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Cylinder A has a constant downward speed of 1 m/s. compute the velocity of cylinder B for θ= 30 and θ= 45. The spring is in tension throughout the motion range of interest, and the pulleys are connected by the cable of fixed length.

Figure:

https://www.dropbox.com/s/laifibai6tuiht1/Engineering%20Mechanics%20Dynamics.png

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Let the distance from A and the break point be $A(t)$ and distance from B and the break point be $B(t)$($t$ is time variable). And the distance between A & B be $l$(which is constant).

Then $A(t)=-l\sin\theta(t)$, $B(t)=l\cos\theta(t)$. ($\theta$ decreases. $90^\circ$~$0^\circ$. That why the A is negative.)

$\frac{\mathrm{d}}{\mathrm{dt}}A(t)=-l\cos\theta(t)\frac{\mathrm{d}\theta}{\mathrm{dt}}$

Where $\mathrm{d}A/dt$ is the speed of A.(1m/s)

Then we know that $\mathrm{d\theta}/\mathrm{d}t=-1/l\cos\theta$

The velocity of B,

$\frac{\mathrm{d}}{\mathrm{dt}}B(t)=-l\sin\theta(t)\frac{\mathrm{d}\theta}{\mathrm{dt}}=l\sin\theta(t)\frac{1}{{l\cos\theta}}=\tan\theta$.

$\tan30^{\circ}=\sqrt{3}/2$ and $\tan45^{\circ}=1$.

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    $\begingroup$ Dear user28936. Welcome to Phys.SE. In the future please be careful not to give away full homework solutions, cf. Phys.SE homework policy. $\endgroup$
    – Qmechanic
    Sep 13, 2013 at 13:45
  • $\begingroup$ @Qmechanic Understood. $\endgroup$
    – user28936
    Sep 13, 2013 at 14:32

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