Question 5 of 6: Cable-Pulley Dependent Motion

Question 5: Cable-Pulley Dependent Motion (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single continuous cable runs from block $A$, up and over a left FIXED pulley, down and around a movable pulley (whose axle carries block $B$), up and over a right FIXED pulley, and down to block $C$; $v_A=5\text{ m/s}$ upward, $v_C=2.5\text{ m/s}$ downward.

Find. The speed (and direction) of block $B$.

Approach. Write the cable's total length as a sum of segment lengths measured downward from the two fixed pulleys, noting the movable pulley $B$ is held up by TWO parallel cable segments, then differentiate the (constant) total length and substitute the two known speeds.

1. Write the constant-length equation. Measuring each segment's length $s$ downward from its fixed pulley (so a downward-moving block has $\dot s>0$), the cable length is $$L = s_A + 2s_B + s_C = \text{const}$$ (the factor 2 on $s_B$ because the movable pulley is supported by two parallel runs of the same cable).
2. Differentiate and substitute. $\dot s_A+2\dot s_B+\dot s_C=0$. Block $A$ moves UP at 5 m/s $\Rightarrow\dot s_A=-5\text{ m/s}$; block $C$ moves DOWN at 2.5 m/s $\Rightarrow \dot s_C=+2.5\text{ m/s}$: $$-5+2\dot s_B+2.5 = 0$$
3. Solve for $\dot s_B$. $$\dot s_B = \frac{5-2.5}{2} = \boxed{1.25\text{ m/s, downward}}$$