As I was standing on a weighing machine inside a lift it recorded 50 kg wt. Suddenly for a few seconds it recorded 45 kg wt. What must have happened during that time? Explain with complete numerical analysis.

Question

Accepted Answer

Concept understanding — Pseudo Force

Pseudo Force – The Trick That Makes Newton's Laws Work Everywhere

Imagine you're sitting in a bus that suddenly lurches forward. What happens? You feel thrown backward against the seat. From the ground outside, a friend sees you stay still while the seat pushes you forward. But inside the bus, it feels like a real force yanked you backward. That feeling is the pseudo force.

The core problem is simple: Newton's laws only work in inertial frames — frames that are either at rest or moving with constant velocity. A bus accelerating forward is a non-inertial frame. Inside it, Newton's first law seems to break: you see an object (your own body) accelerate backward with no visible push. To fix this, we invent a fictitious force — the pseudo force — so that inside the accelerating bus, Newton's laws once again hold.

The Precise Statement

\vec{F}_{\text{pseudo}} = - m \vec{a}_{\text{frame}}

Where:

$m$ is the mass of the object you're analyzing
$\vec{a}_{\text{frame}}$ is the acceleration of the non-inertial frame (measured from an inertial frame)
The minus sign means the pseudo force acts opposite to the frame's acceleration

So if the bus accelerates forward ( $\vec{a}_{\text{frame}}$ forward), the pseudo force on every object inside is backward — exactly what you feel.

How to Use It

When solving a problem from inside a non-inertial frame:

Identify the acceleration of the frame itself.
Add a pseudo force $-m\vec{a}_{\text{frame}}$ on every object you're analyzing.
Now treat the frame as if it were inertial — apply Newton's laws normally.

Watch out

The pseudo force is not real. It has no agent — no wall, no rope, no gravity. It vanishes the moment you step outside to an inertial frame. Never count it as a real force in a free-body diagram meant for an inertial observer.

A Concrete Example

A block of mass $m$ sits on the smooth floor of an accelerating train. The train accelerates rightward with $a$ . From the ground, the block stays still (no horizontal force on it). But inside the train, the block appears to slide leftward.

To explain this from inside the train: apply a pseudo force $m a$ to the left on the block. Now the block has a net leftward force, so it accelerates left relative to the train — matching what you see. Newton's second law works perfectly inside the train once you include this fictitious term.

Tip

Pseudo force is also called inertial force or fictitious force. In rotational frames, it gives rise to the centrifugal and Coriolis forces — same idea, fancier geometry.

The Big Picture

Pseudo force is a mathematical patch — a clever trick that lets you use the familiar $F = ma$ even when your reference frame is accelerating. It doesn't represent a physical interaction; it represents the fact that your frame itself is not a perfect Newtonian stage. Once you understand that, you can solve problems from any frame without confusion.

Normal (stationary or constant-velocity) reading = actual weight = 50 kg wt, i.e. true mass $m=50$ kg (using kgf as a weight unit with g=10 m/s^2, true weight $=50\times10=500$ N). During the anomalous reading of 45 kg wt, the apparent weight $W'=45\times10=450$ N is LESS than the true weight -- this happens exactly when the lift accelerates DOWNWARD (4.5.3), since then $W'=mg-ma$ (weight loss). Solving: $450=50(10)-50a\Rightarrow 450=500-50a\Rightarrow 50a=50\Rightarrow a=1$ m/s^2. So the lift must have been moving with a downward acceleration of 1 m/s^2 during those few seconds -- this could correspond either to the lift SPEEDING UP while moving downward, or SLOWING DOWN while moving upward (both give a downward acceleration vector); either way, the reduced scale reading of 45 kg wt is fully explained by this 1 m/s^2 downward acceleration of the lift (reference frame), which produces an upward pseudo-force reduction on the person's felt weight. [!ANSWER] The lift must have had a downward acceleration of 1 m/s^2 during that interval (either speeding up while descending or slowing down while ascending), reducing the apparent weight from 500 N (50 kgf) to 450 N (45 kgf) via $W'=mg-ma$ .

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$Q. When milk is churned then cream is separated due to (A) Gravitational force (B) Frictional force (C) Centripetal force (D) Centrifugal force$
›Reveal solutionSolution

Cream separates from milk because of centrifugal force acting in the rotating churn.

When milk is churned (spun) rapidly, every particle in it moves in a circle, which requires a centripetal (inward) force actually supplied by the walls of the container. Viewed from the rotating frame of the churn itself, each particle instead appears to experience an outward pseudo-force — the centrifugal force — whose magnitude depends on the particle's mass and density.

Cream (fat globules) is less dense than the rest of the milk, so it experiences a smaller outward centrifugal effect relative to the denser skimmed milk, causing the denser liquid to move outward and the lighter cream to collect near the axis — achieving separation.

✓Final answer
$(D) Centrifugal force.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. When milk is churned, then cream is separated. It happens due to (A) gravitational force (B) frictional force (C) centrifugal force (D) cohesive force$
›Reveal solutionSolution

Cream separates from milk in a churn due to centrifugal force acting in the rotating frame.

When milk is rapidly rotated in a centrifugal churn/separator, in the (non-inertial) rotating frame every particle experiences an outward pseudo-force called the centrifugal force, of magnitude $m\omega^2r$ . Because cream (fat globules) is less dense than the rest of the milk, it experiences a smaller net outward requirement and gets pushed toward the centre/top, while the denser skimmed milk is thrown further outward — separating the cream.

✓Final answer
$(C) Centrifugal force.$
CBSE 2020Set ANNUAL1 markMCQ
$Q. An object of mass 10 kg is hanging from a spring scale which is attached to the roof of a lift. If the lift is in free fall, the reading in the spring scale is:$
(a) 98 N
(b) zero
(c) 49 N
(d) 9.8 N
›Reveal solutionSolution

During free fall, both the lift and the hanging mass accelerate downward at g, so the spring tension (and hence the scale reading) is zero.

The spring scale reads the tension T in the spring, which by Newton's second law applied to the hanging mass (taking downward as positive) is:

mg - T = ma

where a is the downward acceleration of the mass (equal to the lift's acceleration, since the mass moves with the lift).

When the lift is in free fall, a = g (it falls under gravity alone, with no other force acting on the lift-mass system to oppose gravity). Substituting:

mg - T = mg

T = 0

This is the condition of 'apparent weightlessness' — although gravity still acts on the mass, the scale (which measures the contact/spring force, not gravity itself) reads zero because there is no relative acceleration between the mass and the lift.

✓Final answer
$The reading on the spring scale is zero (option b).$

Q.As I was standing on a weighing machine inside a lift it recorded 50 kg wt. Suddenly for a few seconds it recorded 45 kg wt. What must have happened during that time? Explain with complete numerical analysis.

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