Understanding Actual and Apparent Weight in an Elevator

When a person is inside an elevator, their weight is the force exerted on them by gravity. This is often called their actual weight, which is equal to $mg$, where $m$ is the mass of the person and $g$ is the acceleration due to gravity.

However, the feeling of weight we experience is actually the normal force exerted on us by the surface we are standing on. In an elevator, this surface is the elevator floor. This normal force is what is often referred to as apparent weight.

Analyzing Forces in an Elevator

Let's consider the forces acting on a person of mass $m$ inside an elevator:

• The force of gravity acting downwards, with magnitude $mg$.
• The normal force exerted by the elevator floor acting upwards, with magnitude $N$. This is the apparent weight.

According to Newton's second law of motion, the net force acting on the person is equal to their mass times their acceleration ($\vec{F}_{net} = m\vec{a}$). If we take the upward direction as positive, the net force is $N - mg$.

So, the equation of motion is:

\(N - mg = ma\)

Where $a$ is the acceleration of the elevator (and the person inside).

When Apparent Weight Equals Actual Weight

We want to find the condition under which the apparent weight ($N$) is equal to the actual weight ($mg$). So, we set $N = mg$ in the equation above:

\(mg - mg = ma\)

\(0 = ma\)

Since the mass of the person $m$ is not zero, this equation is true only if the acceleration $a$ is zero.

Interpreting Acceleration $a=0$

An acceleration of zero ($a=0$) means that the velocity of the elevator is not changing. This happens in two cases:

• The elevator is stationary (velocity is zero and constant).
• The elevator is moving with a constant velocity (either upwards or downwards).

Both these cases fall under the description of uniform motion.

Evaluating the Options

Let's look at the given options based on our analysis:

• Option 1: elevator is accelerating upwards. If the elevator is accelerating upwards, $a > 0$. From $N - mg = ma$, we get $N = mg + ma$. The apparent weight ($N$) is greater than the actual weight ($mg$).
• Option 2: elevator is accelerating downward. If the elevator is accelerating downwards, $a < 0$ (if upwards is positive), or we can write $mg - N = ma$ if downward is positive, giving $N = mg - ma$ (where $a$ is magnitude of downward accel). In either case, the apparent weight ($N$) is less than the actual weight ($mg$).
• Option 3: elevator is in uniform motion. If the elevator is in uniform motion (constant velocity or at rest), the acceleration $a = 0$. From $N - mg = ma$, we get $N - mg = m(0)$, which means $N - mg = 0$, or $N = mg$. The apparent weight ($N$) is equal to the actual weight ($mg$).
• Option 4: none of the above. This is incorrect because option 3 matches our finding.

Conclusion

The apparent weight of a person in an elevator is equal to their actual weight when the elevator is not accelerating, i.e., when it is in uniform motion (moving at a constant velocity or stationary).