

Weighing the options: A simpler question about pulleys and ropes. - RiderOfGiraffes
http://www.solipsys.co.uk/new/WeighingTheOptions.html?HN2

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phaedrus
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.

~~~
dedward
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.

~~~
modeless
Your first paragraph is wrong. You can't switch frames like that because in a
collision both cars will quickly come to rest relative to the ground, so
kinetic energy in the ground frame is what matters. A car at 120 MPH has 4
times the kinetic energy of a car at 60 MPH, not 2 times, because kinetic
energy increases as the square of velocity.

I'll agree with the projectile thing though.

~~~
RiderOfGiraffes
Not entirely true. After colliding with a stationary car the previously moving
car will retain some of its velocity. That will then dissipate over time and
be less violent.

So to some extent - when simplified, to a first approximation - you really can
switch frames of reference like that.

~~~
modeless
When simplified, yes, but if you start talking about projectiles then you are
explicitly not simplifying things, and should also account for the effect of
the ground because it's probably a larger concern than projectiles in a
collision like this.

------
modeless
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.

------
jazzychad
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.

~~~
exit
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
> 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
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.

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meelash
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.

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RiderOfGiraffes
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>) 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.

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panic
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.

------
meelash
Free body diagrams, people.

~~~
kreiling
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>

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kreiling
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
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
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.

