Solution

Engineering Mechanics (Dynamics)

Chapter 12

Exercise 203

Given: $v_B = 10$ m/s, $a_A = 3$ m/s$^2$. From the picture we can see that the total length of the cord is:

$$\begin{align*} l_1 &= s_B + 2(s_B - s_C) \qquad 0 = 3v_B - 2v_C \\ l_2 &= s_C + s_A \qquad v_A = - v_C \\ v_A &= - \frac{3}{2} v_B \\ a_A &= - 1.5 a_B \\ a_B &= - 0.67 a_A \\ &= -0.67 \cdot 3 \\ &= -2 \, \text{m/s}^2. \end{align*}$$

Answer

$5$ s.