Solution

Engineering Mechanics (Dynamics)

Chapter 12

Exercise 206

Given: $v_A = 6$ m/s. Length of the cord and the velocity of the body B is:

$$\begin{align*} l &= s_A + s_B + (s_B - a) + 2(s_B - a - b) \\ l &= s_A + 4 s_B - 3a - 2b \\ 0 &= v_A + 4 v_B \\ v_B &= -0.25 v_A\\ v_B &= -0.25 \cdot 6 \\ v_B &= -1.5 \, \text{m/s.} \end{align*}$$

Velocity of the body B is equal to $v_B = 1.5$ m/s to the left.

Answer

$1.5$ m/s to the left.