saath) → Ex 6; G (wall) → Ex 7; H (accelerated ramp) → Ex 8. Har sign, har zero, har limit dikhayi gayi hai.

Active Recall

Connections

Newton's Second Law — har row bas $\sum F_{⊥} = m a_{⊥}$ hai.
Apparent Weight & Elevators — Cells D aur E.
Inclined Plane Problems — Cells F aur H.
Free Body Diagrams — sabhi aath ke liye setup tool.
Newton's Third Law — har $N$ ka sahi reaction object ka surface par wapas push karna hai.
Friction — in $N$ values ke milne ke baad, $f = μ N$ follow karta hai.