Question

In the arrangement shown in figure, find the tensions in the rope and accelerations of the masses $m_1$ and $m_2$ and pulleys $P_1$ and $P_2$ when the system is set free to move. Assume the pulleys to be massless and strings are light and inextensible.

Answer 1

$a_{P1} = \frac{g(3m_1 - 2m_2)}{9m_1 + 4m_2} \text{ (down)}, a_{P2} = \frac{g(2m_2 - 3m_1)}{9m_1 + 4m_2} \text{ (down)}, a_1 = \frac{3g(3m_1 - 2m_2)}{9m_1 + 4m_2} \text{ (down)}, a_2 = \frac{2g(2m_2 - 3m_1)}{9m_1 + 4m_2} \text{ (down)}, T_A = \frac{20m_1m_2g}{9m_1 + 4m_2}, T_B = \frac{10m_1m_2g}{9m_1 + 4m_2}, T_C = \frac{15m_1m_2g}{9m_1 + 4m_2}$

Answer 2

The problem involves a system of pulleys and masses. We need to find the tensions in the ropes and the accelerations of the masses and pulleys. We assume the pulleys are massless and the strings are light and inextensible.

Let's identify the distinct strings and their tensions:

1. String A: Passes over the top fixed pulley (let's call it $P_0$) and connects the axles of pulley $P_1$ and pulley $P_2$. Let the tension in this string be $T_A$.

2. String B: Passes over pulley $P_1$. One end is attached to mass $m_1$, and the other end is attached to the axle of pulley $P_2$. Let the tension in this string be $T_B$.

3. String C: Passes over pulley $P_2$. One end is attached to mass $m_2$, and the other end is attached to the fixed support (ceiling). Let the tension in this string be $T_C$.

Let's define accelerations:

Let $a_{P1}$ be the acceleration of pulley $P_1$ downwards.

Let $a_{P2}$ be the acceleration of pulley $P_2$ downwards.

Let $a_1$ be the acceleration of mass $m_1$ downwards.

Let $a_2$ be the acceleration of mass $m_2$ downwards.

1. Free Body Diagrams and Tension Relations (for massless pulleys):

• Pulley $P_1$:

Forces: $T_A$ (upwards), $T_B$ (downwards from left side), $T_B$ (downwards from right side).

Since $P_1$ is massless, the net force on it is zero:

$T_A - T_B - T_B = 0 \implies T_A = 2T_B$ (Eq. 1)

• Pulley $P_2$:

Forces: $T_A$ (upwards from string A), $T_B$ (upwards from string B), $T_C$ (downwards from left side), $T_C$ (downwards from right side).

Since $P_2$ is massless, the net force on it is zero:

$T_A + T_B - T_C - T_C = 0 \implies T_A + T_B = 2T_C$ (Eq. 2)

Substitute $T_A = 2T_B$ from Eq. 1 into Eq. 2:

$2T_B + T_B = 2T_C \implies 3T_B = 2T_C$ (Eq. 3)

From these relations, we can express all tensions in terms of one base tension, say $T_B$:

$T_A = 2T_B$

$T_C = \frac{3}{2}T_B$

2. Constraint Equations (relating accelerations):

• String A (over fixed pulley $P_0$):

The string connects $P_1$ and $P_2$. For an inextensible string over a fixed pulley, the sum of accelerations of the connected ends (in the direction away from the pulley) is zero.

$a_{P1} + a_{P2} = 0$ (Eq. 4)

This means if $P_1$ moves down, $P_2$ moves up with the same magnitude of acceleration. Let $a_{P1} = a$. Then $a_{P2} = -a$. (Here, positive 'a' means downwards).

• String B (over movable pulley $P_1$):

The string connects $m_1$ and $P_2$. For an inextensible string over a movable pulley, the acceleration of the pulley is the average of the accelerations of the two ends of the string.

$a_{P1} = \frac{a_1 + a_{P2}}{2}$

$2a_{P1} = a_1 + a_{P2}$

$a_1 = 2a_{P1} - a_{P2}$ (Eq. 5)

Substitute $a_{P1} = a$ and $a_{P2} = -a$:

$a_1 = 2a - (-a) = 3a$ (Eq. 6)

• String C (over movable pulley $P_2$):

The string connects $m_2$ and the fixed support (acceleration = 0).

$a_{P2} = \frac{a_2 + 0}{2}$

$a_2 = 2a_{P2}$ (Eq. 7)

Substitute $a_{P2} = -a$:

$a_2 = 2(-a) = -2a$ (Eq. 8)

So, we have the accelerations in terms of $a$:

$a_{P1} = a$ (downwards)

$a_{P2} = -a$ (upwards)

$a_1 = 3a$ (downwards)

$a_2 = -2a$ (upwards)

3. Equations of Motion for Masses:

• Mass $m_1$:

Forces: $m_1g$ (downwards), $T_B$ (upwards).

$m_1g - T_B = m_1 a_1$

$m_1g - T_B = m_1 (3a)$ (Eq. 9)

• Mass $m_2$:

Forces: $m_2g$ (downwards), $T_C$ (upwards).

