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Here is a kinematic problem which I solved using a rigorous geometrical approach. However, a high-school student suggested a different quicker approach, which has its own intricacies. I wish to understand how to correctly apply the latter, quicker "high-school", approach.

The Problem

We have this kinematic situation, where the 2 ropes are being pulled down with a speed u. The block moves up with speed v. We have to find the relation between u and v.

enter image description here

The correct answer to the problem is: $$ \textbf{v} = \frac{\textbf{u}}{cos(\theta)} $$

Doubt#1 : The Quicker, High-school Approach -- How does it work?

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In this diagram, the student suggests, first drop a perpendicular from A to OB on C. Now, in $\Delta{ABC}$, we can see that $u = v*cos(\theta)$.

However, the basis for this argument is not clear. I mean, a different student came up with a different answer when he said that if we look at $\Delta{OAB}$, we see that $v = u*cos(\theta)$.

How exactly does this approach work? How do we know which answer is the correct one?

Doubt#2 : Why don't the kinematic quantities add up as the dynamic quantities do?

Another student raised another interesting doubt. The answer that he came up with was this:

$$ \textbf{v} = \frac{2\textbf{u}}{cos(\theta)} $$

You must have guessed why he came with this answer. His argument was that since there are 2 symmetrical ropes, their motion will add up to give the motion of the block. He said that just like forces add up, the displacements/velocities should also add up. Now, the argument is clearly fallacious.

But how do I explain the fallacy to a high-school student?

APPENDIX -- The Rigorous Geometric Approach:

Here is the formal geometric approach which I used to derive the correct answer.

enter image description here

In the diagram shown above, we know that:

$$POB = L \text{ (constant, total length of string)}$$ $$PO + OB = L \label{a} \tag{1}$$ $$PO + \frac{OA}{sin(\theta)} = L$$

Now, differentiate the above expression w.r.t. time (knowing that OA is constant, and $\dot{PO} = \textbf{u}$):

$$\dot{PO} - \frac{OA*cos(\theta)}{sin^2(\theta)}*\dot{\theta} = 0$$ $$\dot{\theta} = \frac{sin^2(\theta)}{OA*cos(\theta)}*\textbf{u} \label{b} \tag{2}$$

Now, take equation \ref{a} again, put $OB = \frac{AB}{cos(\theta)}$, and then differentiate w.r.t. time:

$$ PO + \frac{AB}{cos(\theta)} = L$$ $$ \dot{PO} + \frac{\dot{AB}}{cos(\theta)} + \frac{AB*sin(\theta)}{cos^2(\theta)}*\dot{\theta} = 0$$

Now, $\dot{PO} = \textbf{u}$, $\dot{AB} = - \textbf{v}$ and $AB = OA*tan(\theta)$. Hence: $$ \textbf{u} - \frac{\textbf{v}}{cos(\theta)} + \frac{OA*sin(\theta)}{tan(\theta)*cos^2(\theta)}*\dot{\theta} = 0$$ $$ \textbf{u} - \frac{\textbf{v}}{cos(\theta)} + \frac{OA}{cos(\theta)}*\dot{\theta} = 0 \label{c} \tag{3}$$

Now, using equations \ref{b} and \ref{c} to eliminate $\dot{\theta}$, we get the relation:

$$ \textbf{u} = \textbf{v}*cos(\theta) $$

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    $\begingroup$ Velocity of block upward, along the string is transferred to the whole rope. So when the block moves up with velocity $v$ then along the rope the component is $v\cos(\theta)$ which is same as the velocity of the rope $u$. Moreover the first diagram under doubt 1 is the correct diagram. Always remember that the component of a vector $v$ can never have greater magnitude than itself. $u$ is the component of $v$ along the string. $\endgroup$ – user135951 Dec 8 '16 at 17:35
  • $\begingroup$ Possible duplicate of Doubt in a question related to Newton's laws of motion $\endgroup$ – sammy gerbil Dec 9 '16 at 0:41
  • $\begingroup$ @sammygerbil: Yes, I realized that this is indeed a duplicate of that question. However, while the original one seems like a homework problem where the user has put no effort in solving, my question is much more well-framed. Original one has been CLOSED as it was marked "Unclear". So, in that case, this shouldn't be counted as a Duplicate, should it? $\endgroup$ – shivams Dec 9 '16 at 6:17
  • $\begingroup$ Both questions are asking the same thing, and the other was earlier. You have put in more effort, but your question does not raise any new issue. I have not voted to close it, and so far nobody else has either. Noting duplicates links them for future searches. If you are satisfied with the answers provided here, please select one. $\endgroup$ – sammy gerbil Dec 9 '16 at 17:22
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There's an easier way to get your solution, just call the length $OA$ equal to $1$ (it's constant, so we can scale all lengths to it), and call length AB equal to $z$. Then $u$ equals the negative rate of change of the length of the hypotenuse $OB$ (since $OB$ + $OP$ is constant), and v equals the negative rate of change of $z$. The hypotenuse is the square root of $1+z^2$, and it's negative time derivative by the chain rule is $\cos(\theta)$ times $dz/dt$, so that is $v\cos(\theta)$, and that equals $u$, and you are done.

