Exercises — Weightlessness — true (free fall) vs apparent

What we know: at rest means $a=0$, so $N=mg$. Step — solve for mass. From $N=mg$: $m=N/g=500/10=50\text{kg}$. Acceleration: $a=0$ (the scale is not moving nor accelerating). ==$m=50\text{kg}$, $a=0$.== The scale reads true weight only when $a=0$ — this is the baseline every later problem departs from.

The trap-buster: "moving down" is a velocity, and velocity never appears in $N=m(g+a)$ — only acceleration does. Step — find $a$. Constant velocity ⇒ $a=0$. Step — apply the law. $N=m(g+a)=50(10+0)=500\text{N}$. ==$N=500\text{N}$ (normal weight — the person feels nothing unusual).==

Set $N=m(g+a)=0$. Since $m\ne 0$, we need $g+a=0$, i.e. $a=-g$. ==Answer: (c) $a=-g$.== Check the others: (a) $N=m(2g)=2mg$ (twice as heavy); (b) $N=mg$; (d) $N=m(g-2g)=-mg$ — a negative $N$ is impossible for a floor, meaning the person would leave the floor and press on a ceiling instead.

Step — sign of $a$. Accelerating up ⇒ $a=+30$. Step — apply. $N=70(10+30)=70\times 40=2800\text{N}$. Step — g-force. True weight $mg=700\text{N}$. Ratio $=2800/700=4$. ==$N=2800\text{N}$, i.e. 4 g — the candidate feels four times their weight.==

Step — direction of $a$. Moving up while slowing ⇒ acceleration points down ⇒ $a=-4$. Step — apply. $N=50(10-4)=50\times 6=300\text{N}$. ==$N=300\text{N}$ — lighter than the true $500\text{N}$.== Notice: same "feel light" outcome as an accelerating-down lift, because slowing an upward rise and speeding a downward drop are the same downward acceleration.

Step — use the local $g$. The law is $N=m(g_{\text{local}}+a)$ with $g_{\text{local}}=1.6$. Step — sign. Accelerating down ⇒ $a=-1.6$. Step — apply. $N=50(1.6-1.6)=0\text{N}$. ==$N=0$ — yes, weightless.== The weightless condition is always $a=-g_{\text{local}}$, whatever planet you're on. Free fall on the Moon feels identical to free fall on Earth — zero support force either way.

Step — invert the law. $N=m(g+a)\Rightarrow a=\frac{N}{m}-g=\frac{480}{60}-10=8-10=-2{\text{m/s}}^2$. Direction: negative ⇒ acceleration points downward, magnitude $2{\text{m/s}}^2$. Trick sub-question: No — you cannot tell whether it moves up or down. The scale only reveals $a$, not velocity. It could be accelerating downward while descending, or decelerating while ascending — both give $a=-2$. ==$a=2{\text{m/s}}^2$ downward; direction of motion is undetermined.==

(a) Free fall ⇒ $a=-g=-10$. $N=50(10-10)=0\text{N}$. Weightless feeling. (b) This is Free Fall Kinematics: $v=u+at$ with $u=0$, $a=g=10$ (speed grows, take magnitude): $v=0+10(2.0)=20\text{m/s}$. (c) No. True weight $mg=50\times 10=500\text{N}$ the whole time — gravity is fully on; that's exactly why the elevator accelerates at $g$. Only the support force $N$ is zero. ==(a) $0\text{N}$; (b) $20\text{m/s}$; (c) true weight $=500\text{N}$, not zero.==

Why centripetal? Standing on a rotating Earth is gentle Circular Motion & Centripetal Acceleration: the person travels a circle once a day, so a tiny net downward acceleration $a_c$ is required. That net force comes from gravity slightly out-pulling the normal force. Step — Newton's 2nd law, down as the acceleration direction. The net downward force must equal $m{a}_c$: $mg-N=m{a}_c\Rightarrow N=m(g-a_c).$ Step — numbers. $N=50(9.8-0.034)=50\times 9.766=488.3\text{N}$. Compare: true weight $mg=490\text{N}$. So the scale reads about $1.7\text{N}$ less. ==$N\approx 488.3\text{N}$, i.e. $\approx 1.7\text{N}$ below true weight.== This is the same $N=m(g+a)$ with $a=-a_c$ (downward). Earth's rotation is a mild, permanent "elevator accelerating gently downward."

