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We have the setup as shown in the picture. Two blocks A and B are connected by a string that always remains taut. Block A is on a frictionless surface. Block B is being pulled downwards with a constant velocity, $V_B$. I want to stress that $V_B$ is constant. Obviously block A moves to the left with a velocity $V_A$


I understand the velocity constraint between these two blocks. $V_A\cos\alpha$ is the component of $V_A$ parallel to the string. There is a component of $V_A (= V_A\sin\alpha)$, perpendicular to the string.

If I analyse this relation at $\alpha = 0^\circ$, it makes sense that parallel component of $V_A = V_B$ and perpendicular component of $V_A = 0$.

Likewise at $\alpha = 90^\circ$, parallel component of $V_A = 0$. And all of its velocity is in perpendicular direction to the string.

What I am unable to understand is, how does $V_A$ change as the angle changes from $\alpha$ to zero?

$V_B$ is constant, so if I consider this relation :

$V_A\cos\alpha=V_B\implies V_A=\frac{V_B}{\cos\alpha}$

$V_B$ is constant, say, $2 \ \text{m/s}$. Let's take an angle close to $90^\circ$. Say, at $\alpha=89.99^\circ$, does $V_A$ become too large?

I cannot seem to understand how $V_A$ becomes too large at an angle close to 90, when $V_B$ is constant at $2 \ \text{m/s}$

  • $\begingroup$ @Qmechanic How is this a homework-and-exercise question? Is this a numerical problem and am I looking for a solution? I need help with a specific concept. Sorry, I don't agree that this is a homework-and-exercise question $\endgroup$
    – 4d_
    Sep 10 at 1:31
  • 1
    $\begingroup$ Hi 4d_. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Sep 10 at 1:33
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    $\begingroup$ Sorry, I made an oversight in my previous answer (which I have now deleted). Upon further consideration I think that the constraint $V_A\cos{\alpha}=V_B$ is actually correct. $\endgroup$
    – DanDan0101
    Sep 10 at 4:18
  • $\begingroup$ @DanDan0101 thank you again. If the constraint equation is correct, can you please further explanation why $V_A$ comes out to be a very large value at $\alpha$ = $89.99⁰$ ? It looks very counterintuitive and I feel my understanding of it is wrong at some point. $V_A$ becomes unreasonably large for a constant speed of $V_B$ = $2$$\frac{m}{s}$ $\endgroup$
    – 4d_
    Sep 10 at 5:02

Although it might look counterintuitive at first, block $A$ will indeed become very fast. To see this in reality, you can take a string, maybe 0.5 m long, hold both ends in your hands so that the string is loose and then pull them apart. Even if you do not move your hands extremely fast, the string will start to vibrate when it becomes taut, fast enough to be blurry in your vision, and it's velocity at the equilibrium point of the vibrations will be what is comparable to the velocity of block $A$.

The physical explanation for this phenomenon is, that in your model, you assume that block $A$ is fixed to the surface it slides on. When $A$ approaches the pulley, the force needed to keep it on that surface compensates an increasing fraction of the rope's force. This means, the force needed to keep $B$ at constant $V_B$ increases, too, as does the work one has to put into the system and which eventually ends up in the kinetic energy of $A$, so there is no surprise in that energy becoming very large.

Assuming constant velocities of constraining objects in mechanical systems often leads to this kind of paradoxes. For another example, see the sections "Example" and "Remark" of this other answer of mine.

  • $\begingroup$ Thank you so much. Now it is clear to me $\endgroup$
    – 4d_
    Sep 10 at 15:17

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