The first student is correct, the ABC triangle is a good projection. The second student is wrong because the OAB triangle is not a good projection, the block is not moving along OB. The third student is wrong because velocities don't add like that-- if you have two straight ropes attached to a block, and pull both ropes at speed $u$, the block moves at speed $u$, not speed $2u$ .

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    $\begingroup$ Actually, at any instant we can directly say $u=v\cos(\theta)$. $\theta$ is the instantaneous angle, and that first method isn't wrong. $u$ is simply the component of $v$ along the string. The length of string must remain constant and so the constraint relation holds. $\endgroup$ – user135951 Dec 8 '16 at 17:57
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    $\begingroup$ Good point, that's the easiest solution. I misspoke, the issue is not the rate of change of theta, as the projection is instantaneous. The student who used the ABC triangle is correct because the actual velocity is along AB and can be projected along OA. The student who used the OAB triangle is wrong because that solution treats the actual velocity as if it is along OA and can be projected along AB. It doesn't work to keep projecting like that, one would need to include both the projected components. $\endgroup$ – Ken G Dec 8 '16 at 18:04
  • $\begingroup$ I think you should add your comment as an edit to the answer :). Also, please use LaTeX while writing the math symbols. $\endgroup$ – user135951 Dec 8 '16 at 18:06
  • $\begingroup$ @KenG: Your comment clarifies a lot. My doubt was precisely this that why are we taking the component of v along u and not u along v. Your argument that "the actual velocity is along AB, and hence that velocity should be projected along OA and not the other way round" is clarifying. Thanks for that. +1 $\endgroup$ – shivams Dec 8 '16 at 18:29
  • $\begingroup$ Year back I had this precise doubt ! I need a bit more clarification. You say that it doesn't work to keep projecting like that , one would need to include both the projected components. Could you please show how or which both components should be added to get the correct answer if we use the triangle OAB or u as the velocity and break it into "which all components " to get the correct answer. It would be really helpful if you could add this bit in the answer too. Thanks $\endgroup$ – Shashaank May 11 '17 at 11:10
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I believe this diagram shows clearly how the approach of your first student can be explained:

enter image description here

From similar triangles, you can see that when the rope gets shorter by distance $u$, the load moves vertically by a distance $\frac{u}{\cos\theta}$.

As for the fallacy of the second student's approach: while velocities are vectors, and vectors can be summed, summation only makes sense when you are considering motion in different frames of reference. If I am in a train moving at velocity $\vec v$, and I throw a ball out of the window at velocity $\vec u$, a person on the ground would see the ball moving at $\vec v + \vec u$. But when two people on the train see the same ball moving at $u$, you can't say "well, A saw a velocity of $u$, and B saw a velocity of $u$, so the object is moving at $2u$"...

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  • $\begingroup$ +1 For a different and more clarifying approach. However, you've taken a different $\theta$ as given in the question. Accordingly, your answer should be $u/sin(\theta)$. $\endgroup$ – shivams Dec 8 '16 at 18:49
  • $\begingroup$ @shivams - sorry, sloppy of me... thanks for pointing it out. Have fixed it now. $\endgroup$ – Floris Dec 8 '16 at 18:50
  • $\begingroup$ Something irrelevant, but could I ask which software do you use for such diagrams/illustrations? Your illustration looks good. Currently I use IPE, which I'm loving very much, but I'm always on the lookout. $\endgroup$ – shivams Dec 8 '16 at 20:55
  • $\begingroup$ @shivams - I just use Powerpoint on the Mac... It has a decent set of tools for aligning / rotating etc, and works for most of my diagrams. $\endgroup$ – Floris Dec 8 '16 at 20:57
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    $\begingroup$ Oh... Okay. I always forget about the simple tools for a simple job :) $\endgroup$ – shivams Dec 8 '16 at 21:06
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Doubt #1: A short explanation:

In the right triangle $OAB$, side $OA$ is constant and $\cos\theta = \dfrac{AB}{OB}$. Applying the time derivative to the Pathagorean Theorem for this triangle yields: \begin{align} \frac{d}{dt} \left( (OB)^2 \right) &=& \frac{d}{dt} \left( (OA)^2 + (AB)^2 \right) \\ 2(OB)(\dot{OB}) & = & 2(OA)(\dot{OA}) + 2(AB)(\dot{AB}) \end{align} Simplifying and using $\dot{OA} = 0$, $\dot{OB}=u$ and $\dot{AB}=v$, this becomes $u=v\cos\theta$ which yields the result \begin{equation} v=\frac{u}{\cos\theta} \end{equation}

Doubt #2: The factor of 2

The confusion stems from the nature of the initial conditions stated in the problem; the problem specifies the speed of points $P$ (left fixed pully) and $Q$ (right fixed pulley) $=u$. The problem could have instead stated the forces applied at $P$ and $Q$ by giving a tension $T$ in the ropes and asking you to solve for the acceleration of $M$. If this had been the case, the factor of $2$ would indeed show up in the answer (one from tension on the $P$ side and one from tension on the $Q$ side) because forces applied to $M$ do add as vectors, while the speed of the ropes attached to $M$ do not.

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