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

Free-body diagram (kitaab par arrows):

• $\uparrow N$ (table upar push karti hai)
• $\downarrow mg$ (Earth neeche kheenchti hai)

KAUNSA case: horizontal surface, koi extra vertical force nahi, koi vertical acceleration nahi. KYU: sirf vertical forces hain gravity $mg$ neeche aur $N$ upar, aur kitaab rest mein hai isliye $a_y=0$. $\sum F_y=N-mg=0\Rightarrow N=mg=4\times 10=\boxed{40\text{N}}$ Yeh woh ek case hai jahan $N=mg$ hota hai.

Free-body diagram (person par arrows):

• $\uparrow N$ (floor upar push karta hai)
• $\downarrow mg$ (Earth neeche kheenchti hai)

$N=m(g+a)$ kahan se aata hai? Upar ko positive lo aur $a$ ko person ka upward acceleration lo (toh downward acceleration $a$ ko negative banata hai). Sirf do vertical forces hain, $N$ upar aur $mg$ neeche, isliye Newton ka 2nd law likhta hai $\sum F_y=N-mg=ma.$ Dono sides mein $mg$ add karo: $N=mg+ma=m(g+a).$ Yahi poora derivation hai — elevator ka "formula" sirf $\sum F=ma$ rearrange kiya hua hai. Ab har case padho:

• (a) rest mein, $a=0$: $N=mg$ — barabar.
• (b) upar accelerate, $a>0$: $N=m(g+a)$ — zyada (tumhe heavy feel hota hai).
• (c) neeche accelerate, $a<0$: $N=m(g+a)<mg$ — kam (tumhe light feel hota hai).
• (d) free fall, $a=-g$: $N=m(g-g)=0$ — zero, weightless.

$N$ hamesha surface ke perpendicular hota hai, surface se bahar object ki taraf point karta hua. Ek ramp par yeh seedha upar nahi hota — yeh ramp ki tarah $25^{\circ }$ jhukta hai. Seedha upar gravity ki direction hai, $N$ ki nahi.

Free-body diagram (crate par arrows):

• $\uparrow N$ (floor upar push karta hai)
• $\downarrow mg=60\text{N}$ (Earth neeche kheenchti hai)
• $\downarrow F=25\text{N}$ (tumhara haath neeche press karta hai)

Koi vertical acceleration nahi, isliye up-arrow ko dono down-arrows cancel karne chahiye. $\sum F_y=N-mg-F=0\Rightarrow N=mg+F=60+25=\boxed{85\text{N}}$ Floor $mg$ se zyada push back karta hai — use gravity aur tumhari push dono cancel karni hai.

Free-body diagram (crate par arrows):

• $\uparrow N$ (floor upar push karta hai)
• $\uparrow T=20\text{N}$ (rope upar kheenchta hai)
• $\downarrow mg=60\text{N}$ (Earth neeche kheenchti hai)

Ab do up-arrows gravity balance karne ka kaam share karte hain. $\sum F_y=N+T-mg=0\Rightarrow N=mg-T=60-20=\boxed{40\text{N}}$ Rope load share karta hai, isliye floor kam push karta hai.

Free-body diagram (person par arrows):

• $\uparrow N$ (floor upar push karta hai)
• $\downarrow mg=500\text{N}$ (Earth neeche kheenchti hai)

Upar ko positive lo. Acceleration downward hai, isliye $a=-2{\text{m/s}}^2$. $\sum F_y=N-mg=ma\Rightarrow N=m(g+a)=50(10-2)=\boxed{400\text{N}}$ $mg=500\text{N}$ se kam — tumhe lighter feel hota hai. Dekho Apparent Weight & Elevators.

Figure kya dikhata hai. Kala square ramp par block hai. Kala downward arrow gravity $mg$ hai. Lal arrow seedha ramp surface se bahar point karta hai — yeh normal force $N$ hai, is figure ka key object. Block se ramp mein jaata chhota kala dashed line gravity ka perpendicular component hai, jise $mg\cos \theta$ label kiya gaya hai; $N$ exactly wohi hai jo us dashed piece ko balance karta hai.

