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find velocity of A

2 stands B, with pulleys mounted on them move towards each other with velocity V A rope is passing over the pulleys and a block A is attached to the rope Find the velocity of A when rope is at angle theta Please refer figure for any doubt in the language of question.

MY ATTEMPT this is my attempt

I was able to find 2 equations but there are 3 variables, please help me in finding the 3rd equation.

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  • $\begingroup$ Try energy conservation to find another equation. $\endgroup$
    – Tausif Hossain
    Commented Jul 1, 2017 at 14:28
  • $\begingroup$ @TausifHossain Thanks for the hint, but I haven't been taught energy conservation yet. $\endgroup$
    – Gem
    Commented Jul 1, 2017 at 14:28
  • $\begingroup$ please can anybody help $\endgroup$
    – Gem
    Commented Jul 1, 2017 at 16:10

1 Answer 1

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You should simplify your equations - get rid of $\theta$. Next, use the symmetry of the situation to look at just one pulley moving towards the middle of the room. The other quantity we need is the slack in the rope - the difference between the width of the room and the length of the rope. Let's call that difference $2S$ (S for "slack").

The hypotenuse (rope hanging from pulley to weight) will have length $S+x$, and this results in a height $y$. So now we know

$$(S+x)^2 + x^2 = y^2\\ S^2 + 2Sx + 2x^2 = y^2\\ (2S+4x)dx = 2y dy\\ \frac{dy}{dx}=\frac{S+2x}{y}$$

If we have the rope at a certain height $y$ for a given $x$, we can compute $S$. And now you have what you need so solve this (you don't need to solve for x and y; you just need the relationship - that is, the derivative of y with respect to x).

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  • $\begingroup$ shouldn't the hypotenuse be S-x $\endgroup$
    – Gem
    Commented Jul 1, 2017 at 14:45
  • $\begingroup$ I am not able to understand your solution can you please illustrate it in a figure. $\endgroup$
    – Gem
    Commented Jul 1, 2017 at 14:59

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