Anonymous at Sat, 14 Sep 2024 00:53:32 UTC No. 16379391
100 N
Please go away now.
Anonymous at Sat, 14 Sep 2024 01:02:31 UTC No. 16379394
22 lb, it's not a metric force gauge
Anonymous at Sat, 14 Sep 2024 01:02:40 UTC No. 16379395
This is high school shit. It not read more than 100N.
Anonymous at Sat, 14 Sep 2024 01:07:55 UTC No. 16379398
>>16379389
Middle school at best, elementary...
Anonymous at Sat, 14 Sep 2024 01:20:15 UTC No. 16379405
>>16379398
this post belongs in daycaregen, if i'm being honest
Anonymous at Sat, 14 Sep 2024 01:49:34 UTC No. 16379427
>>16379389
It looks like it reads แแแแแ
Anonymous at Sat, 14 Sep 2024 08:35:08 UTC No. 16379714
>>16379391
>>16379395
Not OP, but can you explain why?
Anonymous at Sat, 14 Sep 2024 08:51:58 UTC No. 16379721
>>16379714
the scale is in equilibrium.
the lefthand weight is pulling on the scale measuring side, while the other is keeping it still.
think of the righthand weight as exerting the force that a wall sustaining the scale would exert
Anonymous at Sat, 14 Sep 2024 09:20:51 UTC No. 16379740
>>16379714
The measuring device only measures the force on one end, not both ends. If the force was measured on both ends it would always be zero because for any force there's always an equal opposite force. Think of how you would throw a baseball with X newtons, you could not do that without the ball also pushing your hand to the opposite direction with the same X newtons.
Or think of how the device works by stretching a string and then measuring how much the string is stretched. You could not stretch it from one end unless you kept the other end attached to something or if no force was acting on it.
Anonymous at Sat, 14 Sep 2024 09:23:54 UTC No. 16379742
>>16379721
>>16379740
But the weights are pulling on both sides?
Anonymous at Sat, 14 Sep 2024 10:32:21 UTC No. 16379777
>>16379742
If they weren't, then the scale would not be in equilibrium, it would be accelerating. F=ma
Anonymous at Sat, 14 Sep 2024 10:50:44 UTC No. 16379796
>>16379389
Is the scale calibrated?
What's the mass of the hook? The string pulls left with a force of 100N
If the hook weighs 5grams then the hook experiences 5N of force vertically
Pythagorean theorem says scale reads 100.1N
If you don't believe me test this with an IOLab device. Just $200.
Anonymous at Sat, 14 Sep 2024 10:51:51 UTC No. 16379798
>>16379391
>>16379740
>>16379796
It's pulling 100 from one side and 100 from the other, and the spring is getting the full force.
It shows 200.
Anonymous at Sat, 14 Sep 2024 11:14:49 UTC No. 16379817
>>16379389
As a LLM I don't know shit about physics but I can recommend you this video:
https://youtu.be/XI7E32BROp0
Anonymous at Sat, 14 Sep 2024 11:21:26 UTC No. 16379825
>/sci/ is failing mechanics for 12 year olds...
sad, but not unexpected
Anonymous at Sat, 14 Sep 2024 13:12:01 UTC No. 16379946
>>16379798
>If you don't believe me test this with an IOLab device. Just $200.
Anonymous at Sat, 14 Sep 2024 15:22:18 UTC No. 16380076
>>16379817
Thank you very much for this explanation. While doing a textbook question, I came across such a scenario, and was under the false impression that the forces would cancel out and the answer would be 0 N of force, but your explanation indeed cleared my incorrect thinking and confusion. Thanks once again!
Anonymous at Sat, 14 Sep 2024 18:25:35 UTC No. 16380395
>>16379798
>IT'S 200!
I'm guessing you grew up with a European public education.
Anonymous at Sat, 14 Sep 2024 18:50:49 UTC No. 16380462
Brainlet analogies in here. It's 100, because simply it wouldn't be 200 if it was hanging from a baloon either.
Anonymous at Sat, 14 Sep 2024 18:53:41 UTC No. 16380467
it's zero you fucking retards. the forces cancel out.
