Q.A man of mass $70\ \text{kg}$ stands on a weighing scale in a lift which is moving

Concept: The scale reading equals the normal force $N$ exerted by the scale on the man (his apparent weight), with $g=9.8$ m s$^{-2}$.

1. For uniform motion (a), acceleration $a=0$, so $N=mg$.
2. For accelerated motion, apply $N-mg=ma$ (upward positive): upward acceleration gives $N=m(g+a)$; downward acceleration gives $N=m(g-a)$.
3. For free fall (d), $a=g$ downward, so $N=m(g-g)=0$.

Calculations ($m=70$ kg, $g=9.8$ m s$^{-2}$):

• (a) $N = 70 \times 9.8 = 686\ \text{N}$
• (b) $N = 70 \times (9.8-5) = 70 \times 4.8 = 336\ \text{N}$ …

The scale in a lift measures the normal force exerted by the floor on the man — the force he feels through his feet. This is not always equal to his weight $mg$; it adjusts to whatever the man's actual vertical acceleration demands.

Newton's second law: the net force on the man equals his mass times his acceleration. The two forces acting on him are his weight $mg$ downward and the normal reaction $N$ upward. Taking upward as positive:

where $a$ is the acceleration of the lift (and the man, since he moves with it). Rearranging:

If the lift accelerates upward, $a > 0$ and $N > mg$ — the man feels heavier. If it accelerates downward, $a < 0$ and $N < mg$ — he feels lighter. For uniform velocity, $a = 0$ and $N = mg$.

Now apply this to each case, with $m = 70\ \text{kg}$ and $g = 9.8\ \text{m/s}^2$.

1. Case (a): Upwards with uniform speed $10\ \text{m/s}$ Uniform speed means zero acceleration: $a = 0$.

2. Case (b): Downwards with uniform acceleration $5\ \text{m/s}^2$ Downward acceleration is negative in our sign convention: $a = -5\ \text{m/s}^2$. $N = m(g + a) = 70 \times (9.8 - 5) = 70 \times 4.8 = 336\ \text{N}$ …

Concept: Apparent Weight — Newton's Second Law Applied to a Person on a Scale

Step 1: Set up the general equation

The scale reads the normal force $N$ the floor exerts on the man — his apparent weight —

not necessarily his true weight $mg$. With upward positive and $a$ the lift's

acceleration:

Step 2: (a) Uniform speed upward ($10\ \text{m s}^{-1}$)

Constant velocity $\Rightarrow a=0$:

Step 3: (b) Downward acceleration $5\ \text{m s}^{-2}$

Downward acceleration is negative in our convention, $a=-5$:

Step 4: (c) Upward acceleration $5\ \text{m s}^{-2}$ …