How can I show a fucking idiot that the tension in the rope/scale is of 100N?

it doesn't take a whole lot to see it's in equilibrium, but that's not what the problem is asking

>How can I show a fucking idiot that the tension in the rope/scale is of 100N?

I will clarify, since I forgot to give his 'numbers', and maybe it is necessary to understand his mental disability.
He believes that the scale will read (which is the tension) 200N.

Because if it was 200 the weight would fall on the floor

Makes sense. It's fundamentally no different from hanging it from a heavier object.

could i get 1 (just one) written out, non picture, sum of forces that shows that this is 100N?

>How can I show a fucking idiot that the tension in the rope/scale is of 100N?
Get him to replicate the experiment.

it reads zero you retards

the system is in equilibrium so the forces balance each other out, 100 on the left and 100 on the right

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how does it feel to be this retarded

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t. Brainded

is the scale moving?

no.

therefore the net force on it is zero.

simple stuff...

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my right arm weights about as much as me left arm therefore they both weigh nothing

cover up the right side of the page and say "pretend the scale is attached to a wall, what is it going to read?", if he's only mildly autistic he'll say 100.

Then add the right side of the page back in and explain that since the system is in equilibrium, it should be no different since instead of a normal tension with a wall, its just replaced with an equal and opposite weight.

you do realize it's asking for Tension right? not Net Force?

if you draw the free body diagram of one weight attached to a wall on top of the table, the force in the scale will still be 100N, and the wall will be pulling on the string with 100N. You can represent that force the wall exerts with another 100N weight.

none of you can provide the result, what are you people on about?

why this board fucking sucks this bad? isn't there like one fucking physicist who can adress the OP's question? what the actual fuck?

Everyone has already answered the question itt. There's really nothing else to it and it will measure 100 because at that force applied to the spring (which ks then applied to the other weight) you get equilibrium. Brainlet may be just a fucking meme, but it's not that difficult to picture even without much knowledge fr mechanics. Saying it's 0 is just being dumb.

This

Jesus christ this board does not exist to help you with your grade 4 homework, you could determine the answer to the ops question any of the following ways: thought experiment, inspection, google, read a book, and finally perform the ops experiment yourself. Given the incredible ease of conducting any of these research methods constructing an argument becomes moot as ops friend is clearly a retard beyond critical thought.

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does that mean if I grab a random person our weights will cancel out and we will float, nice.

honestly the only thing that actually helped me understand it itt wasn't any of the intuition, etc. it was

peak pure mathematician

What about this situation? What will scale read?

50 Newtons.

I'm a physics brainlet, my instinctive answer was 200N. The problem lies in how your friend thinks the scale works, not how forces work.

50, at least once it's in equilibrium assuming the scale doesn't fling onto the floor

Shouldn' the tension of each string be 100N and the scale read 200?

But the 100N answer is still true if you remove the scale.
No information about the scale is provided, anyway.

No. If that was true. It would mean that a free falling system of a string and a weight (ignore atmosphere) would experience 100N of tension. Think of it.
I provide you with that mental experiment

I know. He probably thinks the scale adds the opposite forces into a scalar value.

Let's rewrite what I just said: he thinks that the scale can measure two vectorial forces by just adding the norms.

this is wrong
in one equation you're saying J = ma
in the other you're saying a = 0
and finally you're saying J is somehow 100, even though it is a result of multiplication by 0
lmao

I wrote F = ma as a starting point before resolving in the horizontal direction towards the right. Tension acts towards the left whereas the weight acts towards the right.
As for a = 0. It has to equal 0 since the system is in equilibrium

Think of it this way.
First the right weight is on the floor. You start pulling the string 10N, 20N, 30N. The weight doesn't move until you reach 100N, then you start lifting it and you are stuck at 100N even if you pull harder.

>this board does not exist to help you with your grade 4 homework
a grade 4 homework which apparently can't even properly solve. i'm not here to read your half-arsed answers based on your nonexistent physics skills you retard. nobody here needs your dumbass input on how easy it is to solve a question you can't even fucking explain why.

it wont be 50, it will me
T=2g m1 m2 /(m1+m2) where m1 and m2 are masses of the weights.

Ronnie Coleman grabs your right arm and screams in your ear "ERRYBODY WANNA BE A BODYBUILDER NOBODY WANNA LIFT NO HEAVY ASS WEIGHT" then Jay Cutler approaches you and grabs your left arm and whispers to you "Come on kid, suck it up." They both proceed to pull on either half of your body in opposite directions. You hear the sound of your shoulder bone snapping out its socket. You can finally feel it. Your body is starting to actually rip apart into two pieces. Luckily there is a physicist nearby and he examines the situation. It is obvious to him that the tension from both bodybuilders cancels the other out and just as your spine snaps in half he explains how it is physically impossible that the sum of the tension would be 200 newtons.