Anonymous at Sat, 14 Sep 2024 19:15:29 UTC No. 16380492
The number of people who think it is not 100 is concerning. You can test this yourself with a few dollars of material. It won't be exactly 100 in that case as real ropes have mass and different parts of a rope with mass will have different tensions, but w/e. It will probably be pretty close to 100.
Anonymous at Sat, 14 Sep 2024 19:16:57 UTC No. 16380496
>>16380467
SF =/= T. Just do the math yourself.
Anonymous at Sat, 14 Sep 2024 20:38:35 UTC No. 16380653
100
Same as holding the scale in your hand with the weight hooked onto it.
Anonymous at Sat, 14 Sep 2024 21:24:06 UTC No. 16380729
>>16379389
square root of 100N
Anonymous at Sat, 14 Sep 2024 21:30:12 UTC No. 16380740
>>16379389
it's 100. the spring inside the measuring scale is will extend until the tension in the wire is balanced and it will be balanced (if it is properly calibrated) at whatever weight it equalizes the downward acceleration
Anonymous at Sat, 14 Sep 2024 21:38:26 UTC No. 16380748
>>16379389
wouldn't it be a little less of 100N, considering the spring elasticity constant and the symmetry being off since the spring starts on the right side or whatever, I'm not a physicel
Anonymous at Sat, 14 Sep 2024 21:46:52 UTC No. 16380759
>>16379389
200, each side of the cable pulls its respective weight up with force equal to the side's gravity in order for the weight to remain still.
100+100 = 200 N
Interesting question, OP.
Anonymous at Sat, 14 Sep 2024 22:03:59 UTC No. 16380781
>>16380759
No, but now that I think about it, what you mention will happen if the weight of the counterweight is increased, what you mention will occur.
However, if the weight of the counter was lower, the scale would still show 100.
That aside, imagine if that you tie a weight where the scale gets hooked into a rope/wall/etc and then hold the body (not the hook) with your hand. It will still read 0.
Anonymous at Sat, 14 Sep 2024 23:10:01 UTC No. 16380871
I'm just a midwit tourist here, but how do you end up messing this up? Imagine removing one of the weights. You're just pulling the entire scale, not the hook.
Anonymous at Sat, 14 Sep 2024 23:27:06 UTC No. 16380891
>>16379389
The answer is in Newtons, while a regular scale measures mass. If the answer is in newtons then this is a scale in the broader sense of the word, which is simply a measuring device, in this case it measures tension.
I think this is the trick. If it's newtons then it's a question about tension, what is the tension on the spring?
200 N. If one side was loose, the tension would be 100 N.
Anonymous at Sat, 14 Sep 2024 23:47:40 UTC No. 16380919
>>16380871
>how do you end up messing this up?
People are imagining the scale like a pulley instead of part of the rope.
This should fix it for people.
Anonymous at Sun, 15 Sep 2024 03:50:45 UTC No. 16381234
>>16380919
/thread
Anonymous at Sun, 15 Sep 2024 04:16:36 UTC No. 16381257
Only the hook can be used to measure weight
Anonymous at Sun, 15 Sep 2024 04:31:32 UTC No. 16381278
but how is the measuring device calibrated?
Anonymous at Sun, 15 Sep 2024 04:38:52 UTC No. 16381282
you could replace the counterweight with a wall, the force on the device is the same, 100N to be in equilibrium
imagine taking a section of the scale, what's the force on it?
Anonymous at Sun, 15 Sep 2024 04:38:52 UTC No. 16381283
>>16379389
just imagine grasping the scale, the reading won't change and then it's obviously 100N
Anonymous at Sun, 15 Sep 2024 06:21:21 UTC No. 16381369
>>16380919
>>16381234
The best way to say it with words instead of a rearrangement, is "if it was hanging from a balloon, it would still be an equal opposite 100N".
Anonymous at Sun, 15 Sep 2024 06:25:18 UTC No. 16381374
People pretend this is an easy question because they are narcissists. It may be fundamental at mechanics, but it's still counterintuitive; "opposite and equal reaction" was never an intuitive concept; I've been to structural engineering classes and half the students took months to get it.
