Solution

Engineering Mechanics (Dynamics)

Chapter 12

Exercise 202

Given: $v_A = 3$ m/s. We now set equations for the length of the entire cord:

$$\begin{align*} l_1 &= 2s_B + s_D \qquad 0 = 2v_B + v_D \\ l_2 &= s_A - s_D + L- s_D \qquad 0 = v_A - 2v_D \\ v_A &= 2 v_D \\ v_A &= -4 v_B \\ v_B &= -\frac{v_A}{4} \\ &= -\frac{3}{4} \\ v_B &= -0.75 \, \text{m/s.} \end{align*}$$

Answer

$0.75$ m/s