Which cat will move down?
A kitten is inside each of two boxes, initially held at rest, as shown in the diagram. Kitty together with its box weighs 1 k g , and Petty together with its box weighs 1 . 5 k g .
When the boxes are released, which kitten will move down?
Notes
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Assume that the pulleys and strings are both massless and frictionless.
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The brown line at the bottom of the diagram shows that the other end of Kitty's rope is attached to the ground.
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When the boxes are released, they are simply let go. No ropes are cut!
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8 solutions
dont' get it .. how exactly are the boxes released? I thought that meant the rope from Kitty to the ground was released, presuming the rope from the ceiling is holding the entire thing up? so if the rope from the ground is removed, there is no counterbalance resistance of 1 kg which in turn means the 2 kg pull down is 1 kg pull down .. happy to show my stupidity for better understanding of the original puzzle asked. thanks :)
Understand that really believe that this is one of the questions that does not give a good explanation of what is going on
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I added a drawing. Maybe that will help.
Yeah and thank you for the illustration
Can you elaborate on what problems you have with the wording? This will help me to resolve the issue. Thanks.
Its not worded well.
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What are the concerns you have with the wordings, can you please elaborate?
"When the boxes are released" is the critical statement, and the implied "release" can produce different results. In the initial position, the only way the boxes can be static is if there is a force pulling Petty down with 0.5 kg. There is no additional force holding Kitty in place, because the ground is counteracting Kitty's 1 kg weight.The 2.0 kg of Kitty + ground would pull Petty up, except for the 0.5 kg force holding Petty in place.
That's because Kitty + Kitty's anchor rope exert 2 kg downward on the right hand side of Petty's pulley.
In the 1st scenario, "release of the boxes" means that the 0.5 kg force holding Petty DOWN is released, and NO ROPES ARE CUT! Now, the 2 kg on the right side of Petty's pulley pulls Petty UP, and that causes more rope to transfer to the pulley supporting Kitty. So, Kitty goes down, and eventually reaches the ground. At that point, tension on right hand side of Petty's pulley eases to 1.5 kg, which balances Petty + box somewhere aloft, and the motion ceases. [Or Petty gets jammed into the higher pulley, depending on the length of the ropes.]
In an alternate scenario, "release of the boxes" means cutting both boxes free of their restraining ropes AND removing the additional 0.5 kg pulling down on Petty. In this case, both cats go into freefall and impact the ground. Scratch one life each from both Kitty and Petty.
In the third scenario, I interpreted "release of the boxes" to mean that the rope attached to the ground holding up Kitty is cut, and the 0.5 kg Force pulling down on Petty is removed. Kitty goes into freefall, and all tension on that side of the top pulley is gone.. That 2 kg. weight release allows Petty to go into freefall, too.
Let's face it. The puzzle composition was imprecise.
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Thanks, the problem statement has been updated to include that no ropes are cut and we have simply let go the system from rest.
Don't get it - As the boxes are being held initially, there is no pull being exerted to the left of Kitty's pulley. As Petty is heavier, it will move down, pulling Kitty upwards. If this is incorrect, the puzzle is just poorly presented to try and cause confusion.
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Petty is heavier-- but because Kitty is attached to a moving pulley, her weight counts twice!
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Will that also work with Gold Bars? I'll sell you one for the price of two!
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@Derek Maltby – Of course it works with gold bars. I put 1.5 kg of rocks in the left bucket, and you put 1 kg of gold in the bucket on the right. I get to keep all gold in the bucket that reaches the ground. Deal?
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@Arjen Vreugdenhil – Can you help me understand how the downward force exerted by Kitty's weight is doubled? I don't see it.
And the moment Kitty stops moving Petty goes down as well. So the answer should be both, or the question which moves first
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Actually, when Kitty stops moving, which happens either when Kitty is on the ground or Petty gets stuck against the top pulley, the whole system stops moving.
kg is a mass unit, not a force
Forget the static case (brown string attached). The pulleys and strings are massless and frictionless, so do not contribute forces but merely redirect them. When the string is released, the only forces acting on the boxes and kittens are from gravity. So both kittens will move down.
