So me and a couple of others are having an argument over this.

If the pulley's are massless you would either need to know the angle of inclination + length of the leg or the magnitude and direction of the pivot force on the leg.

>>8524392
you could estimate the angle.

just calculate it, leaving the angle as a variable and then plot the mass over the angle, that gives you a range. the measure the angle from the..paint picture (or original picture OP might have). based on the uncertainty of the angle you can calculate the incertainty of the mass with error propagation.

>>8524405
what is the point of any of that. you dont take experimental data from subjectively portrayed paint pictures. as the problem stands it is not solvable with the given information

>>8524405
Some people claim the answer is 7,5kg, regardless of angle as they just cancel out the angles of both sides of the equation. This is simply false in my opinion.

>>8524408
actually they are right, i didnt even read the thing correctly. it is because the moment arm of the force of gravity and the tension of the string both have a factor of sin(whatever the angle s). you can take this factor out. In fact you dont even need the length of the rod to conclude that, simply from the fact the center of mass is presumably half way between both ends, that the mass of the rod is half that of the block

>>8524415
what i meant was that the block has half the mass of the rod before a critical typo ensued but you get the picture. so m_weight = 7.5 kg

>>8524415
But what if the block was standing upright, in an angle of 90 degrees. Then the mass of the weight would have to equal the mass of the block, meaning that when angle increases, the mass of the weight increases.

Are we assuming the system is in equilibrium?

>>8524408
would need to see the equation for that.

>>8524407
>>8524415
>>8524417
>>8524419
get the mass(angle) function. if the angle crosses out for some reason you get a definite answer, though i don't see it happening (and am not interested enough to calculate it myself)

>>8524427
continued:

also, if the angle changes, will the pulley still be right above the ropes attachment point? because that would change some things aswell

>>8524419
In such a case the pivot force would restore the block's center of mass to equilibrium regardless how heavy the block is. We implicitly assumed that sin theta is not zero so that net torque could only be zero if the force of gravity on the rod was twice that of the string tension. The special case of it standing upright is analagous to changing the setting to anti gravity, it is not the intention of the problem even though it doesnt explicitly forbid it in the premise

>>8524427
Ah so you are that get the protractor out for paint picture guy huh.

Net torque of leg about pivot
= 0
= sin(90-theta)*mg*d/2 - sin(90-theta)*T*d
=cos(theta)*d*(mg/2 - T)
Therefore as long as it isnt "upright" we must have mg = 2T so that the mass of the block is half that of the leg mass.

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quick guess

Not enough information.

It's impossible.

If you could lift the leg at all it wouldn't stop until it was pulling on the fulcrum. It would not hang.

>>8524495
apparently gravity doesnt exist in your world

>>8524505
Hello brainlet

>>8524511
>calls other people brainlets yet thinks a rod on a pivot being held up by a string is physically impossible

>>8524495
Not sure what you are trying to say at all. You can just set the rod at the angle you like and it will remain in that position. Nowhere does it say its pulled up from the ground with no outside assistance.

>>8524495
That's not true. As the leg lifts, the angle changes and so to do the component resultant forces felt by the string.

The leg will lift until the string tension cancels exactly.

>>8524519
No thats also wrong. You're both wrong. Regardless of the angle then net torque on the rod is 0. Lets say you start the system with the rod close to the ground. Even close to the ground the system will be at equilibrium. If you then "pull" on the block slightly because there is no net torque at any angle the rod will continue rotating until the string doesnt even point straight down anymore and the whole thing looks like crap. But thats alot of IF's when any sensible person would just realize the system starts in this equilibrium and no invisible hand is pulling on things.

>>8524514
It is impossible in equilibrium.

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find the reduction in distributed load (due to CoG) at any given angle and you have your mass ratio.

If the mass of the weight is half of the legs mass, then you are assuming an angle. So yeah, impossible.

>>8524540
although it will be unstable.

think of it the other way around, a piece of massless string with a weight hanging off the edge of a cliff balancing this leg at X angle. the ratio will be the inverse of op's picture.

but any movement will tip it one way or the other as the CoG moves.

ok, so:
[math]momentum M=r x F = I\cdot \alpha [/math]
[math]momentum of inertia for such a rod is I = \frac{1}{3}\cdot m_{leg}\cdot l_{leg}^2 [/math]

the rod is in equilibrium, so \sum_{}^{} M_i = 0

the rope pulls upward with [math]F_{rope}=m_m \cdot g [/math]. the tangential component of that is [math]F_t = m_m \cdot g cos( \theta )[/math]

since [math]F=m\cdot a[/math] the tangential acceleration upwards (and downwards, because equilibrium) is [math]g cos(\theta)[/math]
and since the angular acceleration [math]\alpha = \frac{a_T}{r} = \frac{g cos(\theta)}{l_{leg}}[/math]

the sum of the two momentums being zero means that
[math]l_{leg}\cdot m_m \cdot g cos(\theta) = \frac{1}{3} m_{leg} l_{leg}^2 \cdot \frac{g cos(\theta)}{l_{leg}}[/math]
[math]\rightarrow m_m=\frac{1}{3} m_{leg}[/math]

assuming the pulley is always above the rope point. ..who would have thought.

