Mechanics - 3

Based on the given information, $v_A - v_C = v_{AC} = 300 \implies v_C = v_A - 300\\ v_B - v_A = v_{BA} = -200 \implies v_B = v_A - 200$

Now call the lower pulley $P$ and note that $v_P = -v_A \\ v_B - v_P = v_{BP} = -v_{CP} = v_{PC} = v_P - v_C$ so that $(v_A-200) + v_A = -v_A - (v_A - 300) \implies v_A = \frac{500}{4} = \boxed{125}$ where the direction is $\uparrow$ because we have oriented all positive measurements to be up.

Let us denote $\uparrow$ direction as $+$ and $\downarrow$ direction as $-$ . Then we have $\begin{cases} v_{AC} = v_A - v_C = 300 & ...(1) \\ v_{BA} = v_B - v_A = -200 & ...(2) \end{cases}$

From $(1)+(2): v_B - v_C = v_{BC} = 100$ . Since, with respect to the right pulley $P$ , $v_{BP} = - v_{CP}$ , $\implies -2v_{CP} = 100$ or $v_{CP} = - 50$ . We know that the velocity of the right pulley, $v_P = - v_A$ . Then we have:

$\begin{aligned} v_{AC} & = 300 \\ v_A - v_C & = 300 \\ v_A - (v_P + v_{CP}) & = 300 \\ v_A - (-v_A - 50) & = 300 \\ 2v_A & = 300 - 50 \\ \implies v_A & = \boxed{\text{125 m/s}\uparrow}\end{aligned}$

Let us refer all velocities to $A$ . $V_{BA}=200 \text{ m/s} \downarrow$ .

Being $v_{AC}=300 \text{ m/s} \uparrow$ , the veloctity of $C$ referred to $A$ is $v_{CA}=300 \text{ m/s} \downarrow$ , and $v_{AA}=0$ , of course.

The velocity of $P_2$ referred to A must be $v_{P_2A}=\dfrac{v_{BA}+v_{CA}}{2}=\dfrac{300 \text{ m/s} \downarrow+200 \text{ m/s} \downarrow}{2}=250 \text{ m/s} \downarrow$ , since $B$ approaches the pulley at the same speed $C$ receeds from it.

Likewise, the velocity of $P_1$ referred to $A$ must be $v_{P_1A}=\dfrac{v_{AA}+v_{P_2A}}{2}=\dfrac{0+250 \text{ m/s} \downarrow }{2}=125 \text{ m/s} \downarrow$ .

Since we know $P_1$ must be zero in earth's reference frame, we must add $125 \text{ m/s} \uparrow$ to all velocities:

$v_C=175 \text{ m/s} \downarrow$ , $v_B=75 \text{ m/s} \downarrow$ and $\boxed{v_A=125 \text{ m/s} \uparrow}$