NLM O32

Let the tension in the string be $T$. Since the string is massless and the pulley is frictionless, the tension is uniform throughout the string. The external pulling force is applied directly to the free end P, so:

$$T = F$$

Block $3m$: The string is anchored to its right face (pulling right with $T$), and a loop passes around its pulley. The string enters the pulley from the right and leaves to the right. Thus, the pulley exerts a force of $2T$ to the right. Total force:

$$F_{3m} = T + 2T = 3T \quad (\text{Right})$$

$$a_{3m} = \frac{3T}{3m} = \frac{F}{m} \quad (\text{Right})$$

Block $2m$: The string passes around the pulley attached to this block. The string segments go to the left (towards $3m$). Thus, the pulley feels a force of $2T$ to the left.

$$F_{2m} = 2T \quad (\text{Left})$$

$$a_{2m} = \frac{2T}{2m} = \frac{F}{m} \quad (\text{Left})$$

We need the acceleration of the free end P, denoted as $a_P$. Let’s define the positions of the blocks and point P relative to a fixed origin on the left:

• $x_3$: Position of $3m$ block
• $x_2$: Position of $2m$ block
• $x_P$: Position of free end P

The total length of the string $L$ is constant. The string consists of three segments:

1. From $3m$ anchor to $2m$ pulley: length is $(x_2 – x_3)$.
2. From $2m$ pulley back to $3m$ pulley: length is $(x_2 – x_3)$.
3. From $3m$ pulley to free end P: length is $(x_P – x_3)$.

Equation for string length (ignoring pulley radii constants):

$$L = (x_2 – x_3) + (x_2 – x_3) + (x_P – x_3)$$

$$L = 2x_2 – 3x_3 + x_P$$

Differentiating the length equation twice with respect to time (setting $\ddot{L} = 0$):

$$0 = 2a_2 – 3a_3 + a_P$$

$$a_P = 3a_3 – 2a_2$$

We must be careful with signs. Let the positive direction be to the right.

• $a_3 = +\frac{F}{m}$ (Accelerates right)
• $a_2 = -\frac{F}{m}$ (Accelerates left)

Substituting these values:

$$a_P = 3\left(\frac{F}{m}\right) – 2\left(-\frac{F}{m}\right)$$

$$a_P = \frac{3F}{m} + \frac{2F}{m} = \frac{5F}{m}$$