F$12-40.$ Determine the velocity of block $A$ if end $B$ of the

rope is pulled down with a speed of $6\frac{m}{s}.$

Explain why the red part of the

cable is fixed.

Explain why the blue part are

not fixed.

Provide visuals and key

summaries.

Explain why there's no fixed green

cables.

Explain why the red part of the cable

holding $w$ is not accounted for in the

equation.

why do we separate the light green

cables on one side and the dark green

cable on the other side?

Why was there two equations for the

lengths?

why is $vp=1\frac{m}{s}$ and $v1=-1\frac{m}{s}?$

Hint: make assumptions

equation $2$ :

$3s_2+s_1=c_1$

$\frac{d}{dt}(3s_2)+\frac{R}{dt}(s_1)=\frac{d}{dt}(c_1)$

$3V_2+V_1=0$

$3V_2=-V_1$

$3v_2=-(-1)$

$3V_2=1$

$V_2=\frac{1}{3}$

$2s_3+(s_3-s_2)=c_2$

$3s_3-s_2=c_2$

$\frac{d}{dt}(3s_3)-\frac{d}{dt}(s_2)=C_2$

$3V_3-V_2=0$

$3v_3=V_2$

$V_3=\frac{V_2}{3}$

$=\frac{1}{3}*\frac{1}{3}=\frac{1}{9}\frac{m}{s}$