Step 1: Understand the Concept of Apparent Weight in an Accelerating Frame

When an object is in an accelerating frame of reference (like an elevator), the forces acting on it appear differently compared to an inertial frame (a non-accelerating frame). This difference is often described using the concept of apparent weight or by considering an effective gravitational acceleration.

Let’s consider an object of mass $m$ inside an elevator, and analyze the forces using Newton’s Second Law, $\Sigma F = ma$.

• Case 1: Elevator is stationary or moving with constant velocity ($a_e = 0$).

  The net force on the object is zero. If the object is resting on a scale, the normal force ($N$) exerted by the scale on the object is equal to the object’s gravitational force (weight) $mg$.

  $$N – mg = 0 \implies N = mg$$

• Case 2: Elevator accelerates upwards with acceleration $a_e$.

  When the elevator accelerates upwards, the object inside it also accelerates upwards with the same acceleration $a_e$. According to Newton’s Second Law, there must be a net upward force on the object equal to $ma_e$.

  The forces acting on the object are:

  • Normal force $N$ (or tension for a hanging mass) acting upwards.
  • Gravitational force $mg$ acting downwards.

  Applying Newton’s Second Law (taking upward as positive):

  $$N – mg = ma_e$$

  Now, let’s solve for the normal force $N$, which represents the apparent weight:

  $$N = mg + ma_e$$

  We can factor out $m$ from the right side:

  $$N = m(g + a_e)$$

  Comparing this to the stationary case ($N = mg$), it looks as if the gravitational acceleration has effectively increased. We define this new effective gravitational acceleration as $g’$.

  $$\text{Therefore, we get the relation: } g’ = g + a_e$$

  This means any object inside the upward accelerating elevator will behave as if gravity is stronger by an amount $a_e$.

• Case 3: Elevator accelerates downwards with acceleration $a_e$.

  If the elevator accelerates downwards, the net force on the object is $ma_e$ downwards.

  Applying Newton’s Second Law (taking downward as positive in this case, or upward as negative):

  $$mg – N = ma_e$$

  Solving for $N$:

  $$N = mg – ma_e$$

  $$N = m(g – a_e)$$

  Here, $g’ = g – a_e$, meaning objects inside a downward accelerating elevator behave as if gravity is weaker.

In this specific problem, the elevator is accelerating upwards with $a_e = g/10$. So, the effective gravitational acceleration ($g’$) experienced by the masses relative to the elevator’s frame will be:

$$g’ = g + a_e = g + \frac{g}{10} = \frac{10g}{10} + \frac{g}{10} = \frac{11g}{10}$$

Now, we can substitute the value of $g = 9.8 \text{ m/s}^2$:

$$g’ = \frac{11 \times 9.8}{10} = 11 \times 0.98 = 10.78 \text{ m/s}^2$$

We will use this effective $g’$ in our calculations for the tension in the string connecting the masses, treating the elevator’s frame as our non-inertial reference frame.

Step 2: Define Variables and Draw Free Body Diagrams (Relative to the Elevator’s Accelerating Frame)

Let $m_1 = 1.5 \text{ kg}$ and $m_2 = 3.0 \text{ kg}$. Let $T$ be the tension in the string connecting the two masses. The pulley and string are light and smooth, so the tension is uniform throughout the string.

We will analyze the motion of each mass relative to the elevator’s accelerating frame, using the effective gravitational acceleration $g’$. This approach simplifies the problem as we don’t need to consider pseudo forces explicitly if we use $g’$.

• For mass $m_1$ ($1.5 \text{ kg}$):

  The forces acting on $m_1$ are the tension $T$ pulling upwards and its effective weight $m_1 g’$ pulling downwards. Since $m_2 > m_1$, $m_1$ will accelerate upwards relative to the elevator. Let its acceleration relative to the elevator be $a$.