KYA karte hain: gravity ko ramp ke perpendicular part ($mg\cos \theta$) aur ramp ke along part ($mg\sin \theta$) mein split karo. Lal arrow ($N$) ko sirf perpendicular piece balance karna hai. KYU $\cos \theta$: vertical gravity arrow aur ramp ke inward normal ke beech angle exactly $\theta$ hai, isliye adjacent (perpendicular) component $mg\cos \theta$ hai. Ramp ke perpendicular koi acceleration nahi hai (block surface mein nahi, surface ke along slide karta hai): $\sum F_{\perp }=N-mg\cos \theta =0\Rightarrow N=mg\cos \theta =8\times 10\times \cos 40^{\circ }=80\times 0.766=\boxed{61.28\text{N}}$

Free-body diagram (box par arrows — figure dekho):

• $\uparrow N$ (floor upar push karta hai, figure mein lal arrow)
• $\downarrow mg=100\text{N}$ (Earth neeche kheenchti hai)
• push $P$ at $30^{\circ }$ below horizontal split hota hai $\to P\cos 30^{\circ }$ aur $\downarrow P\sin 30^{\circ }$ mein

KYA karte hain: push mein horizontal part $P\cos 30^{\circ }$ aur vertical (downward) part $P\sin 30^{\circ }$ hota hai. Sirf vertical part $N$ ko change karta hai. KYU: $N$ vertical balance se set hota hai; horizontal components horizontal motion mein jaate hain, surface push mein nahi. Vertical downward component: $P\sin 30^{\circ }=50\times 0.5=25\text{N}$. $\sum F_y=N-mg-P\sin 30^{\circ }=0\Rightarrow N=mg+P\sin 30^{\circ }=100+25=\boxed{125\text{N}}$ Angle par floor mein push karna $N$ badhata hai — jo friction bhi badhata hai, kyunki $f=\mu N$ (dekho Friction).

Free-body diagram (box par arrows):

• $\uparrow N$ (floor upar push karta hai)
• $\downarrow mg=100\text{N}$ (Earth neeche kheenchti hai)
• pull $T$ at $30^{\circ }$ above horizontal split hota hai $\to T\cos 30^{\circ }$ aur $\uparrow T\sin 30^{\circ }$ mein

Ab rope ka vertical component upward hai: $T\sin 30^{\circ }=25\text{N}$. $\sum F_y=N+T\sin 30^{\circ }-mg=0\Rightarrow N=mg-T\sin 30^{\circ }=100-25=\boxed{75\text{N}}$ Angle par upar pull karna $N$ kam karta hai — box floor ke against "lighter" ho jaata hai.

Free-body diagram at lift-off (box par arrows):

• $N=0$ (floor ab push nahi karta)
• $\uparrow T\sin 30^{\circ }$ (rope ka vertical pull)
• $\downarrow mg=100\text{N}$ (Earth neeche kheenchti hai)

$N=0$ set karo: $mg-T\sin 30^{\circ }=0\Rightarrow T=\frac{mg}{\sin 30^{\circ }}=\frac{100}{0.5}=\boxed{200\text{N}}$. $T>200\text{N}$ ke liye box lift off karta hai aur $N=0$ (surface sirf push kar sakta hai, kabhi pull nahi — isliye $N$ negative nahi ja sakta).

Free-body diagram (block par arrows):

• $\uparrow N$ (floor upar push karta hai)
• $\downarrow mg=30\text{N}$ (Earth neeche kheenchti hai)
• $\to f$ (floor se friction, horizontal — yahi block ko accelerate karta hai)

Acceleration horizontal hai; abhi bhi koi vertical acceleration nahi hai. Isliye vertical balance unchanged hai: $\sum F_y=N-mg=0\Rightarrow N=mg=3\times 10=\boxed{30\text{N}}$ Horizontal push friction se aata hai, $N$ se nahi. Lesson: sirf acceleration jo surface ke perpendicular ho woh $N$ change karta hai.

Free-body diagram (block par arrows — figure dekho):

• $\downarrow mg$ (Earth neeche kheenchti hai)
• lal arrow $N$ ramp ke perpendicular, jo split hota hai $\to N\sin \theta$ (horizontal) aur $\uparrow N\cos \theta$ (vertical) mein

KYA karte hain: ground axes mein kaam karo (horizontal $x$, vertical $y$). Block par sirf do forces act karte hain: gravity $mg$ neeche aur $N$ incline surface ke perpendicular. Block vertically accelerate nahi karta (yeh wedge ke saath horizontally move karta hai), lekin yeh horizontally $a$ par accelerate karta hai. Vertical (koi vertical acceleration nahi): $N\cos \theta -mg=0\Rightarrow N=\frac{mg}{\cos \theta }(1)$ Horizontal ($a$ par accelerate karta hai): $N\sin \theta =ma(2)$ (2) ko (1) se divide karke given $a$ check karo: $\frac{N\sin \theta }{N\cos \theta }=\frac{ma}{mg}\Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{a}{g}\Rightarrow \tan \theta =\frac{a}{g}\Rightarrow a=g\tan \theta .$ $N$ aur $m$ cleanly cancel ho jaate hain, aur hum exactly wahi acceleration recover karte hain jo problem mein promise ki gayi thi — isliye equation (1) $N$ ka sahi expression hai. Numerically: $N=\frac{mg}{\cos \theta }=\frac{2\times 10}{\cos 30^{\circ }}=\frac{20}{0.866}=\boxed{23.09\text{N}}$ Notice karo $N>mg=20\text{N}$ ab — static incline se bada (jahan $N=mg\cos \theta =17.3\text{N}$). Horizontal acceleration force karta hai surface ko zyada press karne ke liye.

