Question

Determine
the speed with which block $B$ rises in Fig. 6.183 if the end of the cord at
$A$ is pulled down with a speed of 2 ms${}^{-1}$.

Answer 1

Given:

$v=2m/s$

we then write the equation for the total length of the cord

$S_a\left(t\right)-s_c\left(t\right)+s_b)t-s_c(t)+s_b(t)=constant$

$s_c\left(t\right)+s_b\left(t\right)=constant$

When we take derivatives of cord equations:

$\frac{d}{dt(s_a-s_c+s_b-s_c+s_b)}=0$

$\frac{d}{dt(s_c+s_b)}=0$

we get

$v_a+2v_b-2v_c=0$-----(1)

and

$v_b+v_c=0$-----(2)

solve the second of the two equations for velocities: to eliminate$V_c$

$v_a+2v_b-2v_c=0$

and

$vb+vc=0$

we get

$vc=-vb$

substituting this equation in equation we get $v_b=-\frac{1}{4V_a}=-\frac{2}{4}=-\frac{1}{2}$ m/s

$\therefore vb=-0.5m/s$