NLM BYU 15

1. Calculating Tension ($T$)

Consider the massless Pulley A. The upward force $F$ is balanced by the tension $T$ acting downwards on both sides of the pulley string. $$2T = F$$ Given $F = 20$ N: $$T = \frac{20}{2} = 10 \text{ N}$$

2. Acceleration of the Blocks

For Mass $m_2$ (Right Side):
Forces acting on $m_2$ are Tension $T$ (upwards) and Weight $m_2g$ (downwards). $$T – m_2g = m_2 a_2$$ $$10 – (2.0 \times 10) = 2.0 a_2$$ $$10 – 20 = 2 a_2 \implies -10 = 2 a_2$$ $$a_2 = -5 \text{ m/s}^2$$ The negative sign indicates acceleration is downwards.

For Mass $m_1$ (Left Side):
Mass $m_1$ is attached to Pulley B. The string passes under Pulley B, so the tension $T$ pulls Pulley B upwards from both sides.
Total Upward Force = $2T$.
Total Downward Force = $m_1g$. $$2T – m_1g = m_1 a_1$$ $$2(10) – (1.0 \times 10) = 1.0 a_1$$ $$20 – 10 = a_1$$ $$a_1 = 10 \text{ m/s}^2 \quad (\text{upwards})$$

3. Acceleration of Pulley A

We need to find the acceleration of Pulley A ($a_A$). We use the constraint equation derived from the constant length of the string.
Let upward direction be positive. The constraint relation relating the accelerations of the central pulley (A) and the two ends (Block 2 and Pulley B) is: $$2a_{pulley} = a_{left\_branch} + a_{right\_branch}$$ Here, the “pulley” is A.

• Right branch is connected to $m_2$, so $a_{right} = a_2$.
• Left branch is connected to Pulley B. Since Pulley B consumes string from the loop to move up, the effective contribution is $2a_1$. (Alternatively derived via $L = \text{const}$ method: $2a_A = 2a_1 + a_2$).

Substituting the values ($a_1 = +10$, $a_2 = -5$): $$2a_A = 2(10) + (-5)$$ $$2a_A = 20 – 5 = 15$$ $$a_A = 7.5 \text{ m/s}^2$$