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Each rope pulls on the buckets and on the pulleys. They are important forces and cannot be ignored!
When Kitty moves down, the pulley above her must move down (because the rope attached to her cannot stretch). When that pulley moves down, the rope it is attached to must move down; this requires that Petty goes up.
The only way to make both cats go down is by cutting a rope somewhere in the setup.
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You are stuck in the static case. When the rope is released, different case. With no friction and no mass, the pulleys contribute nothing force-wise.
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@Arsenio Ibay – Incorrect. The pulleys produce normal force, which keeps the support the rope so it doesn't fall off. This does not change when the rope or pulley starts moving.
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@Arjen Vreugdenhil – Breaking it down - when the brown rope is released, Kitty is in freefall. The pulley is frictionless, so there is no resistance there. The end of the rope is free and massless, so no opposing force on the other side of the pulley. Moving over to Petty, when the brown rope is released, the force balancing it in the static case is gone, too - no mass or friction in pulley or rope. Petty will go into freefall, ignoring when the two pulleys might make contact. But the question is not how far or how fast Kitty or Petty would move, but which direction. Both would go down.
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@Arsenio Ibay – Do you realize that Kitty's rope remains attached to the ground?
Let F T , K be the tension in Kitty's rope. Then for Kitty we have a net force of F K = m K g − F T K so that her acceleration is m K a K = m K g − F T , K ( 1 ) .
The pulley accelerates downward in such a way that Kitty's rope does not change length. Thus for this pulley we have a P = 2 1 a K . ( 2 ) Since Petty's rope moves along with the pulley and keeps constant length, Perry is moving upward with the same acceleration. If the pulley is massless, the net force on it is zero, i.e. the forces are balanced. Here we find F T , P = 2 F T , K ( 3 ) . Finally, for Petty the equation of motion is m P a P = m P g − F T , P ( 4 ) .
Combining the four equations, we find m P ( g − a P ) = 2 m K ( g − a K ) m P ( g − 2 1 a K ) = 2 m K ( g − a K ) ( 2 1 m P − 2 m K ) a K = ( m P − 2 m K ) g , with the conclusion a K = 2 1 m P − 2 m K m P − 2 m K g = 0 . 4 g a P = 0 . 2 g . The tension in the ropes are F T , K = m K ( g − a K ) = 0 . 6 ( k g ) ⋅ g , F T , P = m P ( g − a P ) = 1 . 2 kg ⋅ g .
From this analysis it is clear that even while the cats move, the tensions in the rope do not become zero, and the cats accelerate more slowly than they would in free fall. In fact, Petty is going upward!
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@Arjen Vreugdenhil – Question asks at the end - "When the boxes are released , which kitten will move down?" If it means the rope holding Kitty is released at the ground, the static case is no longer valid. If the boxes are literally released (at the boxes), the boxes will fall. In both cases, this is no longer a static case . There is no balancing of forces, but accelerating elements.
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@Arsenio Ibay – Here it is presented even simpler. When the rope is released from the ground, Kitty's weight is no longer balanced by the force exerted on the rope from the ground. Kitty will fall down. Agreed? Likewise, Petty is no longer being balanced by the force exerted by the other pulley (again, massless and frictionless), and assuming a uniform gravitational field. Petty will fall down. And they both fall at the same acceleration.
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@Arsenio Ibay – From the problem statement and the posted answer, it appears that the rope is not released from the ground. Rather, "released" means here that the system is allowed to move under the given constraints.
The boxes were "initially held at rest" but now that they are "released" they move--even though Kitty's rope remains attached to the ground.
I will make an edit in the problem statement to make this clearer.
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@Arjen Vreugdenhil – That still would not make sense. With Petty and Kitty having different weights, the system is not at equilibrium as shown until Kitty's box reaches the pulley. Is there an unseen restraint on Kitty's box? It would make more sense if a different-colored rope connecting Kitty's box, or both Kitty's and Petty's boxes, to the ground, and then indicate that the rope(s) is/are released. In that case, Kitty would move up, and Petty would move down.
