Normal force — reaction force, not always = mg: Question bank

1.2.5 · D5 · Physics › Newton's Laws & Dynamics › Normal force — reaction force, not always = mg

Traps try karne se pehle, humein notation aur conventions pin down karni hongi — in mein se zyaadaatar traps bilkul isi step ko skip karne se aate hain.

Neeche figure dekho: yeh teen geometries dikhata hai jismein ye traps rehte hain — flat floor, incline, aur elevator — har ek mein positive $⊥$ axis drawn hai, taaki tumhe hamesha pata rahe ki "positive" ka matlab kya hai.

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

Agla figure woh result derive karta hai jo students sabse zyada bina dikhaye assert karte hain — incline par $N = m g cos θ$ — gravity ko uske perpendicular aur along-slope parts mein decompose karke. Ise incline traps ke liye saamne rakhna.

Aur yeh figure elevator case ko teeno sign scenarios ke saath dikhata hai ( $a$ upar, $a = 0$ , $a$ neeche), taaki tum dekh sako kyun $a$ apni sign ke saath $\sum F_{⊥} = m a_{⊥}$ mein enter karta hai.

Teen anchors yaad rakhne ke liye:

Direction: $N$ hamesha surface ke perpendicular hota hai, surface se door object ki taraf point karta hua.
Value: surface-normal axis ke saath forces resolve karke aur Newton's Second Law apply karke nikali jaati hai.
Reaction pairing: $N$ ka asli partner alag body par act karta hai — dekho Newton's Third Law.

Neeche "push down" items mein, $F$ ka matlab hai ek baahri push ki magnitude $F$ jo vertically neeche ki taraf directed hai ek block ke upar par, jo horizontal floor par rakh hua hai.

True ya false — justify karo

Har item: true/false decide karo, phir physics do.

TF — Kisi bhi surface par, normal force object ke weight ke barabar hoti hai.

False.

N = m g

sirf horizontal surface par hota hai jab koi aur vertical forces na hon aur koi vertical acceleration na ho; general rule hai

\sum F_{⊥} = m a_{⊥}

TF — Normal force hamesha seedha upar ki taraf directed hoti hai.

False.

N

surface ke perpendicular hota hai, ground ke nahi — incline par yeh tilt karta hai, aur ceiling par yeh neeche point karta hai.

TF — Ek heavier object hamesha same surface se zyada normal force feel karta hai.

Generally False. Incline par

N = m g cos θ

, aur agar geometry ya acceleration change ho, toh mass akela

N

decide nahi karta; sirf

\sum F_{⊥} = m a_{⊥}

karta hai.

TF —

N

aur weight

m g

flat floor par Newton's-third-law action–reaction pair banate hain.

False. Dono same body (box) par act karte hain; 3rd-law pair alag bodies par act karna chahiye. Dekho Newton's Third Law.

TF — Floor ka box par normal force ka reaction, box ka floor ko neeche push karna hai.

True. Same magnitude, opposite direction, doosri body (floor) par act karta hua — yahi genuine 3rd-law partner hai.

TF — Free fall mein object abhi bhi neeche floor par press karta hai.

False. Positive

⊥

upar ke saath aur

a_{⊥} = - g

\sum F_{⊥} = N - m g = - m g

N = 0

milta hai — kuch bhi pressed nahi, isliye weightlessness (dekho Apparent Weight & Elevators).

TF — Resting block par neeche zyada push karna (extra downward force

F

) normal force badhata hai.

True. Positive

⊥

upar ke saath,

N - m g - F = 0

N = m g + F

milta hai; surface

a_{⊥} = 0

rakhne ke liye zyada upward push supply karta hai.

TF — Frictionless incline par, normal force slope ke neeche acceleration provide karta hai.

False.

N

slope ke perpendicular hai; down-slope acceleration

m g sin θ

se aata hai, jo surface ke saath ek force hai.

TF — Agar ek block incline par bina slide kiye baitha hai, toh ise rokne ke liye

N

flat se zyada hona chahiye.

False.

N = m g cos θ < m g

; friction (bada

N

nahi) hi ise rokta hai. Dekho Friction.

