Animated Solution for Physics - Laws of Motion: A person standing on a spring balance inside a stationary lift measures $60\text{ kg}$. The weight of that person, if the lift descends with uniform downward acceleration of $1.8{\text{ m/s}}^2$ will be ......... $\text{N}$. $[g=10{\text{ m/s}}^2]$

Have you ever noticed that sudden feeling of weightlessness when an elevator starts moving downwards? It feels as if you've momentarily lost a few kilograms. This isn't just an illusion; it's a beautiful demonstration of Newton's Laws of Motion in action. In physics, the weight you 'feel' is called your apparent weight, and it is directly related to the normal force exerted on you by the floor.

To understand what's happening, let's isolate the person inside the lift and draw a Free Body Diagram (FBD). There are two primary forces at play here. First, the Earth pulls the person downwards with a gravitational force, which is their true weight, calculated as $mg$. Second, the floor of the lift (or the spring balance they are standing on) pushes back upwards with a normal force, $N$.

Crucially, a weighing machine does not measure your true weight ($mg$); it measures the normal force ($N$) it has to exert to support you.

When the lift is stationary or moving at a constant velocity, the upward normal force perfectly balances the downward gravitational force, so $N=mg$. However, in our scenario, the lift is accelerating downwards at $a=1.8{\text{ m/s}}^2$.

Because the person is accelerating downwards along with the lift, the net force acting on them must be directed downwards. According to Newton's Second Law ($F_{\text{net}}=ma$), we can write the equation of motion as:

We want to find the apparent weight, which is the normal force $N$. Rearranging the equation gives us:

This equation reveals something fascinating: the effective acceleration due to gravity is reduced by the acceleration of the lift. Instead of feeling the full $10{\text{ m/s}}^2$, the person feels an effective gravity of $(g-a)$.

Now, we simply substitute the given values into our derived formula. The mass of the person is $m=60\text{ kg}$, the acceleration due to gravity is $g=10{\text{ m/s}}^2$, and the downward acceleration of the lift is $a=1.8{\text{ m/s}}^2$.

Therefore, the spring balance will read $492\text{ N}$. The person's true weight is $600\text{ N}$, but due to the downward acceleration, their apparent weight has dropped significantly. This is the exact physics behind that stomach-dropping sensation in a descending elevator!