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@Arsenio Ibay – Indeed, the system is not in equilibrium. If it were, none of the cats would start moving.
@Arsenio Ibay – Kitty's box will not reach the pulley. Kitty is going to the ground or Petty's box will reach the pulley
Marta Reece - your static diagram is incomplete; you do not show the forces acting on the "ground" and "ceiling".
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But these forces are not relevant. Since we are only interested in the movement of the cats in the buckets, we only need to determine the forces acting on them.
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But they are - when examining an equilibrium, you need to look at all forces in the closed system. Case in point - do you mean to tell me that with the with massless and frictionless rope and pulley, Kitty's 1kg weight will result in 2kg of weight at the pulley rope, as shown?
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@Arsenio Ibay – Yes. Look at the pulley on the right. There are three pieces of rope pulling on it: upward the rope with Petty at the end, downward the rope with Kitty at the end, and downward the rope that is attached to the ground.
If the cats were not accelerating, the upward tension would be m P g ∝ 1 . 5 k g , and the downward tensions would be m K g ∝ 1 k g each . The pulley would be unbalanced and start accelerating downward .
The actual forces are: upward tension is m P ( g + a P ) ∝ 1 . 8 k g , and the downward tensions are m K ( g − a K ) ∝ 0 . 9 k g each. Here, a P = 0 . 2 g upward and a K = 0 . 1 g downward. (It is easy to see from the ropes that the ratio of accelerations must be 2:1.)
It's more like related rates. Kitty has to move twice the distance than petty for the puppy system to hold true.
Is that brown color plane on the bottom cannot be moved? I ask this because the question didn't state that.
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I have assumed that that is the case, based on the illustration. I could be stated explicitly, you are right about that.
"the ground" I don't think it can be moved
This question is not well posed. The key question "When the boxes are released, which kitten will move down?" does not describe what change in the system transpires to cause motion? If the boxes are released from the ropes to which they are attached, of course, both will fall. If the entire system is severed at the top rope, of course, both will fall. If the entire system is severed at the bottom rope, of course, Kitty will fall to the ground, followed by Petty, and so, both will fall.
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I don't see much of a problem. The pulleys, ropes and buckets are free to move, and they will do so because the tensions in the ropes and the gravitational forces are not balanced.
Missing detail pulley holding up Kitty fixed, or is it suspended by the rope holding Petty. Since there is blank space there, it would appear the latter.
One of the issues that I have with the problem is that mass is treated as if it were a force, which it is not. Now gravity is a force, but that's not what the question is asking nor is the given solution saying anything about that. If the strings/ropes are massless, they cannot exert a force, at least not in Newtonian Mechanics. Also, if the pulleys are frictionless, then as soon as the string is released, it will slide over the pulley and Kitty and her box will fall. And because the other string and pulley are also massless and frictionless, that string will also slide over the upper pulley, and Petty and her box will also fall. Thus, the correct answer is both will move down. That is, of course, as long as the entire contraption is on some planet or other object which exerts some gravitational forces on it.
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- The solution talks about force in terms of mass, but that cannot be blamed on the problem.
In Newtonian mechanics, a "massless" string has so little mass μ that it does not significantly alter the outcome. Specifically, if the string accelerates, the net force μ a is much smaller than the other forces involved; and the gravitational force μ g on the string is similarly negligible. A "frictionless" string/pulley system consists either in a string that slides with no friction, or a string that rolls with the pulley without slipping, and the absence of frictional torques in the pulley.
None of this rules out motion of strings or pulleys, nor does it prevent tension forces in the string.
Kitty's box will go down, but will not be in free fall. The reason is that the box can only move down if the nearby pulley moves downward. This pulley is hanging from a string that exerts upward tension. In fact, if Petty's mass had been precisely twice as big as Kitty's, the whole system would remain in (neutral) equilibrium.
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While Kitty will not be in free fall, she will move with a constant downward acceleration and will hit the ground, if that's the final outcome, with some some thump. So in that sense she will be "falling."
