2
$\begingroup$

So, there is this exercise that I remember from way back when that still pops into my head and bugs me. I love the puzzle, it just keeps being absurd no matter how I twist my head around possible solutions. It is taken from memory, it was a bit more precise but this is how I remember it. It was a situation for which we recieved no score or feedback and it is hard to say afterwards if I answered correctly (military aptitude test)

I would love to see some other peoples answer.

The two questions given for this exercise was

Assuming no friction except the knot at each end of the rope, and that the weights in the image rests on the floor;

  1. With how much free hanging mass must the rope be connected to in order for any of the three blocks to move

  2. Upon enough free hanging mass at the ? - will the first block be lifted, the second, the third, or all of them at once?


Mechanics - what will happen, and when?


So - if you wish to have a go yourself and don't want my musings, then stop reading now.

I initially thought the answer to the first question would be something just a smidge heavier than 100.

But - if all the weights were free hanging, and the weight in issue was half of that, 50, there would be balance - all the other weights travel half the distance of the weight at the end when the rope is drawn, so the forces should be equal with half the weight at the end. And if you consider the pulleys, 2 of them carry 100, the ones at the end have to carry 100 between them, so 50. Right?

And as for the second question, they should all move at the same time, given no friction... I have no explanation, it just... feels right. given the frictionless pulleys, there is nothing definite to separate what the first or the third weight "sees" - so the only symmetric solution would be that they all lift. No?

$\endgroup$
1
$\begingroup$

I think you are absolutely correct in your answers. But this is how I attempted to solve the problem: Firstly,

Assuming the rope is weightless, the tension on the rope is the same throughout its length. (Why? Because if the tensions were unequal at two different points, the rope between the points would feel a non-zero net force. And given that its acceleration can only be finite, the product of mass and acceleration would not equal the force. )

So you can straightaway answer your second question. If $T$ is the required tension on the rope to lift first mass then it is also the required tension to lift the second and the third mass(because they have the same weight). So as long as the rope does not have that much tension none of the masses would be lifted and as soon as it has enough tension all of them would be lifted at once.

Now to solve the first question: The value of $T$ must be $50$ because there are effectively two ropes (and therefore two tensions) acting on each block. $T$ will also equal the force we apply on the loose end of the rope.

Or you could apply the work-energy theorem. If you pull the loose end of the string by a distance $6x$ all the masses would be displaced by $x$ (due to the geometry). And $50*6x = 100*x + 100*x + 100*x$

$\endgroup$
  • $\begingroup$ Thank you. I had not thought about actually formalizing the rope tension through the work energy theorem, I feel the urge to waste my work day on wikipedia now. $\endgroup$ – Stian Yttervik May 8 '17 at 7:17
  • $\begingroup$ No problem. I gave a rough sketch though and if you're stuck I can elaborate. $\endgroup$ – Abhijeet Melkani May 8 '17 at 7:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.