Free-body diagram (box par arrows):

• $\uparrow N$ (floor upar push karta hai)
• $\downarrow mg=50\text{N}$ (Earth neeche kheenchti hai)
• push $P$ at $20^{\circ }$ below horizontal $\to$ $\to P\cos 20^{\circ }$ aur $\downarrow P\sin 20^{\circ }$
• $\leftarrow f=\mu N$ (kinetic friction motion ka oppose karta hai)

Vertical: $N=mg+P\sin 20^{\circ }$ (downward push $N$ mein add hota hai). Horizontal, constant velocity ($a=0$): $P\cos 20^{\circ }=f=\mu N$. $N$ substitute karo: $P\cos 20^{\circ }=\mu (mg+P\sin 20^{\circ })$ $P\cos 20^{\circ }-\mu P\sin 20^{\circ }=\mu mg$ $P=\frac{\mu mg}{\cos 20^{\circ }-\mu \sin 20^{\circ }}=\frac{0.4\times 50}{0.9397-0.4\times 0.3420}=\frac{20}{0.8029}=\boxed{24.91\text{N}}$ Phir $N=mg+P\sin 20^{\circ }=50+24.91\times 0.342=\boxed{58.52\text{N}}$. Neeche ki taraf angle par push karna $N$ badhata hai, jo friction badhata hai — ek reason yeh hai ki box ko neeche-aur-aage push karna upar-aur-aage pull karne se zyada mushkil hota hai.

Free-body diagram for $A$ (arrows on $A$):

• $\uparrow N_{AB}$ (block $B$ upar $A$ ko push karta hai)
• $\downarrow m_{A}g$ (Earth $A$ ko neeche kheenchti hai)

Free-body diagram for $B$ (arrows on $B$):

• $\uparrow N_{Bfloor}$ (elevator floor upar $B$ ko push karta hai)
• $\downarrow m_{B}g$ (Earth $B$ ko neeche kheenchti hai)
• $\downarrow N_{AB}$ (block $A$ neeche $B$ par press karta hai — Newton's 3rd-law partner)

(a) Block $A$ akela. Upar $a$ par accelerate karta hai: $N_{AB}-m_{A}g=m_{A}a\Rightarrow N_{AB}=m_A(g+a)=2(10+5)=\boxed{30\text{N}}$ (b) Block $B$ akela. $A$ neeche $B$ par $N_{AB}=30\text{N}$ se press karta hai (dekho Newton's Third Law): $N_{Bfloor}-m_{B}g-N_{AB}=m_{B}a$ $N_{Bfloor}=m_B(g+a)+N_{AB}=3(15)+30=45+30=\boxed{75\text{N}}$ Cross-check: $A+B$ ko ek $5\text{kg}$ system maano — floor push $=(m_A+m_B)(g+a)=5\times 15=75\text{N}$. ✔ Match karta hai.

Free-body diagram for $A$ (arrows on $A$):

• $\uparrow N_{AB}$ (block $B$ upar $A$ ko push karta hai)
• $\downarrow m_{A}g$ (Earth $A$ ko neeche kheenchti hai)

Sign convention set karna. Upar ko positive lo. Block elevator ke andar baitha hai aur uske saath move karta hai, isliye block ka acceleration elevator ke acceleration ke barabar hai. Elevator neeche magnitude $a$ ke saath accelerate karta hai, isliye uska acceleration negative direction mein point karta hai: block ka vertical acceleration $a_y=-a$ hai. Isliye downward $a$ equation mein minus sign ke roop mein enter karta hai.

$A$ ke liye Newton ka 2nd law: $N_{AB}-m_{A}g=m_A{a}_y=m_A(-a)\Rightarrow N_{AB}=m_A(g-a)$ $N_{AB}=0$ set karo: $g-a=0\Rightarrow a=g=\boxed{10{\text{m/s}}^2}$. Yeh exactly free fall hai — us instant mein kuch bhi $A$ ko neeche se push nahi karta, aur woh (aur $B$) weightless hain. Kisi bhi $a>g$ ke liye blocks alag ho jaayenge aur cabin ke andar float kar jaayenge.