Free body diagrams — systematic drawing technique: Question bank

1.2.4 · D5 · Physics › Newton's Laws & Dynamics › Free body diagrams — systematic drawing technique

Intuition Is page ko kaise use karein

Ye conceptual traps hain, number-crunching nahi. Har answer ko cover karo, apni reasoning zabar-dast aawaz mein bolo pehle reveal karne se. Aawaz mein kyun? Silent reading mein brain ek plausible-lagta answer dekh ke sirf "haan" bol deta hai bina test kiye — iska naam hai "illusion of knowing." Because bolne se tumhe poori logic chain commit karni padti hai, jo exact broken link expose karta hai. Example: "kya $N$ hamesha $m g$ hota hai?" ke liye sirf "nahi" mat socho — kaho "Nahi, kyunki $N$ sirf tab $m g$ ke barabar hota hai jab surface flat ho aur koi vertical pull na ho; slope par $N = m g cos θ$ hota hai." Agar us sentence ka koi bhi word tumhe hesitate karaaye, wahi hesitation woh trap hai jo tumhe dhundhna tha. Toh scoring rule ye hai: agar tumhara yes/no sahi tha lekin spoken reason galat ya missing tha, toh use miss count karo — reasoning hi sab kuch hai. Ye deep dive parent Free body diagrams — systematic drawing technique ko extend karta hai, aur Newton's Second Law, Newton's Third Law, Normal Force and Friction, Tension in Strings and Pulleys, Inclined Plane Problems aur Resolving Vectors into Components pe lean karta hai.

Definition Symbol legend — questions se pehle padho

Neeche ke har trap mein same standard symbols reuse hote hain. Ye list saamne rakho:

$m$ — chosen body ki mass, kilograms mein.
$g$ — Earth ke paas gravitational field strength, lagbhag $9.8 m/s^{2}$ ; ye mass ko weight mein convert karta hai.
$θ$ (theta) — incline angle, yaani sloped surface aur horizontal ground ke beech ka angle. Flat ground par $θ = 0^{\circ}$ ; vertical wall $θ = 9 0^{\circ}$ hoti hai.
$N$ — normal force, woh push jo ek surface apne aap ke perpendicular direction mein deti hai.
$T$ — tension, woh pull jo ek string apni length ke saath, body se door direction mein lagati hai.
$μ_{k}$ (mu-k) — coefficient of kinetic friction, woh fraction jo sliding friction ko normal force se jodata hai: $f_{k} = μ_{k} N$ .
$a$ — body ki acceleration (velocity kitni tezi se change hoti hai).
$m g$ — weight, gravity ka body par neeche ki taraf pull.

Figure — Free body diagrams — systematic drawing technique

True or false — justify

Har ek True or False se shuru hota hai, aur answer mein because dena zaroori hai.

Normal force hamesha

m g

ke barabar hoti hai.

False — ye sirf tab valid hai jab surface horizontal ho aur koi vertical push na ho. Incline par

N = m g cos θ

hota hai (jahan

θ

surface-to-horizontal angle hai), aur agar tum block ko neeche dabaao toh

N

badhta hai, agar upar khicho toh ghatta hai. Dekho Normal Force and Friction.

Normal force aur weight ek Newton's-third-law action–reaction pair hain.

False — dono ek hi body (block) par act karte hain, isliye ye 3rd-law pair nahi ho sakte. Weight ka pair hai block ka Earth ko upar khichna;

N

ka pair hai block ka surface ko neeche push karna. Dekho Newton's Third Law.

Agar kisi body par net force zero ho, toh body zaroor rest mein hogi.

False — zero net force ka matlab hai zero acceleration (

a = 0

), yaani constant velocity. Ek body jo steady speed se chal rahi hai uska FBD perfectly balanced hota hai.

Free body diagram mein object ka velocity arrow bhi hona chahiye.

False — FBD sirf forces dikhata hai. Velocity motion hai, koi push ya pull nahi, isliye ye kabhi appear nahi hota; ise draw karna speed aur force ko confuse karne ki invitation deta hai.

Ek massless string jo frictionless pulley par chalti hai, usmein tension poori length par same rehti hai.