TF — Incline angle ko

3 0^{\circ}

6 0^{\circ}

par double karna normal force ko half kar deta hai.

False.

N = m g cos θ

;

cos 6 0^{\circ} / cos 3 0^{\circ} = 0.5/0.866 \approx 0.577

0.5

nahi —

cos

θ

mein linear nahi hai.

TF — Vertical wall par, ek leaning object par normal force zero ho sakti hai chahe gravity act kar rahi ho.

True. Ek bilkul vertical wall sirf horizontally push karta hai; agar kuch object ko wall mein push nahi karta, toh

N = 0

weight chahe kuch bhi ho.

Error pakdo

Har item ek galat reasoning line state karta hai. Batao kya toot gaya.

"Elevator upar ja raha hai, isliye

N = m (g + a)

Error hai velocity aur acceleration ko confuse karna.

N

set karta hai

a_{⊥}

ki sign, motion ki direction nahi — ek elevator jo upar ja raha ho lekin decelerate ho raha ho uska

N = m (g - a)

hoga.

"Gravity block ko incline par

m g

force se push karta hai, isliye

N = m g

Gravity ka sirf woh component jo surface ke perpendicular ho surface mein press karta hai:

m g cos θ

. Baaki,

m g sin θ

, slope ke saath act karta hai aur

N

ko use balance nahi karna hota.

"Slope par,

N = m g sin θ

kyunki slope angle

θ

banata hai."

Galat component.

m g

ka perpendicular projection block ke angle mein adjacent side use karta hai,

cos θ

deta hai;

sin θ

along-slope part hai.

"Kyunki

N

m g

ko balance karta hai, resting box par net force

N + m g

hai."

Balance karne ka matlab hai wo oppose karte hain: positive

⊥

upar ke saath, net force

N - m g = 0

hai,

N + m g

nahi. Magnitudes add karna direction ignore karta hai.

"Elevator floor par box ka

N = m g

hai kyunki box slide nahi kar raha."

Slide na karna horizontal/along-surface motion ke baare mein hai;

N

vertical acceleration

a_{⊥}

se set hota hai. Agar elevator accelerate kare, toh

N = m (g \pm a) \neq = m g

chahe slide na ho.

"Ceiling ke against rakha object ka upward normal force hai kyunki

N

upar point karta hai."

N

surface se door object ki taraf point karta hai, isliye ceiling object ko neeche push karti hai. Yeh kabhi "upar" fix nahi hota.

"Free fall mein

N = 0

, isliye gravity bhi zero ho jaati hai."

Free fall mein gravity

m g

unchanged rehti hai; sirf contact force

N

khatam hota hai. Felt weightlessness

N

ki absence hai, gravity ki nahi.

Why questions

Normal force surface ke perpendicular kyun point karta hai aur kisi aur direction mein kyun nahi?

Ek smooth solid surface sirf penetration resist kar sakti hai, jo perpendicular direction hai; koi bhi along-surface resistance alag force (friction) hai, isliye

N

purely perpendicular hai.

N

ko "constraint force" kyun kaha jaata hai instead of fixed formula dene ke?

Yeh exactly woh value leta hai jo constraint enforce karne ke liye zaroori hai "surface mein sink mat ho," isliye iska magnitude solve kiya jaata hai Newton's Second Law se, lookup nahi kiya jaata.

Elevator mein upar accelerate karne par floor zyada hard kyun push karta hai?

Upward acceleration produce karne ke liye net upward force positive honi chahiye, isliye

⊥

axis ke saath

N

m g

se zyada hona chahiye:

N = m (g + a)

— extra push hi tumhe accelerate karta hai.

N = m g cos θ

incline steep hone par kyun shrink karta hai?

Jaise

θ \to 9 0^{\circ}

, gravity ka less hissa surface mein point karta hai aur zyada along it, isliye

cos θ \to 0

aur surface ke paas almost kuch bhi perpendicular resist karne ke liye nahi hota. Dekho Inclined Plane Problems.

N

gravity ka third-law reaction kyun nahi hai, chahe wo equal aur opposite dikhte hain?