>Hello academic advisor this is user, another user called my physics skills nonexistent on the internet i'm dropping out.
You cant identify f=ma and you cant perform basic algebra, is this the power of american education?

think about how that scale works.

First you have to hold it, only then you can hang a weight and measure it.

Therefore on the right side there is 100N weight or there could be 200, 500, 2000N weight or it could just be attached to wall, or you could be holding it. t doesnt matter, what matters is the weight on the left
the answer is 100N

No, that's the tension of the rope in an atwood machine. It will read 50. Jesus Christ the brainlets.

look here brainlet, both weights are 1 kg, scale reads 9.8 which is in accord with what i said,
IF YOU CAN PROVE ME WRONG POST EXPERIMENTAL PROOF AND CALCULATIONS

yeah 1kg is 10N(or 9.8N if you are autistically precise) are you retarded?

obviosuly 250 brainlets
first imagine hanging a 300N weight on a wall
then imagine adding a 50N weight on the opposite side
the forces counteract, so the reading is 250N

of all incorrect answers possible 250N is the most brainlet one

i think saying "0 since it's in equilibrium" woul dbe the most brainlet response

If the scale is free to move, it won't show the real weight of the object, since it's moving. If the scale is isn't free to move, the scale will show the weight of the object that is on the side which measures it.

Nah, it's fixed. You set up the scale beforehand to read -10.

>t. I spend half of my pension on tin foil hats

>hurr durr the equilibrium position is in acording with my shitty formula
No shit retard, so does the other models, the point is off when one weight is heavier.

Well done Sr. Isaac

Fuck the rope and use symmetry.

this retards had the same problem, and not happy with obviuos lab results, they destroyed some cars :^^^)
youtube.com/watch?v=-W937NM11o8

Long time since A level physics, come at me bros:

For every action there is an equal and opposite reaction, right? So if one end of the scale were attached to a wall not a weight, (and the scale would then read 100N, yes?) the wall would be resisting the pull of the weight with a force of 100N. Am I right so far?

So if you replace the wall with a 100N weight, you are not making any material difference to the force acting on that end of the scale. So the scale still reads 100N.

How did I do?

the scale will move untill it is stoped somewhere (sbd sshould calculate) and there it is going to have a different tenson probably

Better than the average engineer. Grats user, you are correct.

if the weights are not equal the system is not in equilibrium and thus accelerates toward the side with more weight which will reflect on the measured force

the scale obviously works symmetrically as you can attach a weight to either end

my hypothesis is that the scale divides the sum of opposing forces on both ends by 2 so if you attach 100 N to one end and 200 N to the other end the system will accelerate toward the end with 200 N and the scale will measure 150 N

I am curious, can anyone here confirm or deny my guess?

Tell me does the scale reads something when you don't have a weight in one end?
This should have ended the debate.

in an idealized scenario it will not

in a realistic scenario it will due to the acceleration and the inherent mass of the measuring device

I just want to mention there are obviously other factors too but that would be the most significant in a realistic scenario

The spring may oscilate a bit because of friction and aberrations in the floor yes, but if you can actually calculate that without using an idealized friction (which in turn will only end afecting the acceleration of the apparatus and not the ideal measurement) then by all means, give a better model. But if you aren't autistic and want a concrete and measurable answer, the dynamometer will read 50 newtons if the apparatus holds to just slide at a constant acceleration .

Fucking brainlets, it's all because of relativity.
>attach first 100N
>attach second 100N
>The moving part of the scale moves relative to the stationary part of the scale.
>Only one end of the rope experiences a force (relative to the other)
>So only 100N is applied to the scale.

>repeat from the perspective of the moving block
>see that the scale still reads only 100N

Your answer is correct but you cannot use relativity per se because you still need to explain why a part of the scale is stationary which is not a hypthoesis or can be considered a priori. It's just balance of forces considering the scale has a fucking spring inside it. Your method would break for
As you cannot fix your frame to the base as it will not be inertial.

Being a time-traveler I can solve this easily.
The scale will read 100N, as it's identical to a merchant's scale used from ancient times up to the late 20th century. The main types were the scales for goods where you had goods on one side and weights on another. The other type was used till the 90s for weight measurements (the last time I encountered it was at a pediatric clinic in 1996): to weigh a person you had them stand on the scale and you had to move a weight along a nearly horizontal bar to adjust the lever-force against the person standing (the lever allows for far lower counter-weight, as a multiplicator is introduced). The second type was later also used for kitchen scales, my grandmother still used one around 2005, she has switched to a digital scale since then, though.

The forces are not equal, so there will be acceleration. Acelleration depends on the masses.

The net force is 250N.
F = m * a
250 = (m1+m2) * a
a = 250/(m1+m2)
This is the acceleration of each block.

