4 ms·

Weighing the options: A simpler question about pulleys and ropes.

RiderOfGiraffes 16y ago
The answers people give include, but are not limited to, 0, 1 and 2. I find people's reasoning interesting, and after the discussion on the item about the weights and the pulleys (http://news.ycombinator.com/item?id=1659326 http://news.ycombinator.com/item?id=1659326) I'll be interested to see what people say about this, if anything. There is a one-off experiment hidden under this submission - email me if you want to know more, or have concerns and I'll answer them. Alternatively I'll post more in a day or so. Thanks.
jazzychad 16y ago
POTENTIAL SPOILER ALERT: I'm giving my answer below. I'm going to say the balance reads 1kg. Since the setup on the right is in stasis, you could easily remove one of the weights and anchor the string to the ground and remain in stasis. You would then be recreating the setup on the left (equipment anchored to ceiling) except the string is just being redirected by pulleys. Since each bend in the string is 90 degrees you multiply all the forces by 1, still giving you a 1kg reading on the balance. Thinking aloud about why 0kg and 2kg don't make sense. 0kg: the spring in the balance is going to stretch to some extent because of gravity pulling the weights down, so 0kg doesn't make sense. 2kg: I'm not sure how to disprove this one except by saying that I'm so convinced that 1kg is the right answer that 2kg doesn't make sense either. That's my reasoning, anyway.
- jules 16y ago
  That was exactly my line of thought. Another way to see it is to look at what the spring-balance feels. In the picture on the left it feels the lower rope pulling down with 10N, and because it is at rest it feels the upper rope pulling up with 10N. In the picture on the right it feels exactly the same forces: 10N pulling on one side and 10N pulling on the other side. So the reading has to be the same. Yet another and perhaps the simplest explanation is that you pick up the left mass and you don't hang its rope over its pulley, instead you just hang it in mid air. Now we just have the rope hanging over one pulley, with the mass and the spring-balance on the left and the other mass on the right. Because everything but the masses is massless it's still in rest, and the situation is exactly the same as when the rope was attached to the ceiling instead of to the other mass.
- exit 16y ago
  but what you're saying is that the spring can't distinguish between the setup on the right and the setup on the left. that doesn't seem right. if you detach one of the weights and anchor it on the other weight, the balance should read 2kg. and if you anchor weight A on the string of weight B, it would also read 2kg. as you move weight A's anchor point closer and closer to the balance, the reading remains 2kg. what happens when the anchoring point reaches the balance?
  - jazzychad 16y ago
    > you're saying is that the spring can't distinguish between the setup on the right and the setup on the left. Correct, because according to the balance, both setups are indistinguishable. The dark secret of spring balances is that they measure (and display) force in only one direction. In stasis, the net forces on the balance are exactly zero N since there are equal and opposite forces acting upon the balance (true in each setup), yet the needle moves because it is measuring the movement (stretching) of the spring inside the balance (which at one end is anchored internally to the balance casing, and the other end is anchored to the string/mass). So, as for the setup on the right, the left-hand weight is pulling on the balance with weight 1kg (9.8N), and the right-hand weight is pulling on the balance with weight 1kg (9.8N), but since the balance only displays forces in one direction, it will display 1kg (in this case it doesn't matter which direction it is measuring because the forces are 180 degrees from each other).
    - dedward 16y ago
      I think that's reading a bit too much one-directionality into it. While true, it's the amount of force that's at issue here - not the direction. If a second 2kg weight were added to the "anchor" end of the spring balance - it would either register -1kg (if it's built to do that) or break or lock up givinging invalid or useless readings. Of course - assuming the balance isn't anchored anywhere else, this would set the whole system in motion, and it would all fall apart ratherquickly.
meelash 16y ago
Free body diagrams, people.
- kreiling 16y ago
  In case you, like me don't know what free body diagrams are, then: "The only rule for drawing free-body diagrams is to depict all the forces that exist for that object in the given situation. Thus, to construct free-body diagrams, it is extremely important to know the various types of forces. If given a description of a physical situation, begin by using your understanding of the force types to identify which forces are present. Then determine the direction in which each force is acting. Finally, draw a box and add arrows for each existing force in the appropriate direction; label each force arrow according to its type. If necessary, refer to the list of forces and their description in order to understand the various force types and their appropriate symbols." http://www.physicsclassroom.com/class/newtlaws/u2l2c.cfm http://www.physicsclassroom.com/class/newtlaws/u2l2c.cfm
