In the illustrated arrangement, the ends P and Q of an inextensible...

Let the vertical distance of mass $M$ from the horizontal line joining the pulleys be $y and the horizontal distance between the vertical line passing through$M$and either pulley be$x$.<br><br>The length of the string segment between the pulley and the mass$M$is$l$, given by:<br>$l^2 = x^2 + y^2$<br><br>Differentiating with respect to time$t$:<br>$2l \dfrac{dl}{dt} = 2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt}$<br><br>Since the pulleys are fixed,$x$is constant, so$\dfrac{dx}{dt} = 0$.<br><br>$l \dfrac{dl}{dt} = y \dfrac{dy}{dt}$<br><br>The ends$P$and$Q$are pulled downwards with speed$u$, meaning the length$l$decreases at the rate$u$. Thus,$\dfrac{dl}{dt} = -u$.<br><br>The mass$M$moves upwards, so its speed is$v = -\dfrac{dy}{dt}$.<br><br>Substituting these rates into the equation:<br>$l(-u) = y(-v)$<br><br>$v = u \left(\dfrac{l}{y}\right)

From the given figure, the angle $\theta$ is between the vertical and the string, which gives $\cos\theta =$\dfrac{y}{l}.

Therefore, the upward speed of the mass $M$ is:
$v =$\dfrac{u}{\cos\theta}