$m_2g - T_C = m_2 a_2$

$m_2g - T_C = m_2 (-2a)$ (Eq. 10)

4. Solving for $a$ and Tensions:

From Eq. 9: $T_B = m_1g - 3m_1a$

Substitute $T_C = \frac{3}{2}T_B$ into Eq. 10:

$m_2g - \frac{3}{2}T_B = -2m_2a$

Substitute $T_B$:

$m_2g - \frac{3}{2}(m_1g - 3m_1a) = -2m_2a$

$m_2g - \frac{3}{2}m_1g + \frac{9}{2}m_1a = -2m_2a$

Multiply by 2 to clear fractions:

$2m_2g - 3m_1g + 9m_1a = -4m_2a$

$g(2m_2 - 3m_1) = -a(4m_2 + 9m_1)$

$a = \frac{g(3m_1 - 2m_2)}{9m_1 + 4m_2}$

Accelerations:

• Acceleration of pulley $P_1$ ($a_{P1}$):

$a_{P1} = a = \frac{g(3m_1 - 2m_2)}{9m_1 + 4m_2}$ (downwards)

• Acceleration of pulley $P_2$ ($a_{P2}$):

$a_{P2} = -a = \frac{g(2m_2 - 3m_1)}{9m_1 + 4m_2}$ (downwards, meaning upwards if $3m_1 > 2m_2$)

• Acceleration of mass $m_1$ ($a_1$):

$a_1 = 3a = \frac{3g(3m_1 - 2m_2)}{9m_1 + 4m_2}$ (downwards)

• Acceleration of mass $m_2$ ($a_2$):

$a_2 = -2a = \frac{2g(2m_2 - 3m_1)}{9m_1 + 4m_2}$ (downwards, meaning upwards if $3m_1 > 2m_2$)

Tensions:

• Tension $T_B$ (in string over $P_1$):

$T_B = m_1g - 3m_1a = m_1g - 3m_1 \frac{g(3m_1 - 2m_2)}{9m_1 + 4m_2}$

$T_B = m_1g \left( 1 - \frac{3(3m_1 - 2m_2)}{9m_1 + 4m_2} \right)$

$T_B = m_1g \left( \frac{9m_1 + 4m_2 - 9m_1 + 6m_2}{9m_1 + 4m_2} \right)$

$T_B = \frac{10m_1m_2g}{9m_1 + 4m_2}$

• Tension $T_A$ (in string over $P_0$):

$T_A = 2T_B = \frac{20m_1m_2g}{9m_1 + 4m_2}$

• Tension $T_C$ (in string over $P_2$):

$T_C = \frac{3}{2}T_B = \frac{3}{2} \frac{10m_1m_2g}{9m_1 + 4m_2} = \frac{15m_1m_2g}{9m_1 + 4m_2}$

Explanation of the solution:

1. Identify Tensions: The system consists of three distinct string segments with different tensions ($T_A$, $T_B$, $T_C$) due to the arrangement of movable pulleys.

2. Free Body Diagrams for Pulleys: Apply Newton's second law for massless pulleys (net force = 0) to relate the tensions. This yields $T_A = 2T_B$ and $T_A + T_B = 2T_C$, which simplify to $T_A = 2T_B$ and $T_C = \frac{3}{2}T_B$.

3. Kinematic Constraints (Acceleration Relations): Define a consistent direction for accelerations (e.g., downwards positive). Use the property of inextensible strings over pulleys to relate the accelerations of connected objects.

   • For the string over the fixed pulley $P_0$ connecting $P_1$ and $P_2$: $a_{P1} + a_{P2} = 0$. Let $a_{P1} = a$, so $a_{P2} = -a$.

   • For the string over movable pulley $P_1$ connecting $m_1$ and $P_2$: $a_{P1} = \frac{a_1 + a_{P2}}{2}$, leading to $a_1 = 3a$.

   • For the string over movable pulley $P_2$ connecting $m_2$ and the fixed support (zero acceleration): $a_{P2} = \frac{a_2 + 0}{2}$, leading to $a_2 = -2a$.

4. Free Body Diagrams for Masses: Apply Newton's second law ($F = ma$) for each mass.

   • For $m_1$: $m_1g - T_B = m_1 a_1$.

   • For $m_2$: $m_2g - T_C = m_2 a_2$.

5. Solve the System of Equations: Substitute the acceleration relations and tension relations into the equations of motion for the masses. This results in two equations with two unknowns (e.g., $T_B$ and $a$). Solve for $a$ and then back-substitute to find all accelerations and tensions.

Answer:

The accelerations are (downwards positive):

• $a_{P1} = \frac{g(3m_1 - 2m_2)}{9m_1 + 4m_2}$

• $a_{P2} = \frac{g(2m_2 - 3m_1)}{9m_1 + 4m_2}$

• $a_1 = \frac{3g(3m_1 - 2m_2)}{9m_1 + 4m_2}$

• $a_2 = \frac{2g(2m_2 - 3m_1)}{9m_1 + 4m_2}$

The tensions are:

• $T_A = \frac{20m_1m_2g}{9m_1 + 4m_2}$ (Tension in the string over the top fixed pulley, connecting $P_1$ and $P_2$)

• $T_B = \frac{10m_1m_2g}{9m_1 + 4m_2}$ (Tension in the string over $P_1$, connecting $m_1$ and $P_2$)

• $T_C = \frac{15m_1m_2g}{9m_1 + 4m_2}$ (Tension in the string over $P_2$, connecting $m_2$ and the fixed support)