Movable Pulleys (Leaving Cert Applied Maths): Revision Notes

Worked Example: Basic Movable Pulley Analysis

Step 1: Determine system motion First, we determine the system motion by comparing forces:

• Pulley A has two strings pulling it upward
• Mass B has one string pulling it upward
• Since both masses are 10 kg, we compare: $10 ÷ 2 = 5$ versus $10 ÷ 1 = 10$
• As $10 > 5$, mass B will move downward and pulley A will move upward

Step 2: Force analysis

For Pulley A:

• Upward forces: $2T$ (two string sections)
• Downward force: $10g$ (weight)
• Using $F = ma$: $2T - 10g = 10a$

For Mass B:

• Upward force: $T$ (tension)
• Downward force: $10g$ (weight)
• Using $F = ma$: $10g - T = 10(2a)$

Note that if pulley A accelerates upward at rate a, then mass B accelerates downward at rate 2a due to string constraints.

Step 3: Mathematical solution

From the two equations:

• Equation 1: $T - 5a = 5g$
• Equation 2: $-T - 20a = -10g$

Solving simultaneously:

• Adding equations: $-25a = -5g$
• Therefore: $a = \frac{g}{5}$ m/s²
• Substituting back: $T = 6g$ N

Worked Example: Movable Pulley with Friction

Step 1: Force analysis for each component

6 kg Mass:

• Horizontal: $T - F = 6a$ (where $F = \text{friction} = \mu R\_1 = 0.5 × 6g = 3g$)
• Vertical: $R\_1 = 6g$
• Therefore: $T - 3g = 6a$ ... (Equation I)

10 kg Mass:

• Horizontal: $T - F = 10b$ (where $F = \text{friction} = \mu R\_2 = 0.5 × 10g = 5g$)
• Vertical: $R\_2 = 10g$
• Therefore: $T - 5g = 10b$ ... (Equation II)

20 kg Pulley:

• Using $F = ma$: $20g - 2T = 20\left(\frac{a+b}{2}\right)$
• Simplifying: $10g - T = 5a + 5b$ ... (Equation III)

Step 2: Mathematical solution

Substituting Equations I and II into Equation III:

• $10g - T = 5a + 5b$
• $10g - T = \frac{5(T-3g)}{6} + \frac{5(T-5g)}{10}$

Solving this system of equations:

• $T = 63$ N
• $a = 5.6$ m/s²
• $b = 1.4$ m/s²
• Pulley acceleration = $\frac{a+b}{2} = 3.5$ m/s²