Anonymous at Sun, 15 Sep 2024 10:30:32 UTC No. 16381660
>>16379389
are you telling me that two horses would do the same job as four horses?
Anonymous at Sun, 15 Sep 2024 10:56:23 UTC No. 16381684
>>16379742
If it measured the sum of both forces it would be 100N + -100N = 0N. Because if you think about F=ma, you have a=0 because the device does not accelerate, therefore you have F=0. So what the device does is that it pretends it was accelerating with just the right "a" to get 100N with its mass because it pretends there was no force to the opposite direction making the sum of the forces zero.
Anonymous at Sun, 15 Sep 2024 11:30:56 UTC No. 16381708
This question almost broke me but I think I got the right answer.
The scale system is a two points system linked with a spring that obeys Hooke's law. We can regard each endpoint as a subsystem under contact forces only.
This leads to a very simple system of equations:
F1 + k (x2 - x1) = 0
F2 + k (x1 - x2) = 0
F1 and F2 are the forces acted on each endpoint, x1 and x2 are x coordinates and k is the spring constant. These equations assume that the scale is at static equilibrium which is safe to assume since the forces are opposite to each other and equal in intensity. Otherwise one must add at each line a dynamic component at the right side of each equation which is non other but the
mass multiplied by the newtonian acceleration. The reading of the scale translates as the
k (x2 - x1) term which intensity is the same as F1 thus 100N.
A lot of anons are misled with the natural assumption that the scale has to be static in order to make a reading. Back to the above equations:
F1 + k (x2 - x1) = m a1
F2 + k (x2 - x1) = m a2
Subtracting the two lines gives :
2k (x2 - x1) + (F1 - F2) = m (a1 - a2)
Or u + (F1 - F2)/2 = m/2 d^2/dt^2 u
u being the reading of the scale. In the ideal case when m = 0 we finally obtain :
u = (F2 - F1)/2
Which yields in this case :
u = (100 - (-100))/2 = 100N
In another case where the scale may not be well attached (F2 = 100N, F1 = 0N) the reading must be :
u = 100/2 = 50N
This is the case where the scale is at free fall and one mass is hanging on one of its sides. Note this is again the ideal case where there is no mass inertia acting on the mouvement of the spring. Otherwise the reading would be off by a sinuse over time with period sqrt(m/2k).
Anonymous at Sun, 15 Sep 2024 11:56:11 UTC No. 16381736
scale just measures the tension of the wire. tension of non elonanting wire is same everywhere
Anonymous at Sun, 15 Sep 2024 12:06:23 UTC No. 16381745
Its 50N, because the pulleys take each a quarter of the tension
Anonymous at Sun, 15 Sep 2024 12:30:54 UTC No. 16381770
>>16381745
Anon, I have no words.
Anonymous at Sun, 15 Sep 2024 12:40:57 UTC No. 16381778
>>16379389
The answer is 0 because this is equivalent to running the experiment in outer space where there's no gravity.
Anonymous at Sun, 15 Sep 2024 13:05:27 UTC No. 16381801
>>16379389
Someone explain why this is wrong: When you hang a weight on a scale hook, it puts stress on the hook, no? When you release the weight, the tension should be relieved and the hook should go back to its rest state. Sort of like a spring on a much smaller scale (if theres no restoring force, the scale would be ruined after the first use). Which means, on the weight you're measuring, there should be 3 forces acting on it: it's own weight, tension from stressing the hook, and tension from the rope. Without this, whatever mechanism it uses to translate received stress into weight, whether displacement or piezoelectricity etc. would have no way to differentiate it from it sitting on a desk and would not function
If we assume the tension on the rope is the same on the weighing side as on the counter balance, then both the forces from the weighed object and the counter balance cancel, and there's no stress on the scale to register it has a load, therefore it can't read any weight. If it did feel any stress it would begin accelerating upwards because the weights on both sides would no longer be equal.
Anonymous at Sun, 15 Sep 2024 13:10:18 UTC No. 16381805
Since the hook is in stati equilibrium it experiences no net force and reports 0N. Tricky question, I like it.