True — ek ideal (massless, inextensible) string ek frictionless pulley par same

T

transmit karti hai har jagah, kyunki string ke saath accelerate karne ke liye koi mass nahi hai aur force bleed karne ke liye koi friction nahi hai. Dekho Tension in Strings and Pulleys.

Frictionless incline par ek bhaari block ek halke block se zyaada tezi se slide karta hai.

False —

a = g sin θ

mein mass hai hi nahi; mass bilkul waise cancel ho jaata hai jaise free fall mein, isliye dono same acceleration se slide karte hain.

Friction hamesha object ki motion ko oppose karta hai.

False — friction relative slipping (ya uski tendency) ko oppose karta hai. Ye motion ko drive bhi kar sakta hai: road se friction hi hai jo ek walking foot ya driving tyre ko aage push karta hai.

Tum wall par jo force lagate ho woh force wall tumhare par laati hai uske barabar hoti hai.

True — ye Newton's 3rd law hai: dono magnitude mein equal hain, direction mein opposite hain, aur alag bodies par act karte hain (tum aur wall). Dekho Newton's Third Law.

Kyunki slope par ek block ruka rehta hai, toh koi force hogi jo use slope ke upar

m g sin θ

ke barabar push kar rahi hogi.

True agar ye static hai — woh up-slope force static friction hai, ek real contact force, koi phantom nahi. Ye

m g sin θ

ke barabar tab hi hoti hai jab block slip na kare.

Zyaada applied force ka matlab hamesha zyaada friction force hota hai.

False — kinetic friction slide shuru hone ke baad roughly constant rehti hai (

μ_{k} N

); zyaada push karna sirf acceleration badhata hai, friction nahi. Static friction tumhare push ke saath zaroor badhti hai, lekin sirf apni maximum tak.

Spot the error

Galati kya hai aur use kaise theek karein. (Har ek ko upar ke reference FBDs se compare karo.)

"Maine ek

ma

arrow acceleration ke along draw kiya."

ma

forces ka result hai, force nahi hai. Ye

\sum F = ma

ke right side par belong karta hai, diagram par arrow ke roop mein kabhi nahi. Dekho Newton's Second Law.

"Incline par maine normal force ko seedha upar draw kiya."

Normal force surface ke perpendicular hoti hai, isliye slope par ye vertical se tilted hoti hai (Figure 1 ka correct panel dekho). Ye seedha upar tab hi point karta hai jab surface horizontal ho.

"Hanging block ke liye maine tension ko neeche ki taraf neeche wale block ki taraf draw kiya."

String mein tension hamesha body se door, rope ke along khichti hai. Ek hanging block par string upar jaati hai, isliye

T

upar point karta hai, neeche nahi (Figure 2).

"Maine block ka weight aur table par block ka push, dono block ke FBD mein daale."

Table par block ka push table par act karta hai, block par nahi. Sirf woh forces jo chosen body par act karti hain, uske FBD par aati hain.

"Do blocks ek string se jude hain, toh maine dono ko alag tension di."

Ek ideal string ka matlab ek tension hai. Unhe alag

T

dena massless-string assumption ko todta hai aur unknowns over-count karta hai (Figure 2 shared

T

dikhata hai).

"Kitaab table par ruki hai, toh normal force use upar push karti hai aur table ka weight bhi use thaame rakhta hai."

Table ka weight table par act karta hai, kitaab par nahi. Kitaab ke FBD mein sirf uska apna weight neeche aur table ki normal force upar hoti hai.

"Incline par maine gravity ko

m g cos θ

down-slope aur

m g sin θ

into-surface resolve kiya."

Functions swap ho gaye hain. Down-slope (along) component

m g sin θ

hota hai; into-surface (perpendicular) component

m g cos θ

hota hai. Dekho Resolving Vectors into Components.

"Object daayein move kar raha hai, toh maine ek daayein taraf 'force of motion' arrow add kiya."

Koi "force of motion" nahi hoti. Ek body bina kisi forward force ke chalti rehti hai (inertia); sirf real pushes/pulls FBD par jaate hain (Figure 1 ka wrong panel dekho).

