So if there's an upward force of N, shouldn't the tension in the two strings each b 1/2 N?

so if there's an upward force of N, shouldn't the tension in the two strings each b 1/2 N?
no bully

Yes, you brainlet.

that's not what my physics professor said, see file for his name (missing the II)

then wtf is this? incorrect?

OP would there be 2 strings or really just 1 string? The Tension would then be 1N, making it 0. Why do you think there's two strings?


The vertical string is applying the upward N force, this does not count as "two strings"

I mean, I know how to test it:
you take identical string and test till it breaks vertically, then you set up the initial picture and add force till they break, but we had a fucking quiz on it today and we haven't even done it in lab

two strings, the non-verticals, connected where the 3 meet

this actually makes sense but I have to test it to be sure, although the two strings will ALWAYS be oscillating back and forth so the entire tension is always in one string

because if you hang a bar between two ropes and sit on it, even if the rope would snap if it was just one, two will hold u.
right? I don't have the means to test this

OP, what is it your professor asserts? It's kind of hard to prove or disprove what claim it is you're making when we aren't really sure what it is you're claiming.

that the force isn't 1/2 N, that it's calculated using trigonometric vector algebra
now my argument:at the instant of jerk,if the two ropes were not connected and then become tensioned, if they are exactly equal (impossible), the tension should be 1/2

Which force? Tension? Tension has an equation and yes, it is not 1/2 of the force applied upward. It's mass of an object * 9.8m/s^2

I guess there's no point in argument, to claim it wouldn't be the full force would be a belief in macrosuperposition

The vertical component of tension in each individual rope will be half the total vertical tension that the force imparts.

But there are two horizontal component vectors in opposite directions for each of the two separate ropes. This contributes to the resultant vector that each is represented by. Thus you need trigonometric algebra to find the total tension in the ropes, not just how much they contribute to the vertical component.

but tension is NOT m*a, it is indeed JERK, the definition of which I forgot, maybe the derivative of acceleration

see, I'm OP, and we've got different answers in this thread; we need Aristotle

>net force
>force along a given vector

you guys know they don't have to be equivalent right

OP, what is it your professor is telling you? I still don't get why you are saying he is saying you are wrong. Plz give explanation and maybe we can help you along those lines, perhaps you are confused.


Trying to be no bully here but we can't help u when u withhold information buddy

Jerk is not mentioned on the Wikipedia page

It's fucking astounding that it tool Veeky Forums this fucking long to answer the question even though I'm pretty sure it's bait. Holy shit this place really is fucked.

this is the last time i'm ever going to post; physics is not in the realm of Wikipedia, schrodeigngers cat or w/e applies to EVERYTHING that's moving, you CANNOT measure it's velocity & it's position at the same time, BUT both can be calculated

Ok OP stop calling everyone jerks or were not gonna want to help you.

You don't need to be rude to get your point across.

>if they are exactly equal (impossible)

The two forces would fluctuate and be constant only for each moment in time.
As well, if they were close enough to equality in the first place then both oscillated, it is reasonable to expect them to at one moment in time be "exactly" equal.
It seems to me there might be extra stress on the two cables since they are no longer vertical, but rather at an angle. I leave for phys to call me retarded on this one though.

Only the y components of each force will be 1/2 N. The x components will depend on the angle they make.