kreiling 16y ago
The scale reads zero. Or, if you pull up on it, bringing it above the "horizon" (or bringing the pulleys closer together and below the horizon) causes the scale to approach 2kg. The force of each weight is converted to a horizontal direction.
- DougBTX 16y ago
  If scale reads zero, then that means there must be no force on the scale. If there's no force acting on it, that's much the same as if it wasn't there to start with. So, if you were to remove the scale (effectively cutting the string) the weights would stay in the same place, floating in mid air.
  - dedward 16y ago
    I think he's assuming you are attaching the scale with the middle of the connecting-cable as the mass to be weighted and then pulling upwards on it..... which would make his description correct for that problem - but that's not the puzzle posted here.
modeless 16y ago
The scale really measures force, not mass. On the left the scale measures 9.81 Newtons applied by the weight, which is labeled 1kg because Earth gravity is assumed. The ceiling applies an equal but opposite force on the top of the scale through the suspending string, otherwise the scale would accelerate downward. On the right the force from the ceiling is replaced by force from another weight; this doesn't change the reading on the scale.
meelash 16y ago
Okay, so I thought back to my physics days some more, and the real understanding of the problem lies in understanding the definition of tension. The scale is actually measuring tension in the rope. An informal definition of tension: cut the rope at any point, and determine what force needs to be applied to the one of the cut ends to keep that part of the rope in equilibrium. That force is the tension in the rope. So in the second example, just measure the tension in the rope.
phaedrus 16y ago
Coincidently I'm reading Ernst Mach's "The Science of Mechanics" at this very moment. It includes a lot of analysis of setups like this. As for this problem: consider first the weights. Each one's downward force is 1 kg, due to gravity. There must also be exactly 1 kg's force's worth of tension in the string, because if it were less the weights would be moving down and if it were more the weights would be pulled up. Because the system (is assumed to be) not moving, we may replace the right side weight and pulley wheel with a fixed anchor for the string, in which case there would be no doubt that the spring balance displays 1 kg. Argument from symmetry says the scale would also read 1 kg if the left side were anchored and the right side free. Because the system in unmoving, it really makes no difference whether the opposite side of the scale is fixed to an anchor or balanced by a free weight: it will read 1 kg in all cases. This is very similar to a Car Talk puzzler that their physicist consultant clarified: is it worse to be in a car going 60 mph that hits an unmovable highway overpass wall, or to be in a car going 60 mph that hits an identical car perfectly head-on that is also going 60 mph? The physicist's answer was that there is no difference: in both cases you experience an identical impulse going from 60 mph to a stop in the same time. You may be inclined disbelieve this on the idea that the crumple zones in the car-to-car collision make for a "softer" collision, but consider this: define the point at which the bumpers of the two cars as "the origin". Assuming the cars and the collision are perfectly symmetrical, each car is going to crumple up on its respective side of the origin identically. Suppose the force starts small and ramps up gradually as the crumpling occurs; each car will be pushing the other away with the same force. An unmoving highway overpass barrier would resist with the same force.
- Ralith 16y ago
  Pedantry: The kg is a unit of mass, not force. You want Newtons.
- dedward 16y ago
  Moving some vectors around to make it look more similar - would I like to hit a nice solid wall at 60MPH or a parked car head-on at 120MPH. The head-on collision has twice as much energy to dissipate, and introduces all kinds of projectiles and other stuff heading in my direction. I'll take the wall over the car.
panic 16y ago
In the first picture, the weight pulls down with 1kg of force, and the ceiling pulls up with 1kg of force. If it didn't, there would be an imbalance, causing a net acceleration downwards. In the second picture, the first weight pulls rightward with 1kg of force, and the second weight pulls leftward with 1kg of force. The situation is exactly the same as the first, except with gravity providing the second force rather than the inter-molecular bonds responsible for maintaining the rigidity of the ceiling.