Considering the block on the left:
F = m * a
T-50 = m1 * a = m1 * 250/(m1+m2)
T = m1 * 250/(m1+m2) + 50

Considering the block on the right:
F = m * a
300-T = m2 * a = m2 * 250/(m1+m2)
T = 300 - m2 * 250/(m1+m2)

The scale will read T.

It's not a three block systen, there's a fucking spring in between them.

Does the spring have a non-negligible mass? If it doesn't have a mass, you can ignore it and consider the system as two blocks connected by a rope.

Thanks to your input I have realized my hypothesis is bullshit. Updated hypothesis: the measured force is [math]x\cdot (1+\frac{x}{x+y})[/math] whereas x is the smaller and y is the larger of the forces attached to the scale.

Continuing, considering that:

m1 = 50/g
m2 = 300/g

We have that T = 85,71 (approximately)

No, the spring as it streches mantains the left block tense and in equilibrium while the other block just pulls everything (the first box and the measuring apparatus) with a resultanr force of 250. But the scale reads 50. Again a spring is only tense if it's being pulled by both sides.

Scratch that, I fucked up my formula.

There is no spring. It is just an ideal device to measure force. You are over-thinking it.

>pic

Neither, the pulley's remove some of the weight. A tiny amount, but its there. Seems like this shows that,

You guys are over-thinking the problem. It is a textbook problem. Everything is idealized. So, there is no friction, rope and scale have no mass and don't stretch.

>let's us assume some bullshit dynamometer that is physically
Tell me how would you built an appartus that can measure forces (mecanical without) without some pivot or rest? Also considering is was a continuation of OPs and you are basically throwing the explanation of why it's 100 out of the window. The point of these problems is to get a proper conceptual understanding of Newtons law, not to show us that you can solve (incorrectly considering that it's not the same tension in each rope for a three block system) some shitty intro problem.

see

Here my updated hypothesis formula, this time it does not matter which force you choose for x and for y.
[math]\frac{2xy}{x+y}[/math]

If x = 100 N and y = 100 N then the output is 100 N. If x is arbitrary and y is 0 N then the output is 0 N. If x is 50 N and y is 300 N the output is ~85.714N which is the same as

I think this time my hypothesis is accurate. Fuck me, how did it take me 3 tries to get it right? I really have become rusty and retarded.

Then it's a laughably ill posed problem. What the hell is an "idealized" measuring apparatus? If you go to the lab and try with a dynamometer you will get 50 in that configuration. Forget the blocks, just consider you ar holding it with your hands, no gravity or friction or whatever, and you are some sort of god an can pull with one end 50 and the other 300, what do you expect to read? Obviously if have no understanding of how the measuring apparatus works you fan have different answers. If you said a ok, I have some bullshit that let's me measure the tension in rope in a direct non invasive way then ok it's just a retarded way to pose the problem, but if you tell me dynamometer that uses a spring is in between the two blocks, then the apparatus will measure 50 which a lot of people find difficult to understand. Otherwise the problem wouls be solve the atwoos machine.

Generalization:

F = m * a
F2 - F1 = (m1+m2) * a
F2 - F1 = (F1+F2) * a/g

a/g = (F2-F1) / (F1+F2)

F2-T = m2 * a = F2 * a/g = F2 * (F2-F1) / (F1+F2)
T = F2 - F2 * (F2-F1) / (F1+F2)

T-F1 = m1 * a = F1 * a/g = F1 * (F2-F1) / (F1+F2)
T = F1 + F1 * (F2-F1) / (F1+F2)

So, if F1 = F2 = 100, T = 100
if F1 = 50, F2 = 300, T = 85,71

Let's say you have a scale and a 50N block on it.
If it is at rest, it will read 50N.
If it is accelerating up, it will read more than 50N.
If it is accelerating down, it will read less than 50N.

this post makes me want to smash your face

No, that's the whole point. To make it easy, that's why you can think of a basic dynamometer as a spring with known hooks coeficient etc etc. The spring streches only if it has a force being applied at both sides. Usually the second force is not considered because you just hang it from a ceiling, but the ceiling acts as a way to apply an oposite force. In this case you have 300 newtons going right and 50 going left so the spring will strech till the point where there's an equilibrium with the left forces, then you have 250 newtons still pulling the whole thing but not stretching it. You can try it, go see if you can strech a spring by just pulling from one side.

Nice quads.
If you pull anything 300N going right and 50N going left, it will accelerate to the right. It may stretch, it may oscillate, but it will accelerate.

This is bait

Yea, bur the question is what's going to be measured by the fucking apparatus. As I mentioned, ita painfully obvious that it's a dynamometer in OPs pic. And yes, the thing will accelerate to the right being pulled by a force of 250 newtons. Now, you can even add mass to the apparatus and solve it.

Yes, I solved it. See: