A More Intuitive Solution for the Kinematics of the Pulley-Rope System

Question

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.

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?

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.

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:

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

Answer 1

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 $2u$ .

Answer 2

I believe this diagram shows clearly how the approach of your first student can be explained:

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 $2u$"...

Answer 3

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.