Free body diagrams — systematic drawing technique: Exercises

1.2.4 · D4 · Physics › Newton's Laws & Dynamics › Free body diagrams — systematic drawing technique

Definition Is page par use ki gayi notation (ek baar padho)

$m g$ — weight, gravity ki kheench, hamesha seedha neeche.
$N$ — normal force, hamesha ⊥ (perpendicular) touching surface ke, usse door push karti hui.
$T$ — tension, rope ki kheench rope ke saath, body se door.
$F$ — ek applied push/pull jo problem deta hai.
$f_{k} = μ_{k} N$ — kinetic friction: sliding-friction force. Subscript $k$ ka matlab hai kinetic (chalte waqt). $μ_{k}$ coefficient of kinetic friction hai, ek pure number (koi units nahi) jo batata hai ki normal force ka kaunsa fraction drag ke roop mein wapas aata hai. Toh friction kabhi independent nahi hoti — ye $N$ se $f_{k} = μ_{k} N$ ke zariye bandi hoti hai (dekho Normal Force and Friction).
double-equal markup neeche ek key term highlight karta hai jo tumhe recall karni chahiye — mentally cover karo aur khud ko test karo.

Neeche use ki gayi difficulty ladder:

L1 Recognition — forces spot karo, koi algebra nahi.
L2 Application — ek axis equation mein plug in karo.
L3 Analysis — do axes, angles, ya friction.
L4 Synthesis — connected bodies, multiple FBDs.
L5 Mastery — sab kuch combine karo, edge cases.

Level 1 — Recognition

Exercise 1.2

Ek ball ek single vertical string se motionless hang karti hai. Ball par har force list karo.

Recall Solution 1.2

Gravity $m g$ neeche.
Contact: sirf string use touch karti hai → ==tension $T$ upar==, ball se string ke saath door point karta hua (strings sirf kheenchti hain).

Koi normal force nahi (koi cheez use push nahi karti), koi friction nahi. Kyunki ye still hang karti hai, $T = m g$ .

Level 2 — Application

Exercise 2.1

Ek $4.0 kg$ ka box floor par baitha hai. Ek insaan rope se seedha upar pull karta hai, tension $T = 25 N$ , lekin box lift off nahi karta. Floor se normal force $N$ find karo.

Recall Solution 2.1

Box par forces (sirf vertical): gravity $m g$ neeche, tension $T$ upar, normal $N$ upar. Box floor par rehta hai ⇒ $a_{y} = 0$ , toh $\sum F_{y} = 0$ : $N + T - m g = 0 \Rightarrow N = m g - T .$ $N = (4.0) (9.8) - 25 = 39.2 - 25 = 14.2 N .$ Upar ki pull box ko hold karne ka kaam share karti hai, toh floor poore weight se kam push karta hai. Sanity check: agar $T$ $39.2 N$ tak pahunche, toh $N = 0$ aur box floor chhodne wala hai. ✓

Exercise 2.2

Ek $2.0 kg$ block frictionless horizontal table par $F = 6.0 N$ se horizontally push kiya gaya hai. Iska acceleration find karo.

Recall Solution 2.2

Horizontal axis: sirf horizontal force $F$ hai. Newton's Second Law se, $F = ma \Rightarrow a = \frac{F}{m} = \frac{6.0}{2.0} = 3.0 m/s^{2},$ $F$ ki direction mein. Vertically $N = m g = 19.6 N$ , lekin koi role nahi kyunki $a_{y} = 0$ .

Level 3 — Analysis

Exercise 3.1

Ek $5.0 kg$ block $3 0^{\circ}$ frictionless incline par baitha hai. (a) normal force aur (b) slope ke neeche acceleration find karo. $g = 9.8 m/s^{2}$ use karo.

Recall Solution 3.1

Axes tilt karo: $x$ down-slope, $y$ ⊥ slope ke (dekho Inclined Plane Problems). Gravity seedha neeche point karti hai, lekin hamare dono tilted axes mein se koi bhi vertical nahi hai, isliye gravity ka ek piece har axis ke saath hoga — hume resolve karna hoga. Kyun $sin$ aur $cos$ aate hain (figure dekho): gravity ka arrow dono axes par drop karo aur ye ek right triangle banata hai jiska acute angle slope angle $θ$ ke barabar hota hai (incline angle vertical aur perpendicular-to-slope direction ke beech wapas aata hai). Us triangle mein:

slope ke saath wali side opposite hai $θ$ ke, aur opposite/hypotenuse $= sin θ$ , deta hai $m g sin θ$ ;
surface mein wali side adjacent hai $θ$ ke, aur adjacent/hypotenuse $= cos θ$ , deta hai $m g cos θ$ .

Kaun sa kaunsa hai iska sanity check: jaise $θ \to 9 0^{\circ}$ (vertical wall) slope vertical ho jaata hai, toh gravity ka sab slope ke saath act karna chahiye — aur sach mein $sin 9 0^{\circ} = 1$ hai. ✓

Figure — Free body diagrams — systematic drawing technique

Figure padhna: solid orange arrow poora weight

m g

hai; do dashed orange arrows iske resolved pieces hain

m g sin θ

(down-slope) aur

m g cos θ

(surface mein); teal arrow

N

surface se bahar point karta hai, exactly

m g cos θ

ko balance karta hua. Plum arc

θ = 3 0^{\circ}

mark karta hai.

(a) Surface se koi motion nahi ⇒ $a_{y} = 0$ : $N - m g cos θ = 0 \Rightarrow N = m g cos θ = (5.0) (9.8) cos 3 0^{\circ} = 42.44 N .$ (b) Slope ke saath: $m g sin θ = ma \Rightarrow a = g sin θ = 9.8 sin 3 0^{\circ} = 4.9 m/s^{2} .$ Mass cancel ho jaata hai — bhaare blocks same $a$ se slide karte hain.

Exercise 3.2

Wohi $5.0 kg$ block wohi $3 0^{\circ}$ incline par, ab friction ke saath. Ye constant velocity se neeche slide karta hai. Friction force aur kinetic friction coefficient $μ_{k}$ find karo.

Recall Solution 3.2

Constant velocity ⇒ $a = 0$ ⇒ forces dono axes par balance karte hain. Yahan $f_{k}$ page ke upar define ki gayi kinetic friction force hai, surface se $f_{k} = μ_{k} N$ se bandi hui.

$y : N = m g cos θ = 42.44 N$ (Ex 3.1 jaisa hi perpendicular balance).
$x :$ block neeche slide karta hai, toh kinetic friction up-slope act karti hai, slip ko oppose karti hui: $m g sin θ - f_{k} = 0 \Rightarrow f_{k} = m g sin θ = (5.0) (9.8) sin 3 0^{\circ} = 24.5 N .$ Phir, $f_{k} = μ_{k} N$ use karte hue (toh $μ_{k} = f_{k} / N$ ): $μ_{k} = \frac{f _{k}}{N} = \frac{24.5}{42.44} = 0.577.$ Note karo $μ_{k} = \frac{m g sin θ}{m g cos θ} = tan θ$ exactly constant velocity par — ek classic result.

Level 4 — Synthesis

Exercise 4.1

Block $A$ ( $m_{1} = 3.0 kg$ ) frictionless table par ek light string se frictionless pulley ke upar hanging block $B$ ( $m_{2} = 2.0 kg$ ) se connected hai. Acceleration aur string tension find karo.

Recall Solution 4.1

Do FBDs — ek har body ke liye (Tension in Strings and Pulleys). Same $T$ (ideal string), same $∣ a ∣$ (inextensible).

$A$ (horizontal): $T = m_{1} a$ .
$B$ (vertical, neeche ko positive lete hue): $m_{2} g - T = m_{2} a$ .

$T$ cancel karne ke liye add karo: $m_{2} g = (m_{1} + m_{2}) a \Rightarrow a = \frac{m _{2}}{m _{1} + m _{2}} g = \frac{2.0}{5.0} (9.8) = 3.92 m/s^{2} .$ Phir $T = m_{1} a = (3.0) (3.92) = 11.76 N$ . Check: $T < m_{2} g = 19.6 N$ — string ko $B$ ke weight se kam pull karna chahiye, warna $B$ nahi gir sakta. ✓

Exercise 4.2

4.1 jaisa hi setup, lekin table mein block $A$ ke neeche friction $μ_{k} = 0.20$ hai. Naya acceleration find karo.

Recall Solution 4.2

Ab $A$ ko kinetic friction apni motion oppose karte hue feel hoti hai (ye pulley ki taraf move karta hai, toh friction peeche act karti hai):

$A$ vertical: $N = m_{1} g = (3.0) (9.8) = 29.4 N$ , toh $f_{k} = μ_{k} N = 0.20 (29.4) = 5.88 N$ .
$A$ horizontal: $T - f_{k} = m_{1} a$ .
$B$ : $m_{2} g - T = m_{2} a$ .

