For the given figure,find the speed of block | vedclass

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(A) Let $h$ be the vertical distance from the pulley to the horizontal path of the ring,and $x$ be the horizontal distance of the ring from the vertic

Vedclass · Accepted Answer

$2\sqrt{3} \, m/s$

Vedclass · Answer

(A) Let $h$ be the vertical distance from the pulley to the horizontal path of the ring,and $x$ be the horizontal distance of the ring from the vertical line passing through the pulley.The length of the string segment from the ring to the pulley is $L = \sqrt{x^2 + h^2}$.The total length of the string is constant,so the speed of block $A$ is equal to the rate at which the length of the string segment $L$ decreases.$v_A = -\frac{dL}{dt} = -\frac{d}{dt}(\sqrt{x^2 + h^2}) = -\frac{1}{2\sqrt{x^2 + h^2}} \cdot 2x \cdot \frac{dx}{dt} = -\frac{x}{L} \cdot v_{ring}$.From the geometry,$\sin 	heta = \frac{x}{L}$.Therefore,$v_A = v_{ring} \cdot \sin 	heta$.Given $v_{ring} = 4 \, m/s$ and $	heta = 60^\circ$,we have:$v_A = 4 \cdot \sin(60^\circ) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \, m/s$.

For the given figure,find the speed of block $A$ when $\theta = 60^\circ$. The ring moves horizontally with a constant speed of $4 \, m/s$.

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