In the arrangement shown in figure the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed
Text Solution
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As P and Q fall down, the length l decreases at the rate of U m/s.
From the figure, l^{2} = b^{2} + y^{2}
Differentiating with respect to time
\(2l \times \frac{dl}{dt} = 2b \times \frac{db}{dt} + 2y \times \frac{dy}{dt}\) \(\left( \mathrm{As} \frac{db}{dt} = 0, \frac{dl}{dt} = U \right)\)
⇒ ⇒ \(\frac{dy}{dt} = \left(\frac{y}{l}\right) \times \frac{dl}{dt}\) \(\Rightarrow \frac{dy}{dt} = \left( \frac{1}{\cos \theta} \right) \times U = \frac{U}{\cos \theta}\)
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