Question

Blocks B and C are connected by a single inextensible cable, with this cable being wrapped around pulleys at D and E. In addition, the cable is wrapped around a pulley attached to block A as shown. Assume the radii of the pulleys to be small. Blocks B and C move downward with speeds of $V_B=6ft/s$ and $V_C=18ft/s,$ respectively. Determine the velocity of block A when $S_A=4ft.$

• A. $6ft/s$
• B. $15ft/s$
• C. $18ft/s$
• D. $20ft/s$

Answer 1

The correct option is B $15ft/s$

Since the cable is inextensible, the length of the cable remains constant.
$\Rightarrow l_1+l_2+l_3+l_4=\text{constant}$ ---- (1)
$\Rightarrow \frac{d{l}_1}{dt}+\frac{d{l}_2}{dt}+\frac{d{l}_3}{dt}+\frac{d{l}_4}{dt}=0$ --- (2)

From the question diagram, $l_2=l_3=\sqrt{S_A^2+3^2}$
Substituting in eqn (1),
$S_C+\sqrt{S_A^2+3^2}+\sqrt{S_A^2+3^2}+S_B=\text{constant}$
Equation $\left(2\right)$ becomes,
$\Rightarrow \frac{d{S}_c}{dt}-2\frac{d}{dt}\sqrt{S_A^2+3^2}+\frac{d{S}_B}{at}=0$
(negative sign because $S_A$ is decreasing as A is moving up)
$\Rightarrow V_C-\frac{2.S_A}{\sqrt{S_A^2+3^2}}\frac{d{S}_A}{dt}+V_B=0$
$[\because \frac{d{S}_c}{dt}=V_c\frac{d{S}_B}{dt}=V_B,\&\frac{d{S}_A}{dt}=V_A]$

Putting values $V_C=6ft/s,V_B=18ft/s$
$S_A=4ft$, we get
$6-\frac{8}{5}{V}_A+18=0$
$V_A=15ft/s$