Anonymous at Sun, 15 Sep 2024 17:36:02 UTC No. 16382115
>>16381257
>Only the hook can be used to measure weight
If a single weight were hanging from the nonhook end of the scale and the hook was hanging from a ceiling, then the scale would still read 100N
Anonymous at Sun, 15 Sep 2024 17:40:36 UTC No. 16382120
>>16379798
ahahahah hahahahaha hahahahahaaaaaaa
if it were fixed to a wall? what would the force on the fixed end be? 0 ?
ahahah
now point me to the medfag thread I need to discuss purely hypothetical shit
Anonymous at Sun, 15 Sep 2024 18:11:29 UTC No. 16382157
>>16381801
>>16381805
Since I am exerting a force of 900N on the earth, and the earth is exerting a force of 900N back, the resultant force is 0N; therefore I weigh nothing, and can float away like a butterfly!!!!
Anonymous at Sun, 15 Sep 2024 18:15:30 UTC No. 16382162
>>16382157
you are literally not moving
fast fuck.
what more do you want? for you to lift off one force would have to be in excess of the other.
Anonymous at Mon, 16 Sep 2024 00:45:00 UTC No. 16382497
>>16379389
Would there be an equillibrium that is dependent on the lengths of the ropes?
Anonymous at Mon, 16 Sep 2024 00:49:31 UTC No. 16382502
>>16382157
You can because you ARE a beautiful butterfly.
Anonymous at Mon, 16 Sep 2024 02:48:22 UTC No. 16382602
>>16382157
If you floated away there would be a net force on you. As you demonstrated, the net force on you is zero. Therefore you're not moving up or down. This isn't the contradiction you think it is boyo.
Anonymous at Mon, 16 Sep 2024 02:53:50 UTC No. 16382606
Hammock with shitty straps
how fat can you be, assuming the hammock was made to hold, and the straps have some minimal tension (it's not strung up on an extremely taut line by itself)?
Reminder that you're not necessarily in the center of the hammock and the lengths are never going to be the exact same.
Anonymous at Mon, 16 Sep 2024 03:06:31 UTC No. 16382617
>>16382606
isnt that the same base problem but phrased stupidly backwards
Anonymous at Mon, 16 Sep 2024 03:25:14 UTC No. 16382630
The real issue is people don't understand how scales work. Keep taking shit off the self checkout scale then immediately throwing it back on, then get annoyed the system calls for support.
Anonymous at Mon, 16 Sep 2024 09:11:13 UTC No. 16382831
>>16379389
It's statically indeterminate. Question requires more information on the sectional and material properties of the cable.
Anonymous at Mon, 16 Sep 2024 12:06:30 UTC No. 16383030
>>16379389
it will reach equilibrium. Put it this way, imagine that the cable is constant and throught it there will always be a constant force everywhere. Meaning you will have a force of 100N throughout the whole cable.
Anonymous at Mon, 16 Sep 2024 20:54:15 UTC No. 16383803
>>16382162
By the principles of GR, you are moving as long as you aren't in freefall.
Anonymous at Mon, 16 Sep 2024 21:28:33 UTC No. 16383823
>>16379714
The easiest way to understand it is to draw a free body diagram of one of the masses. You have 100N going down and the tension force pointing up. Since the system is at rest, the net force on the mass (and on everything else) must be zero.
Since the net force must be zero, the tension force in the rope must be equal to the downward force acting on the weight, 100N. That tension force is all the scale is reading. And the tension in both ropes must be identical for the exact same reason - the system would accelerate if they were not.
This setup is identical to one with the scale anchored to a ceiling with a single 100N weight pulling on the bottom, or to one where one side of the scale is anchored to a wall. It is counterintuitive but true, and a great example of why free body diagrams are important CONTRARY to all of the smarmy engineering undergrads who say they just 'get it' right away until they get into their first statics class and start crying.
>>16379740
this anon is a drooling retard who needs to come to my office hours.
Anonymous at Mon, 16 Sep 2024 21:30:35 UTC No. 16383824
>>16383823
But if all forces are balanced shouldn't it read 0?
Anonymous at Mon, 16 Sep 2024 21:32:27 UTC No. 16383825
>>16383824
The NET force is zero. there are still internal forces in the system, namely the tension force that the scale is actually reading.