  Equation of motion (net force in the direction of acceleration): $$T – m_1 g’ = m_1 a \quad \text{(1)}$$

• For mass $m_2$ ($3.0 \text{ kg}$):

  The forces acting on $m_2$ are its effective weight $m_2 g’$ pulling downwards and the tension $T$ pulling upwards. Since $m_2 > m_1$, $m_2$ will accelerate downwards relative to the elevator. Let its acceleration relative to the elevator be $a$ (same magnitude as $m_1$ due to the string and pulley).

  Equation of motion (net force in the direction of acceleration): $$m_2 g’ – T = m_2 a \quad \text{(2)}$$

Step 3: Calculate the Acceleration of the Masses Relative to the Elevator ($a$)

We have a system of two linear equations:

(1) $T – m_1 g’ = m_1 a$

(2) $m_2 g’ – T = m_2 a$

To find the acceleration $a$, we can add equation (1) and equation (2) to eliminate the tension $T$:

$$(T – m_1 g’) + (m_2 g’ – T) = m_1 a + m_2 a$$

$$T – m_1 g’ + m_2 g’ – T = (m_1 + m_2) a$$

$$(m_2 – m_1) g’ = (m_1 + m_2) a$$

Now, we can solve for $a$:

$$a = \frac{(m_2 – m_1) g’}{(m_1 + m_2)}$$

Substitute the known values:

• $m_1 = 1.5 \text{ kg}$
• $m_2 = 3.0 \text{ kg}$
• $g’ = 10.78 \text{ m/s}^2$ (calculated in Step 1)

$$a = \frac{(3.0 \text{ kg} – 1.5 \text{ kg}) \times 10.78 \text{ m/s}^2}{(1.5 \text{ kg} + 3.0 \text{ kg})}$$

$$a = \frac{1.5 \times 10.78}{4.5}$$

$$a = \frac{16.17}{4.5}$$

$$a \approx 3.593 \text{ m/s}^2$$ (This matches the approximation in the provided solution of $3.59 \text{ m/s}^2$)

Step 4: Calculate the Tension in the String ($T$)

Now that we have the acceleration $a$, we can substitute its value back into either equation (1) or (2) to find the tension $T$. Let’s use equation (1):

$$T – m_1 g’ = m_1 a$$

Rearrange to solve for $T$:

$$T = m_1 a + m_1 g’$$

Factor out $m_1$:

$$T = m_1 (a + g’)$$

Substitute the values:

• $m_1 = 1.5 \text{ kg}$
• $a = 3.593 \text{ m/s}^2$
• $g’ = 10.78 \text{ m/s}^2$

$$T = 1.5 \text{ kg} \times (3.593 \text{ m/s}^2 + 10.78 \text{ m/s}^2)$$

$$T = 1.5 \times 14.373$$

$$T \approx 21.5595 \text{ N}$$ (This matches the approximation in the provided solution of $21.55 \text{ N}$)

Step 5: Determine the Reading of the Spring Balance

The spring balance is attached to the ceiling of the elevator and supports the pulley system. The tension in the string passing over the pulley exerts a downward force on the pulley from both sides. Since the tension in the string is $T$ everywhere, the total downward force exerted on the pulley by the string is $2T$.

The spring balance measures the total force exerted on it. Assuming the pulley’s mass is negligible (it’s “light”), the reading of the spring balance will be equal to this total downward force.

$$\text{Reading of Spring Balance (Force)} = 2T$$

Substitute the calculated value of $T$:

$$\text{Reading (Force)} = 2 \times 21.5595 \text{ N}$$

$$\text{Reading (Force)} = 43.119 \text{ N}$$

Spring balances are often calibrated to show readings in kilograms (mass units) by effectively dividing the measured force by the standard gravitational acceleration ($g = 9.8 \text{ m/s}^2$). It’s important to use the standard $g$ here, not $g’$, because the balance itself is giving a reading that corresponds to a mass under Earth’s standard gravity.

$$\text{Reading in kg (Mass Equivalent)} = \frac{\text{Force}}{g}$$

$$\text{Reading in kg} = \frac{43.119 \text{ N}}{9.8 \text{ m/s}^2}$$

$$\text{Reading in kg} \approx 4.399 \text{ kg}$$ (This matches the approximation in the provided solution of $4.39 \text{ kg}$)