Third-law pairs alag bodies par act karte hain; dono

N

aur

m g

same box par act karte hain, isliye wo sirf do forces hain jo balance karte hain, coincidentally equal sirf flat, non-accelerating case mein.

Agar tumhara actual weight

m g

nahi badalta toh elevator mein "weight feel karna" kyun change hota hai?

Jo tum feel karte ho woh floor ka push

N

hai tumhare pair par, gravity nahi; kyunki

N = m (g \pm a)

acceleration ke saath vary karta hai, tumhara apparent weight change hota hai jabki

m g

fixed rehta hai.

Free body diagram draw karna

\sum F_{⊥}

likhne se pehle zyaadaatar

N

mistakes kyun prevent karta hai?

Yeh tumhe har contact aur field force list karne, surface-normal axis choose karne, aur positive direction fix karne par force karta hai, toh tum galti se

N = m g

assume nahi kar sakte ya extra push drop nahi kar sakte.

Edge cases

Ek object surface par rest karta hai lekin kuch use neeche press nahi karta (jaise floating just touching).

N

kya hai?

N = 0

. Normal force tab aata hai jab constraint active hoti hai; bina press kiye touch karne ke liye koi push zaroori nahi.

Ek block elevator floor par exactly

a = g

(free fall) neeche accelerate karta hai.

N

kya hai?

N = 0

: positive

⊥

upar ke saath,

\sum F_{⊥} = N - m g = m (- g)

N = 0

milta hai — Apparent Weight & Elevators mein classic weightless case.

Ek elevator

a > g

neeche accelerate karti hai (cable ise gravity se tez neeche kheencha). Box ko floor par rakhne ke liye kya hona chahiye?

Plain floor pull nahi kar sakta, isliye

N

negative nahi ho sakta — box floor se lift off ho jaayega unless kuch use clamp kare;

N = 0

aur box separate ho jaata hai.

Vertical wall par (

θ = 9 0^{\circ}

N = m g cos θ

kya deta hai aur kya yeh sense banata hai?

cos 9 0^{\circ} = 0

isliye

N = 0

— sahi hai, kyunki vertical wall koi vertical support nahi deti; wahan koi bhi

N

wall mein horizontal push se aana chahiye, jo alag setup hai.

Bilkul flat ground par (

θ = 0

), kya

N = m g cos θ

sahi reduce karta hai?

Haan:

cos 0^{\circ} = 1

isliye

N = m g

, flat-ground result ko special case ke roop mein recover karta hai jaise hamesha tha.

Kya normal force kabhi object ko surface ki taraf pull kar sakti hai (negative ho sakti hai)?

Nahi. Surface sirf push kar sakti hai, isliye

N \geq 0

; tumhare algebra mein "negative

N

" ka matlab hai object already surface chhod chuka hai aur equation ab apply nahi hoti.

Curved ya bumpy (non-planar) surface par, jahaan single tilt angle

θ

poori surface describe nahi karta, hum abhi bhi

N

kaise nikalte hain?

Har contact point par

N

local tangent plane ke perpendicular hota hai (flat surface jo wahan curve ko just touch karti hai); tum

\sum F_{⊥} = m a_{⊥}

apply karte ho us local normal ke saath, isliye single-angle formula sirf flat-tangent special case hai.

Do identical boxes floor par stack hain. Floor bottom box par kya normal force exert karta hai?

N_{floor} = 2 m g

(yeh dono weights support karta hai), jabki bottom box top box par

N = m g

upar exert karta hai — har surface apna khud ka

\sum F_{⊥}

solve karta hai.

Connections

Newton's Second Law — woh equation jo $N$ produce karta hai
Newton's Third Law — $N$ ke true reaction partner ko untangle karta hai
Friction — $N$ par depend karta hai $f = μ N$ ke zariye, isliye ye traps propagate hote hain
Inclined Plane Problems — $cos θ$ trap ka ghar
Apparent Weight & Elevators — $m (g \pm a)$ traps ka ghar
Free Body Diagrams — woh tool jo yahan har trap defuse karta hai