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In this question, I assumed the two side strings will also move with velocity $u$ and broke the vectors along the direction of motion of M and I got the velocity of the mass as $u\cos\theta +u\cos\theta = 2u\cos\theta$. (Both velocities of the two strings vector resolved and added)

But the answer was wrong, as they broke the velocity of the mass, say $v$, into $v\cos\theta$ along the direction of the string and equated $$v\cos\theta = u$$ $$v=u\sec\theta$$

However, I do not find why my method of approach is wrong and why theirs should be correct. Please explain. Thank You!

Vishnu
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rash
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1 Answers1

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The answers is the related links were quite complicated so I have provided a simple explanation here.

The mistake here is that you are adding the velocity components as if they were a force. If two of your friends ram into you from behind with velocity $x$, you would also go at $x$ and not $2x$. In such types of problems, try to find the invariant i.e value which stays constant throughout the problem.

Here, the distance between the two pulleys remains constant. Let us assume it to be $2d$. Let distance between ceiling and mass be $h$ and length of string from the mass till pulley be $l$. From Pythagoras theorem, we have

$$d^2+h^2=l^2$$ Implicitly differentiating, we get $$h\cdot {\frac{dh}{dt}} = l\cdot {\frac{dl}{dt}}$$ We know that ${\frac{dl}{dt}} = u$ and ${\frac{dh}{dt}}$ is being asked by the question.

Implicitly differentiating, we get $$\frac{dh}{dt} = \frac{l}{h}\cdot u=u\sec\theta $$

Sam
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