Q.Explain in detail the idea of weightlessness using a lift as an example.

Step 1. Setup. A person of mass $m$ stands on a scale inside a lift. Two forces act: gravity $\vec F_G=-mg\hat j$ (downward) and the scale's normal reaction $\vec N=N\hat j$ (upward). The scale reading equals $N$, the apparent weight.

Step 2. Case (i): lift at rest / constant velocity. Net acceleration is zero, so applying Newton's second law: $N-mg=0\Rightarrow N=mg$. Apparent weight equals true weight.

Step 3. Case (ii): lift accelerating upward with acceleration $a$. Applying $\vec F_G+\vec N=m\vec a$ along the vertical: $-mg+N=ma\Rightarrow N=m(g+a)$. Apparent weight is greater than $mg$ -- the "heavy" feeling as an elevator starts moving up.

Step 4. Case (iii): lift accelerating downward with acceleration $a$. Similarly, $-mg+N=-ma\Rightarrow N=m(g-a)$. Apparent weight is less than $mg$ -- the "light" feeling as a lift starts moving down.

Step 5. Case (iv): free fall (e.g. cable cut), $a=g$ downward. Substituting $a=g$ into Case (iii)'s result: $N=m(g-g)=0$. The scale reads exactly zero -- weightlessness. The person and the lift accelerate downward at exactly the same rate, so the person never presses on the floor, and the floor never pushes back. …

Apply Newton's second law to the person for each of the four …

• Confusing which case (up vs down acceleration) increases vs decreases apparent weight. …