(a) $g=g_0{\left(\frac{R}{R+h}\right)}^2=9.8{\left(\frac{6.37\times 10^6}{6.77\times 10^6}\right)}^2=9.8(0.9409{)}^2\approx 9.8\times 0.8853\approx 8.68{\text{m/s}}^2.$ (b) The astronaut and station share the same centripetal acceleration $a=g$ (directed toward Earth). In the station's frame there is nothing left for the floor to supply: $N=m(g-g)=0\text{N}.$ More carefully in the ground frame: both need $ma=mg$ inward, provided entirely by gravity, so the floor contributes nothing. (c) ==They float because astronaut and station free-fall together at $g\approx 8.7{\text{m/s}}^2$; equal acceleration ⇒ no mutual push ⇒ $N=0$. Gravity is ~89% of surface, not zero.==

(a) Here $g=0$ but the floor must accelerate the astronaut at $a=+9.8$. $N=m(g+a)=70(0+9.8)=686\text{N}$. (b) They feel a normal $686\text{N}$ push — exactly Earth-surface weight ($70\times 9.8=686$). With no window, they'd swear they were standing still on Earth. (This is the seed of Non-inertial Frames & Pseudo-forces and Einstein's equivalence principle.) (c) ISS: $g\approx 8.7$ but free-falling, so $N=0$ — gravity present, no support. Deep-space ship: $g=0$ but thrusting, so $N=686$ — no gravity, full support. The scale can't distinguish "real gravity" from "acceleration" — it only ever reports $N$. ==(a) $686\text{N}$; (b) feels like standing on Earth; (c) opposite causes, and the scale — reporting only $N$ — cannot tell them apart.==

(a) Weightless ⇒ $N=0\Rightarrow a=-g=-10{\text{m/s}}^2$ (the plane must accelerate downward at $g$, i.e. arc over the top of the parabola like a thrown ball). (b) $N=m(g+a)\Rightarrow a=\frac{N}{m}-g=\frac{1200}{60}-10=20-10=+10{\text{m/s}}^2$ (upward). g-force $=N/(mg)=1200/600=2\text{g}$. (c) No — true weight $mg=600\text{N}$ throughout. Only $N$ swings between $0$ and $1200\text{N}$. ==(a) $a=-10{\text{m/s}}^2$; (b) $a=+10{\text{m/s}}^2$, $2\text{g}$; (c) true weight constant at $600\text{N}$.==

Phase (i): $a=+2$. $N=50(10+2)=600\text{N}$ (heavy). Phase (ii): $a=0$. $N=50(10)=500\text{N}$ (normal). Phase (iii): moving up while stopping ⇒ acceleration downward ⇒ $a=-2$. $N=50(10-2)=400\text{N}$ (light). Max speed: reached at end of phase (i): $v=u+at=0+2(3)=6\text{m/s}$. (Constant through phase (ii), then decelerates to $0$: check $6-2(3)=0$ ✓, consistent.) ==(i) $600\text{N}$, (ii) $500\text{N}$, (iii) $400\text{N}$; max speed $6\text{m/s}$.==

(a) $a=-12$. $N=50(10-12)=50(-2)=-100\text{N}$. (b) A floor can only push up ($N\ge 0$); it cannot pull down. The formula's $-100\text{N}$ signals the floor has "run out" — the person actually lifts off the floor and, if there's a ceiling, gets pressed against it. The ceiling then supplies a downward $100\text{N}$ on the person (so the ceiling's contact force plays the role, with magnitude $|N|=100\text{N}$). Without a ceiling, the person floats up relative to the falling lift. ==$N_{\text{floor}}=0$ (person leaves floor); the ceiling pushes down with $100\text{N}$ once contact is made.== This is the full picture: $a<-g$ isn't "extra-weightless," it's inverted support — the ground stops being the thing that holds you.