The answers people give include, but are not limited to, 0, 1 and 2. I find people's reasoning interesting, and after the discussion on the item about the weights and the pulleys (http://news.ycombinator.com/item?id=1659326 http://news.ycombinator.com/item?id=1659326) I'll be interested to see what people say about this, if anything. There is a one-off experiment hidden under this submission - email me if you want to know more, or have concerns and I'll answer them. Alternatively I'll post more in a day or so. Thanks.

jazzychad 16y ago

POTENTIAL SPOILER ALERT: I'm giving my answer below. I'm going to say the balance reads 1kg. Since the setup on the right is in stasis, you could easily remove one of the weights and anchor the string to the ground and remain in stasis. You would then be recreating the setup on the left (equipment anchored to ceiling) except the string is just being redirected by pulleys. Since each bend in the string is 90 degrees you multiply all the forces by 1, still giving you a 1kg reading on the balance. Thinking aloud about why 0kg and 2kg don't make sense. 0kg: the spring in the balance is going to stretch to some extent because of gravity pulling the weights down, so 0kg doesn't make sense. 2kg: I'm not sure how to disprove this one except by saying that I'm so convinced that 1kg is the right answer that 2kg doesn't make sense either. That's my reasoning, anyway.

jules 16y ago
That was exactly my line of thought. Another way to see it is to look at what the spring-balance feels. In the picture on the left it feels the lower rope pulling down with 10N, and because it is at rest it feels the upper rope pulling up with 10N. In the picture on the right it feels exactly the same forces: 10N pulling on one side and 10N pulling on the other side. So the reading has to be the same. Yet another and perhaps the simplest explanation is that you pick up the left mass and you don't hang its rope over its pulley, instead you just hang it in mid air. Now we just have the rope hanging over one pulley, with the mass and the spring-balance on the left and the other mass on the right. Because everything but the masses is massless it's still in rest, and the situation is exactly the same as when the rope was attached to the ceiling instead of to the other mass.
exit 16y ago
but what you're saying is that the spring can't distinguish between the setup on the right and the setup on the left. that doesn't seem right. if you detach one of the weights and anchor it on the other weight, the balance should read 2kg. and if you anchor weight A on the string of weight B, it would also read 2kg. as you move weight A's anchor point closer and closer to the balance, the reading remains 2kg. what happens when the anchoring point reaches the balance?
- jazzychad 16y ago
  > you're saying is that the spring can't distinguish between the setup on the right and the setup on the left. Correct, because according to the balance, both setups are indistinguishable. The dark secret of spring balances is that they measure (and display) force in only one direction. In stasis, the net forces on the balance are exactly zero N since there are equal and opposite forces acting upon the balance (true in each setup), yet the needle moves because it is measuring the movement (stretching) of the spring inside the balance (which at one end is anchored internally to the balance casing, and the other end is anchored to the string/mass). So, as for the setup on the right, the left-hand weight is pulling on the balance with weight 1kg (9.8N), and the right-hand weight is pulling on the balance with weight 1kg (9.8N), but since the balance only displays forces in one direction, it will display 1kg (in this case it doesn't matter which direction it is measuring because the forces are 180 degrees from each other).
  - dedward 16y ago
    I think that's reading a bit too much one-directionality into it. While true, it's the amount of force that's at issue here - not the direction. If a second 2kg weight were added to the "anchor" end of the spring balance - it would either register -1kg (if it's built to do that) or break or lock up givinging invalid or useless readings. Of course - assuming the balance isn't anchored anywhere else, this would set the whole system in motion, and it would all fall apart ratherquickly.

meelash 16y ago

Free body diagrams, people.

kreiling 16y ago
In case you, like me don't know what free body diagrams are, then: "The only rule for drawing free-body diagrams is to depict all the forces that exist for that object in the given situation. Thus, to construct free-body diagrams, it is extremely important to know the various types of forces. If given a description of a physical situation, begin by using your understanding of the force types to identify which forces are present. Then determine the direction in which each force is acting. Finally, draw a box and add arrows for each existing force in the appropriate direction; label each force arrow according to its type. If necessary, refer to the list of forces and their description in order to understand the various force types and their appropriate symbols." http://www.physicsclassroom.com/class/newtlaws/u2l2c.cfm http://www.physicsclassroom.com/class/newtlaws/u2l2c.cfm

kreiling 16y ago

The scale reads zero. Or, if you pull up on it, bringing it above the "horizon" (or bringing the pulleys closer together and below the horizon) causes the scale to approach 2kg. The force of each weight is converted to a horizontal direction.

