Question

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring balance reads $49\text{ N}$, when the lift is stationary. If the lift moves downward with an acceleration of $5\text{ m/s}^2$, the reading of the spring balance will be ($g = 9.8\text{ m/s}^2$)

• A. $74\text{ N}$
• B. $15\text{ N}$
• C. $24\text{ N}$
• D. $49\text{ N}$

Hint:

When an elevator moves downward, you feel lighter because gravity is partially offset by the downward acceleration. You can write this ratio directly to avoid finding mass: $\frac{W_{new}}{W_{old}} = \frac{g - a}{g} = \frac{9.8 - 5}{9.8} = \frac{4.8}{9.8} \approx \frac{1}{2}$. Half of $49\text{ N}$ is roughly $24.5\text{ N}$, pointing instantly to $24\text{ N}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A spring balance is attached to the ceiling of an elevator lift. When the lift is stationary (at rest), the balance measures a bag's true weight as $W = 49\text{ N}$.
When the lift accelerates downward at $a = 5\text{ m/s}^2$, we need to find the new apparent weight reading recorded by the scale.

Step 2: Key Formula or Approach:
1. The true weight of an object when resting in an inertial frame is:
$$W = mg \implies m = \frac{W}{g}$$ 2. When an elevator frame accelerates downward with a rate $a$, an observer inside experiences a pseudo-force pointing opposite to the acceleration (upward). This yields an apparent weight ($W_{app}$) of:
$$W_{app} = m(g - a)$$

Step 3: Detailed Explanation:
First, calculate the actual mass $m$ of the bag using its stationary weight ($49\text{ N}$) and the acceleration due to gravity ($g = 9.8\text{ m/s}^2$):
$$m = \frac{49}{9.8} = 5\text{ kg}$$ Now, calculate the apparent weight when the lift accelerates downward at $a = 5\text{ m/s}^2$:
$$W_{app} = m(g - a)$$ Substitute the values ($m = 5\text{ kg}$, $g = 9.8\text{ m/s}^2$, $a = 5\text{ m/s}^2$):
$$W_{app} = 5 \times (9.8 - 5)$$ $$W_{app} = 5 \times 4.8 = 24\text{ N}$$ Therefore, the spring balance will record a reduced reading of $24\text{ N}$.

Step 4: Final Answer:
The reading of the spring balance will be $24\text{ N}$, which corresponds to option (C).