Question

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

• A. $u\cos \theta$
• B. $\frac{u}{\cos \theta }$
• C. $\frac{u[1+\sin \theta ]}{\cos \theta }$
• D. $\frac{u[1+\cos \theta ]}{\sin \theta }$

Answer 1

The correct option is C $\frac{u[1+\sin \theta ]}{\cos \theta }$

Let in a small time $dt$, displacement of $A$ is $dx$ and displacement of $B$ is $dy$. Increase in length $d{l}_1=dx$.

$l_2^2=y^2+l_1^2$

differentiating both sides

$2l_{2}d{l}_2+2ydy+2l_{1}d{l}_1$

$d{l}_2=\frac{y}{l_2}dy+\frac{l_1}{l_2}d{l}_1$

Hence, Increase in the length $l_2$

$d{l}_2=dx\sin \theta +dy\cos \theta$
But net increase in the length should be zero, so
$dx+dx\sin \theta +dy\cos \theta =0$
$\Rightarrow dy=-\frac{(1+\sin \theta )}{\cos \theta }$
$\Rightarrow \frac{dy}{dt}=-\left(\frac{1+\sin \theta }{\cos \theta }\right)\frac{dx}{dt}$
$\Rightarrow v=-\left(\frac{1+\sin \theta }{\cos \theta }\right)u$

The negative sign indicates that $y$ is decreasing