why is velocity of the mass v/ cos θ ? Why not 2v cos θ? If we call the velocity of the mass $u$, then we can say: $u \cos θ = v$ and hence $u= \frac v{\cos θ}$. But why can't we write $v \cos θ= u$ and add the contribution from both of the strings: $u= \frac{2v}{\cos θ}$ ? Why is it only $u= \frac v{\cos θ}$?

Please note that it is not a homework based question. I just want someone to solve the doubt arising in my mind.

• Imagine that there were 1000 masses handing from pulleys, just as $p$, and $q$, in a circle around the central axis (above m). Would you expect the speed to be 1000 v cos θ ? – Alexander Jul 5 '17 at 3:16
• Possible duplicate of Resolving Velocity Components in Constrained Motion or Resolving vectors along axes – sammy gerbil Jul 5 '17 at 11:04
• Kinda looks like homework, why else would there be four answer candidates listed... – Communisty Jul 5 '17 at 12:09

Since there are four choices, we can tell the answer immediately by letting $\theta$ go to $0$ and $\frac{\pi}{2}$. By letting $\theta$ go to $0$, we can eliminate 3) and 4), as is mentioned by @Dawood ibn Kareem; by letting $\theta$ go to $\frac{\pi}{2}$, we can eliminate 1).

Otherwise we can get this result with some calculation. In order to simplify the description, let's assume the pulleys are infinitesimal (this makes no difference to the problem). Denote the distance between $A$ and $B$ by $2d$, the distance between $A$ and $P$ by $y$, the vertical distance between $A$ and the suspension point by $x$. Now from that the string is unstretchable, we know $$\sqrt{x^2+d^2}+y=const.$$ Taking derivative with respect to time $t$, we have $$\frac{x}{\sqrt{x^2+d^2}}\dot{x}+\dot{y}=0.$$ That is $$|\dot{x}|\cos\theta=|\dot{y}|,$$ where $u=|\dot{x}|$, $v=|\dot{y}|$.

Imagine what would happen if the two pulleys were very close together, so that $\theta$ is very close to zero, and $\cos\theta$ is very close to one.

In that case, the mass in the middle will rise with a speed very close to $v$ - and it rises at the same rate no matter how many of the other masses are present. Therefore, its speed can't possibly be $\frac{2v}{\cos\theta}$. It must be $\frac{v}{\cos\theta}$.

• Thank you but I would like to have a simple mathematical answer without using intuition. – Rahul Raman Jul 4 '17 at 9:30
• I'm not sure what mathematics you expect. There is simply no reason to multiply anything by 2 - and if you think there is, then it is your intuition that is amiss, not your mathematics. Consider this (very similar) question. A team of 6 dogs pulls a sled. Each dog runs at 10 m/s. Explain using only mathematics, and no intuition why the sled doesn't go at 60 m/s. You can't give such an answer, because the expectation that the sled would go at 60 m/s is based on incorrect intuition, not on incorrect mathematics. – Dawood ibn Kareem Jul 4 '17 at 10:16
• OK I got your point. But then please tell me, why couldn't 'v' be resolved as v cos θ and hence u= v cos θ( cancelling the sine components of the velocities of the two strings) . I mean, how can we say at a glance that the answer would be v/ cos θ ? – Rahul Raman Jul 4 '17 at 12:15
• I guess the answer to that involves comparing the change in height of one of the outer masses to the change in height of the centre mass, over a short time period. It's not hard to see that the centre mass will always move more than the outer masses, and since $\cos\theta$ is between 0 and 1, we must have to divide rather than multiplying. – Dawood ibn Kareem Jul 5 '17 at 3:08
• Of course, if the question were not multiple choice, we'd actually have to do some mathematics. But the beauty of multiple choice questions is that you can usually survive on intuition alone. – Dawood ibn Kareem Jul 5 '17 at 3:09

(a) $u \cos\theta = v$ is correct. The point of intersection of the 3 strings (which I shall call Y) is moving upwards with speed $u$. This point Y is at the end of the two strings attached to P and Q. Each string AY and BY is shortening at the speed $v$, so the component of velocity of Y along the direction of each string must be $v$. Therefore $v=u\cos\theta$ is correct.

(b) $u=v\cos\theta$ is not correct. The point at which the string is attached to pulley P (I shall call this point A) is moving with velocity $v$ at angle $\theta$ to the vertical. However, the motion of Y is not the same as the motion of A. As well as moving with speed $v$ in the direction AY, the string AY is also rotating about point A. The angle $\theta$ is changing. So the motion of Y (at the other end of the string AY) is the sum of the radial component along AY and a tangential component perpendicular to AY. Writing $u=v\cos\theta$ ignores this tangential motion of Y perpendicular to AY.

The difference between (a) and (b) is that in (a) the point Y has no motion perpendicular to the vertical, whereas in (b) the point Y does have a component of motion perpendicular to AY.

If the section of string AY did not rotate but kept a constant angle $\theta$ with the vertical, then the points A and Y would have the same velocity, in terms of direction as well as magnitude. In this case $u=v\cos\theta$ would be correct. However, Y would not be moving vertically in this case. It would be moving along AY. It cannot move along BY and AY at the same time, because then Y would have to split apart.

Note that if the masses P and Q do not move with the same speed $v$ (which could happen if they have different weights), or if the angles $\theta$ on each side of the vertical are different (which could happen if Y is not on the mid-line between A and B), then the velocity of Y will not be vertical. As a result, neither $v=u\cos\theta$ nor $u=v\cos\theta$ will be correct.