Why questions

Sirf fact nahi, reasoning explain karo.

Recipe mein pehle gravity kyun draw karte hain?

Ye ek long-range force hai jo Earth ke paas hamesha present rehti hai, contact se independent. Ise pehle draw karna guarantee karta hai ki ye kabhi bhula na jaye.

Har touching surface par exactly ek set of contact forces kyun hoti hain?

Objects sirf wahan contact forces exchange karte hain jahan wo physically touch karte hain. Boundary ko touches ke liye scan karna har contact force dhundhta hai aur koi nayi invent nahi karta.

Incline par axes ko horizontal/vertical rakhne ki jagah tilt kyun karte hain?

Taaki acceleration poori tarah ek axis par ho aur doosre par zero ho, ek unknown khatam ho aur equations simplify ho jayein. Dekho Inclined Plane Problems.

a = g sin θ

mein mass cancel kyun ho jaata hai?

Driving force

m g sin θ

mass ke saath badhti hai, lekin inertia

m

bhi jo ise resist karta hai badhti hai. Ek ko doosre se divide karne par

m

poori tarah hat jaata hai.

Action aur reaction dono ek hi FBD par kabhi kyun nahi aane chahiye?

3rd-law pair do alag bodies par act karta hai. Ek FBD ek body ko isolate karta hai, isliye kisi bhi pair ka zyaada se zyaada ek member us par belong karta hai. Dekho Newton's Third Law.

Pulley system ki do equations add karne par tension kyun cancel ho jaati hai?

Tension dono bodies ki equations mein opposite signs ke saath appear hoti hai (ye dono ko pulley ki taraf khinchti hai), isliye summing karne par ye cancel ho jaati hai, sirf external forces aur total mass bachte hain.

Normal force apni size apne aap kyun adjust karti hai?

Ye ek passive constraint force hai: ye utni ho jaati hai jitni zaroorat ho surfaces ko interpenetrate hone se rokne ke liye, isliye ye weight, applied pushes, aur incline angle ke respond karti hai.

Edge cases

Boundary aur degenerate situations jo ye topic invite karta hai.

Jab incline angle

9 0^{\circ}

(ek vertical wall) ho jaata hai toh normal force kya hoti hai?

N = m g cos 9 0^{\circ} = 0

. Surface ab block ko gravity ke against support nahi karti, isliye block effectively wall ke saath free fall mein hota hai.

Jab incline angle

0^{\circ}

(flat ground) ho toh down-slope acceleration kya hoti hai?

a = g sin 0^{\circ} = 0

. Level ground par gravity ka koi along-surface component nahi hota, isliye kuch bhi block ko drive nahi karta.

Pulley system mein, agar hanging mass

m_{2} \to 0

ho toh acceleration ka kya hoga?

a = \frac{m _{2}}{m _{1} + m _{2}} g \to 0

. Koi hanging weight nahi toh system ko pull karne wala kuch nahi, isliye ye rest mein rehta hai.

Agar table mass

m_{1} \to 0

ho toh acceleration ka kya hoga?

a = \frac{m _{2}}{m _{1} + m _{2}} g \to g

. Table par negligible mass ke saath, hanging block essentially free fall mein hai, ek weightless partner ko drag karta hua.

Agar ek block table par baitha hai aur tum use upar ki taraf exactly

m g

ke barabar force se khicho, toh normal force kya hogi?

N = 0

— upward pull already gravity ko balance kar deta hai, isliye table ko push karne ki zaroorat hi nahi. Thoda aur khicho aur block lift off kar lega (

N

negative nahi ho sakta).

Free fall mein ek block (koi surface nahi, koi string nahi) — uske FBD par kitni forces hain?

Exactly ek: gravity

m g

neeche. Kuch bhi use touch nahi karta, isliye koi contact forces bilkul nahi hain.

Ek box rough floor par constant velocity se push kiya ja raha hai — kya FBD balanced hai?

Haan — constant velocity ka matlab zero acceleration hai, isliye applied force kinetic friction ke barabar hai aur normal weight ke barabar hai. Balanced forces, lekin box phir bhi move kar raha hai.