Question

In the figure shown, lower pulley is free to move in vertical direction only. Block A is given a uniform velocity u as shown, what is velocity of block B as a function of angle $\theta$.

• A. u cosθ
• B.
• C.
• D.

Answer 1

The correct option is C

2 strings to 2 constraints:

$x_A+(x_{p_2}-x_{p_1})=L_1$

differentiating,

$U+0-V_{p_1}=0$

Distance of pulley 2 is fixed with respect to our reference point.

$v_{p_1}=u$

so this means pulley 1 goes upwith velocity $u$.

Now for string 2

$x_{p_1}+\sqrt{x_B^2+x_{p_1}^2}=l^2$

Differentiating

$u+\frac{1}{2}\frac{(2xB\frac{d{x}_B}{dt}+2x_{P_1}\frac{d{x}_{P_1}}{dt})}{\sqrt{x_B^2+x_{P_1}^2}}=0$

$\frac{d{x}_B}{dt}=-v_B$ $\frac{d{x}_{P_1}}{dt}=u$

$u-v_B\frac{x_B}{\sqrt{x_B^2+x_{P_1}^2}}+\frac{u{x}_{P_1}}{\sqrt{x_B^2+x_{P_1}^2}}=0$

$u+-v_{B}cos\theta +usin\theta =0$

$u(1+sin\theta )=v_{B}cos\theta$

$v_B=u\frac{(1+sin\theta )}{cos\theta }$