$T$ cancel karne ke liye do driving equations add karo: $m_{2} g - f_{k} = (m_{1} + m_{2}) a \Rightarrow a = \frac{m _{2} g - μ _{k} m _{1} g}{m _{1} + m _{2}} .$ $a = \frac{19.6 - 5.88}{5.0} = \frac{13.72}{5.0} = 2.744 m/s^{2} .$ Friction ne drive ka kuch hissa chura liya, toh $a$ $3.92$ se $2.744 m/s^{2}$ tak drop ho gaya. ✓

Level 5 — Mastery

Exercise 5.1

$3 0^{\circ}$ incline par do blocks: block $C$ ( $m_{C} = 4.0 kg$ , neeche) aur block $D$ ( $m_{D} = 2.0 kg$ , upar) contact mein hain, ek force $F = 60 N$ jo $D$ par slope ke upar direction mein apply hoti hai unhe saath push karti hai. Incline frictionless hai. (a) common acceleration aur (b) $C$ aur $D$ ke beech contact force find karo. $g = 9.8 m/s^{2}$ use karo.

Recall Solution 5.1

Sign convention: up-slope ko positive $x$ choose karo (push blocks ko jis direction drive karta hai); down-slope negative hai. Gravity ka along-slope piece $m g sin θ$ isliye minus sign ke saath enter karta hai, exactly Ex 3.1 mein picture ki tarah.

$a$ find karne ke liye dono blocks ko ek system treat karo ( $C$ – $D$ contact force internal hai, toh drop ho jaati hai): $F - (m_{C} + m_{D}) g sin θ = (m_{C} + m_{D}) a .$ $a = \frac{F}{m _{C} + m _{D}} - g sin θ = \frac{60}{6.0} - 9.8 sin 3 0^{\circ} = 10 - 4.9 = 5.1 m/s^{2} (up-slope, positive) .$ (b) Ab $C$ (lower block) ko isolate karo. Newton's Third Law se, upper block $D$ $C$ ko slope ke upar push karta hai (ye $C$ ko motion ki direction mein aage shove kar raha hai), toh contact force $P$ $C$ par positive $x$ direction mein point karti hai — isliye $P$ up-slope positive liya jaata hai. Gravity phir bhi $C$ ko down-slope kheenchti hai ( $- m_{C} g sin θ$ ). $C$ par $x$ ke saath Newton's 2nd law: $P - m_{C} g sin θ = m_{C} a \Rightarrow P = m_{C} (a + g sin θ) = 4.0 (5.1 + 4.9) = 40.0 N .$ $D$ ko isolate karke check karo: $F - P - m_{D} g sin θ = m_{D} a \Rightarrow 60 - 40 - 2.0 (4.9) = 20 - 9.8 = 10.2 = m_{D} a = 2.0 (5.1)$ . ✓

Exercise 5.2 (edge case)

Ek $1.0 kg$ block Ex 5.1 ke frictionless $3 0^{\circ}$ incline par hai, lekin $F$ tab tak reduce ki jaati hai jab tak block wapas neeche slide karne ke verge par na ho. Sabse chota $F$ (phir bhi up-slope) kya hai jo ise neeche accelerate hone se rokta hai, aur $N$ kya hai?

Recall Solution 5.2

"Verge, frictionless" ⇒ sirf slope ke saath balance karo, $a = 0$ (up-slope positive): $F - m g sin θ = 0 \Rightarrow F = m g sin θ = (1.0) (9.8) sin 3 0^{\circ} = 4.9 N .$ Perpendicular axis $F$ se untouched hai (ye slope ke saath hai), toh Ex 3.1 ki tarah: $N = m g cos θ = (1.0) (9.8) cos 3 0^{\circ} = 8.487 N .$ Koi bhi $F < 4.9 N$ aur block neeche accelerate karta hai; koi bhi $F > 4.9 N$ aur ye upar accelerate karta hai. Ye threshold exactly down-slope gravity component hai. ✓

Active Recall

Isolated-block Atwood:

a < g

hamesha kyun hota hai?

Hanging weight

m_{2} g

ko total mass

m_{1} + m_{2}

accelerate karni hoti hai, toh

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

Ex 4.2 mein table par friction add karne se

a

kyun kam hoti hai?

Kinetic friction

f_{k} = μ_{k} N

motion oppose karti hai, driving weight se subtract karti hai pehle ki wo same total mass accelerate kare.

Exercises — Free body diagrams — systematic drawing technique