A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spr

Q: Why does the reading of the spring balance change if the bag's mass is constant?

The spring balance doesn't measure mass directly; it measures the tension force pulling on its hook. When the lift accelerates, the tension required to accelerate the bag changes, altering the reading.

Q: What would the spring balance read if the lift was moving downwards with a constant velocity?

If the velocity is constant, the acceleration is zero. Therefore, the net force is zero, and the tension would equal the true weight ($49 \text{ N}$), just like when it's stationary.

Q: Why did we use $g = 9.8 \text{ m/s}^2$ instead of $10 \text{ m/s}^2$?

The initial weight is given as $49 \text{ N}$. Dividing $49$ by $9.8$ gives a clean integer ($5 \text{ kg}$). Using $10$ would give $4.9 \text{ kg}$, which makes subsequent calculations unnecessarily messy. Always look for these numerical hints!

Q: Could we solve this using the concept of pseudo force?

Absolutely! If you observe from the accelerating lift's frame (a non-inertial frame), you must apply an upward pseudo force of $ma$. The balanced forces would then be $T + ma = mg$, which gives the exact same equation: $T = m(g - a)$.

Q: What would happen if the lift's downward acceleration was equal to $g$ (free fall)?

If $a = g$, then the apparent weight $T = m(g - g) = 0$. The bag would experience weightlessness, and the spring balance would read zero!

Question

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads $49	ext{ N}$, when the lift is stationary. If the lift moves downward with an acceleration of $5	ext{ ms}^{-2}$, the reading of the spring balance will be

Options:

(A) $24	ext{ N}$

(B) $74	ext{ N}$

(C) $15	ext{ N}$

(D) $49	ext{ N}$

Sigma Percentile · Accepted Answer

### The Elevator Experience

Have you ever felt your stomach drop when an elevator suddenly starts moving downwards? That fleeting sensation of feeling lighter is not just in your head—it is a fundamental consequence of physics!

In this problem, we are exploring exactly that phenomenon using a spring balance and a bag inside an accelerating lift.

### The Stationary Phase: Finding the True Mass

Before the lift starts moving, it is perfectly stationary. In this state of equilibrium, the spring balance reads $49 	ext{ N}$.

But what does a spring balance actually measure? It does not measure mass directly; it measures the **tension** in the spring.

Since the bag is at rest, the upward tension $T$ perfectly balances the downward gravitational force $mg$.

$$T_0 = mg = 49 	ext{ N}$$

From this, we can easily extract the true mass of the bag. By dividing the weight by the acceleration due to gravity ($g = 9.8 	ext{ m/s}^2$), we get:

$$m = \frac{49}{9.8} = 5 	ext{ kg}$$

### The Accelerating Phase: Newton's Second Law

Now, the lift begins its descent, accelerating downwards at $a = 5 	ext{ m/s}^2$.

Because the bag is inside the lift, it must also accelerate downwards at this exact same rate. According to Newton's Second Law, a downward acceleration requires a net downward force.

This means the downward pull of gravity must now be **greater** than the upward pull of the spring's tension.

$$F_{	ext{net}} = ma$$

$$mg - T = ma$$

### The Final Calculation: Apparent Weight

We want to find the new reading on the spring balance, which is the new tension $T$. Let's rearrange our equation to solve for $T$:

$$T = mg - ma$$

$$T = m(g - a)$$

This equation beautifully captures the concept of **apparent weight**. When you accelerate downwards, your apparent weight decreases!

Let's plug in our known values:

$$T = 5(9.8 - 5)$$

$$T = 5 	imes 4.8$$

$$T = 24 	ext{ N}$$

The spring balance will read $24 	ext{ N}$. The bag appears to have lost more than half of its weight simply because the frame of reference is accelerating downwards.

This elegant application of Newton's laws perfectly explains why we feel lighter in a descending elevator!

Select Answer:

Visualized Solution

$Visualizing the System$

$Stationary Lift$

$Finding the Mass$

$Downward Acceleration$

$Newton’s Second Law$

$Apparent Weight Equation$

$Final Calculation$

$Conclusion$

The Sigma Insight: Newton's Laws of Motion

The Elevator Experience

The Stationary Phase

The Accelerating Phase

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Stationary Lift

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Downward Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Newton’s Second Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Apparent Weight Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion