1. Problem Analysis & Physics Principles

We have a system of two blocks with masses $m$ and $M$ ($M > 2m$) connected by ideal pulleys and threads inside a lift. We need to find the condition for the acceleration of the lift such that:

• The thread remains taut.
• Both blocks accelerate in the same direction relative to the ground.

Key Concepts:

• Constraint Motion: The motion of block $m$ and block $M$ is linked by the geometry of the string.
• Non-Inertial Reference Frame: It is convenient to solve the dynamics relative to the lift. This introduces a pseudo-force ($m\vec{a}_{lift}$) acting opposite to the lift’s acceleration.
• Effective Gravity: Inside the lift, the effective gravitational acceleration is $\vec{g}_{eff} = \vec{g} – \vec{a}_{lift}$.

2. System Diagram and Setup

Constraint Relation:
Let $x_m$ be the displacement of block $m$ (upwards relative to lift) and $x_M$ be the displacement of block $M$ (downwards relative to lift). Since the movable pulley is supported by two segments of the string, for every unit distance block $M$ moves down, it requires 2 units of string length. $$ a_{rel, m} = 2 \cdot a_{rel, M} $$ Let $a_{rel}$ be the acceleration of $m$ upwards relative to the lift. Then the acceleration of $M$ downwards relative to the lift is $a_{rel}/2$.

3. Equations of Motion (Lift Frame)

Let $g’$ be the effective gravity inside the lift. The tension in the string is $T$.

For block $m$: (Forces: Tension $T$ up, $mg’$ down)

$$ T – m g’ = m a_{rel} \quad \dots \text{(1)} $$

For block $M$: (Forces: $2T$ up, $Mg’$ down)

$$ M g’ – 2T = M \left( \frac{a_{rel}}{2} \right) \quad \dots \text{(2)} $$

Multiplying equation (1) by 2 and adding to equation (2) eliminates $T$:

$$ 2(T – mg’) + (Mg’ – 2T) = 2m a_{rel} + \frac{M a_{rel}}{2} $$ $$ Mg’ – 2mg’ = a_{rel} \left( 2m + \frac{M}{2} \right) $$ $$ (M – 2m)g’ = a_{rel} \left( \frac{4m + M}{2} \right) $$

Solving for the relative acceleration:

$$ a_{rel} = \frac{2(M – 2m)g’}{M + 4m} $$

Since $M > 2m$, $a_{rel}$ is positive, meaning $m$ accelerates up and $M$ accelerates down relative to the lift.

4. Analyzing Ground Frame Acceleration

We need both blocks to accelerate in the same direction relative to the ground. We consider two cases for the lift’s acceleration $a$.

Case A: Lift accelerates Upwards ($a \uparrow$)

• Effective gravity: $g’ = g + a$.
• Acceleration of $m$ relative to ground: $\vec{a}_{m,G} = \vec{a} + \vec{a}_{rel}$. (Both are upward)
• Acceleration of $M$ relative to ground: $\vec{a}_{M,G} = \vec{a} + \vec{a}_{rel, M}$. Since $\vec{a}_{rel, M}$ is downward, this becomes $a – \frac{a_{rel}}{2}$ (upward positive).

Since $a_{m,G}$ is definitely upward, for both to be in the same direction, $a_{M,G}$ must also be upward ($>0$).

$$ a – \frac{a_{rel}}{2} > 0 \implies 2a > a_{rel} $$

Substitute $a_{rel}$:

$$ 2a > \frac{2(M – 2m)(g + a)}{M + 4m} $$ $$ a(M + 4m) > (M – 2m)(g + a) $$ $$ aM + 4am > Mg + Ma – 2mg – 2ma $$ $$ 4am + 2ma > Mg – 2mg $$ $$ 6ma > (M – 2m)g $$

Case B: Lift accelerates Downwards ($a \downarrow$)

• Effective gravity: $g’ = g – a$. (Note: For string to be taut, $g’ > 0 \implies a < g$).
• Acceleration of $M$ relative to ground: Vector sum of lift (down) and relative motion (down). It is definitely downward.
• Therefore, for the condition to hold, block $m$ must also accelerate downward.

Taking downward as positive for this calculation:

• $a_{lift} = a$ (down)
• $a_{rel}$ is upward relative to lift.
• $a_{m,G} = a_{lift} – a_{rel} = a – a_{rel}$ (downward).

We require $a_{m,G} > 0$ (downward):

$$ a > a_{rel} $$ $$ a > \frac{2(M – 2m)(g – a)}{M + 4m} $$ $$ a(M + 4m) > 2(M – 2m)(g – a) $$ $$ aM + 4am > 2Mg – 2Ma – 4mg + 4ma $$ $$ aM + 2Ma > 2g(M – 2m) $$ $$ 3Ma > 2g(M – 2m) $$