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@Marta Reece – That is correct. Kitty accelerates downward ("falls") at constant 0 . 4 g , and Petty accelerates upward at 0 . 2 g .
Old joke: If you call a dog's tail a leg, how many legs does a dog have? the answer is, four: calling a tail a leg doesn't make it a leg. Similarly, calling mass a force doesn't make it a force. I blame this miscommunication completely on the problem, because at no time was gravity (or any other force acting upon the system) mentioned. Also, in typical problems of this sort, it is conventional to say "the mass of the string is neglible, as is the friction of the pulley wheel." However, you did not write that: you wrote that the strings and pulleys are massless and frictionless. Not "massless", not "frictionless", but massless and frictionless, meaning that there is absolutely no mass and no friction. I understand your point, but you did not write your problem to say what you meant. I and the others who are quibbling with you cannot read minds; at least, I cannot. We can only go with what was written, and as written, I still stand with my answer: both boxes will fall, assuming that there is a force acting downwards relative to the boxes.
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@Jim Carleton – I did not write the problem. I only added some notes for clarification :)
There are different traditions of asking physics questions, but they all employ a certain amount of common sense. Cats typically live on earth, and on earth gravity is ubiquitous. The drawing also suggests the presence of a gravitational force.
As for the language of "massless" and "frictionless"-- they are rather common in mechanics to describe the ideal (or, if you will, limiting situation).
Still, the main point is that both cats cannot go downward because of the ropes to which they are attached. If Petty goes down, then her rope pulls up on Kitty's pulley, causing Kitty to move upward. "Massless" or "frictionless" is entirely irrelevant to this conclusion.
It appears to me that Kitty would have her whiskers tangled up in the pulley and the box she is in would be wedged in the pulley to start with. From there, anywhere a rope gets released, both cats will fall. The puzzle isn't well worded.
TMc
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They were held as shown in the diagram and then just let go. Strings were not initially slacked.
why isn't petty pulled down by 3 kg to balance her?
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It's a tension of 1 kg weight on two of the ropes below the pulley on the right, and that adds up to tension of 2 kg weight on the rope above that pulley. So it is 2 kg worth not 3 kg. And she is pulled up by this not down.
I think you're forgetting that the top pulley is also pulling up with a "force" of 1.5 Kg.
If there is a rope I mean Kitty's rope is attached to the ground. Then how can they move???
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The pulleys are free to move, so when the goes down, Kitty goes down too.
The answer is that one pulley is stationary and one isn't. The pulley that moves gets the advantage.
By thinking carefully at the way the pulleys are joined you can see that if Petty moves a distance x towards the ground, Kitty moves away from the ground a distance 2x. So the change in potential energy ( m g Δ h for each cat) is Δ P E = g x ( 2 × 1 − 1 × 1 . 5 ) = 2 g x . Now if there is any movement at all, the kinetic energy increases, and so the change in potential energy must be negative.
Looking at our formula we see that for Δ P E < 0 , x must be negative, and so petty moves away from the ground, and (since you have already looked at how the pulleys work you will know this!) K i t t y moves towards the ground.
thing is if you release both the boxes both the kitten fall because they are no longer attached to the pullies
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That's the trick. Both are hung direct by the rope without a supporting/sitting box. For science!
No ropes are cut and they are still attached to the pulleys, we simply let the boxes go.
My question is, are the pullies attached to the wall? If they are yo can't assure that any cat will fall if you don't know the lenght of the strings. If you know that they move (well, just the second pulley can move since the first is attached to the ceiling) , happens the same, if you don't know which is the lenght of Petty's string then you can't know that Kitty is going to touch the ground. So, my answer would be that you can't know if any of them will touch the floor.
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The initial situation is shown in the diagram, in which none of the boxes touch the ground. So, if we let go the system (without cutting any ropes, obviously), one of the box is bound to move down and other must move up.
I am sorry to overthink or underthink this but the question says if the boxes are released which one will move down? The boxes released without cutting a rope or presumably untying a rope. Then both boxes will move down. If they are released from the roof then both will move down, If the roof is left in place and the boxes are released from the bottom then both will also move down. In fact releasing the boxes from any point of the machine will allow both kittens to move down. The weight of the kitties is irrelevant. In fact if you release either kitty and not the other one they will both move down.