DougBTX 16y ago
If scale reads zero, then that means there must be no force on the scale. If there's no force acting on it, that's much the same as if it wasn't there to start with. So, if you were to remove the scale (effectively cutting the string) the weights would stay in the same place, floating in mid air.
- dedward 16y ago
  I think he's assuming you are attaching the scale with the middle of the connecting-cable as the mass to be weighted and then pulling upwards on it..... which would make his description correct for that problem - but that's not the puzzle posted here.

modeless 16y ago

The scale really measures force, not mass. On the left the scale measures 9.81 Newtons applied by the weight, which is labeled 1kg because Earth gravity is assumed. The ceiling applies an equal but opposite force on the top of the scale through the suspending string, otherwise the scale would accelerate downward. On the right the force from the ceiling is replaced by force from another weight; this doesn't change the reading on the scale.

meelash 16y ago

Okay, so I thought back to my physics days some more, and the real understanding of the problem lies in understanding the definition of tension. The scale is actually measuring tension in the rope. An informal definition of tension: cut the rope at any point, and determine what force needs to be applied to the one of the cut ends to keep that part of the rope in equilibrium. That force is the tension in the rope. So in the second example, just measure the tension in the rope.

phaedrus 16y ago

Coincidently I'm reading Ernst Mach's "The Science of Mechanics" at this very moment. It includes a lot of analysis of setups like this. As for this problem: consider first the weights. Each one's downward force is 1 kg, due to gravity. There must also be exactly 1 kg's force's worth of tension in the string, because if it were less the weights would be moving down and if it were more the weights would be pulled up. Because the system (is assumed to be) not moving, we may replace the right side weight and pulley wheel with a fixed anchor for the string, in which case there would be no doubt that the spring balance displays 1 kg. Argument from symmetry says the scale would also read 1 kg if the left side were anchored and the right side free. Because the system in unmoving, it really makes no difference whether the opposite side of the scale is fixed to an anchor or balanced by a free weight: it will read 1 kg in all cases. This is very similar to a Car Talk puzzler that their physicist consultant clarified: is it worse to be in a car going 60 mph that hits an unmovable highway overpass wall, or to be in a car going 60 mph that hits an identical car perfectly head-on that is also going 60 mph? The physicist's answer was that there is no difference: in both cases you experience an identical impulse going from 60 mph to a stop in the same time. You may be inclined disbelieve this on the idea that the crumple zones in the car-to-car collision make for a "softer" collision, but consider this: define the point at which the bumpers of the two cars as "the origin". Assuming the cars and the collision are perfectly symmetrical, each car is going to crumple up on its respective side of the origin identically. Suppose the force starts small and ramps up gradually as the crumpling occurs; each car will be pushing the other away with the same force. An unmoving highway overpass barrier would resist with the same force.

Ralith 16y ago
Pedantry: The kg is a unit of mass, not force. You want Newtons.
dedward 16y ago
Moving some vectors around to make it look more similar - would I like to hit a nice solid wall at 60MPH or a parked car head-on at 120MPH. The head-on collision has twice as much energy to dissipate, and introduces all kinds of projectiles and other stuff heading in my direction. I'll take the wall over the car.

panic 16y ago

In the first picture, the weight pulls down with 1kg of force, and the ceiling pulls up with 1kg of force. If it didn't, there would be an imbalance, causing a net acceleration downwards. In the second picture, the first weight pulls rightward with 1kg of force, and the second weight pulls leftward with 1kg of force. The situation is exactly the same as the first, except with gravity providing the second force rather than the inter-molecular bonds responsible for maintaining the rigidity of the ceiling.