## Engineering Mechanics: Statics & Dynamics (14th Edition)

$V_{C} = 0.5\frac{m}{s}$ Block B is traveling upwards.
Formulate two equations for both lines: (1) $l_{1} = S_{A}+2\times S_{B}$ (2) $l_{2} = 2\times (S_{C}-S_{B})+S_{B}$ Take the derivative of both equations to obtain the respective velocities of components $S_{A}$, $S_{B}$, and $S_{C}$: (3) $0 = V_{A} + 2\times V_{B}$ (4) $0 = 2 \times (V_{C} - V_{B}) + V_{B} = 2\times V_{C}-V_{B}$ We are given that $V_{A}$ is $2\frac{m}{s}$. Plug this value into equation (3) to obtain $V_{B}$: $0 = 2 + 2\times V_{B}$ $-2 = 2\times V_{B}$ $V_{B} = -1\frac{m}{s}$ Since we plugged in a positive $V_{A}$ into equation (3), we are considering the downward direction to be positive. Our negative answer for $V_{B}$ means that the velocity of rope segment B is going upwards. The problem requires us to find $V_{C}$. Plug the value found for $V_{B}$ into equation (4): $0=2\times V_{C}-(-1)$ $-2\times V_{C} = 1$ $V_{C} = -0.5\frac{m}{s}$ Once again, the negative answer means that rope segment C is going upwards.