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Well, initially the ropes were not slacked up and were held tight in the position shown in the diagram. Once they were released, then, it is obvious that both the boxes can not move down because that will result in elongation in the rope.
Alternative solution using energy
If Petty moves over distance Δ y , Kitty moves down over distance − 2 Δ y . Thus the change in gravitational potential energy U = − m g ⋅ y is Δ U = ( − m P g ⋅ Δ y ) + ( − m K g ⋅ ( − 2 Δ y ) ) = ( 2 m K − m P ) g ⋅ Δ y Since the system tends to lower potential energy, Δ U < 0 . This means that
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Δ y is in the same direction as g if 2 m K < 2 m P
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Δ y is in the opposite direction of g if 2 m K > m P .
With the given values, we have the latter case; Petty moves upward and Kitty downward.
The wording of the question got me - I thought it meant the rope to the ground was released.
But the statement was " the boxes are released----------no ropes are cut". In that case, BOTH boxes are freed and BOTH will fall. This mathematical exercise does NOT bear on the problem AS STATED.!!
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Look at the string connecting Petty and the other pulley. The length of the string is fixed, so if one side goes down, the other comes up. In this case, Kitty goes down and Petty moves up.
There's a small issue with this problem. Given the diagram, even if petty WAS the one to be moving down, petty wouldn't hit the floor due to estimated rope distances. (no distances are provided).
Due to the pulley system attached to kitty being attached to the floor, that box need only exert half the force in order to achieve twice the force to Petty's Pulley. To demonstrate this point, if Kitty weighed 0.75kg, or Petty weighed 2kg, the whole system would remain in equilibrium. Having the string anchored to the ground is the equivalent of having another "kitty" there, as that pulley on it's own would remain in equilibrium if the pulley itself was anchored to the ceiling.
We just need to tell which box will move down and the question does not care if the box moving down after the release will eventually reach the ground.
Martha's solution is sort of OK. It's a pity the question is stated with wrong physics. Kg measure mass not force. If we take g as 10N/Kg then the tension in Kitty's string is 10N and this exists in the same string on the other side of her pully thus the tension in the string above her pully is 20N which is also the tension in the string above Petty, easily overcoming the downward force of 15N. Therefore Petty moves up and therefore Kitty goes down. Some of the other solutions seem to complicate things excessively.
The gravitational field is uniform, so force is directly proportional to mass. While it is more accurate to say Kitty weighs 15N, It is not totally unreasonable to use kilograms.
Kitty pulling the string with 1Kg. Every action has an equal and opposite reaction. So the string creates another 1Kg pull to balance Kitty, but is also acting downward on Kitty's pulley, for a total of 2Kg. The 2Kg pull on Kitty's side of Petty's pulley is greater than Petty's 1.5Kg.
If the string carrying Petty goes up by
x
distance, the string carrying Petty will go down by
2
x
distance, hence, if acceleration of Petty is
a
, then Acceleration of kitty is
2
a
. We can assume any direction for their acceleration.
We take acceleration if Petty to be
a
units downward and that of kitty to be
2
a
units downwards.
Tension in the string carrying kitty is
T
, and hence the tension in the string carrying Petty is
2
T
.
Mass of Petty =
M
.
Mass of Kitty=
m
.
The two equations of motion derived are
M
g
−
2
T
=
M
a
And,
T
−
m
g
=
2
m
a
Solving these two equations, we get
a
=
−
0
.
9
0
9
0
Since the value is negative, the motion us in the opposite direction of note assumed direction.
Hence, Petty goes upwards, while kitty comes downwards.
Kitty plus pully most likely weighed more than petty, that's how I got my answer anyway
Pulley is massless
Kitty pulls down with 1 kg, and the ground to the left of it also one 1 kg, to balance her.
That makes 2 kg pull on the right with only 1.5 kg on the left.
The right side, and Kitty, will go down.