>>8524438
>get the protractor out for paint picture guy
the one and only.

>>8524553
???

the mass is somewhere between 1 and 1/2.

the only way to get less than 1/2 is to raise the leg above 45 deg at which case youre in tension so it doesn't matter.

>>8524576
are you dissin' my math, bruh?

>>8524553
You say the tangential aceleration divided by r is the angular acceleration is equal to gcos(theta)/L, that is wrong because you contradict yourself in assuming the net tangential acceleration is gcos(theta), its not, its 0. Following that you invoke the torque law, but in reality it should properly be expressed as
L(leg)*mass(block)*0 = Inertia * 0
where the 0's are placed where you erroneously inserted non-zero accelerations. The reason you get no information about the masses from this expression is obvious.

>>8524392
No

>>8524405
No (wtf)

>>8524419
Not what the question is.

>>8524438
>>8524472
These are the sole correct answers to this extremely simple problem.

>>8524485
Wrong.

>>8524495
Wrong. The initial state is at rest obviously.

>>8524519
Wrong. They cancel at every angle.

>>8524527
Wrong.

>>8524546
Wrong.

>>8524553
Wrong beyond redemption.

>>8524552
You're right, but same as that other guy, there is no reason to assume that the initial state is not already in equilibrium. Its like saying the set up is impossible because we cannot exactly measure the correct block mass we need. Its irrelevant to the question.

>>8524576
Wrong

Tremendous amount of retards in here.

>>8524705
wtf, i'm not saying the net-acceleration is a_t?
the up acceleration is a_t which is why the down acceleration must the the same value, because the net_acceleration is zero.

that's why i use gcos(theta) is alpha=a/r

>>8524721
you can't just come in and just say "LUL WRONG!" to everyone.
100% it's this >>8524553

>>8524721
Yeah so like I said if we assume it's balanced it's just a simple distributed load reduction.

But, You would still have to assume the height and angle of the leg to know the CoG and it's relation to it in order to get an actual answer.

>>8524438
why factor 1,2?
why no momentum of intertia?

>>8524806
Because he wants to assume gravity doesn't exist. In which case the entire model wouldn't work so he's retarded.

>>8524766
All you need to assume is that its a uniform bar... The angle is irrelevant.

>>8524758
Referring to tangential acceleration as acceleration in the "up" direction is asking for problems. You unfortunately thoroughly dont know shit. I suggest you retake mechanics.

>>8524806
Why do you need to even consider moment of inertia? If rod is in static equilibrium, the torque about its pivot is 0. Thats all you need to know.

>>8524758
I can say LUL WRONG to everybody because almost everybody is literally wrong. I can't believe you typed all that shit in latex but don't even know basic dynamics.

>>8524829
>referring to tangential acceleration in the "up" direction is asking for problems
sry that english isn't my native language. you still know what i mean!

>typed all that shit in latex, but don't even know basic dynamics.

fucking correct it then.
"science and math", where nobody does math, but instead calls each other assholes all day.

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>>8524388
This is just an elementary statics problem.

assumptions:
>frictionless pulleys
>static equilibrium (implied by the OP)

setting moment about fulcrum equal to zero
m = [(4cosθ)(15g)]/[(8cosθ)(g)]
m = 7.5 kg

theta is irrelevant

>>8524863
Its difficult to correct because it seems incomprehensible. Tangential has a very clear meaning. If you have "defined" tangential direction as going up, then "tangential" acceleration would refer to the component of the rods acceleration in the up direction. You cannot equate "up" linear acceleration with angular acceleration using the equation r*(angular acceleration)=linear acceleration unless you are specifically referring to linear acceleration that is tangent hence why people use the words TANGENT in this context to the circular path of the rod end. Overall your shit just makes no sense. I recommend you instead read this >>8524438 very simple answer which for the record I did not write, because it is actually correct and if you read the first couple posts you will notice OP's "associates" non-coincidentally got the same answer.

mfw /sci/ can't do calculus

>>8524829
The angle is irrelevant for a formula, but not irrelevant for an actual answer.

Since OP included the mass of the bar we can assume he wanted an actual answer. Since the mass of the bar is just as irrelevant as anything else if you aren't going to solve.

>>8524874
No, this is a calculus problem. It's not elementary statics. M changes depending on theta. In fact at some theta before 90 M is zero (obviously)

You can't just assume there's no moment otherwise it would be a boring, needlessly complicated question with no basis in reality.

>>8524944
>The angle is irrelevant for a formula, but not irrelevant for an actual answer.
This makes no sense. An actual answer has been provided, using a "formula" Its 7.5 kg.

>>8524944
>You can't just assume there's no moment
You mean the moment of force or the torque? You assume that because its synonymous with static equilibrium, and gives the solution to the problem.

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At first I incorrectly thought that the angle is required, but upon closer inspection this is what I got.

>>8525009
But what if it's almost upright? Surely the weight must have more mass if the angle is, say, 89 degrees.

>>8525030
Nope. Your misconception may be because you think this system can be altered to cases of different thetas just by pulling the block down. Thats actually not how it would work, because then the rod would rotate so that its end is not directly below the pulley anymore. In other words the system is uniquely configured for each theta, but no matter which theta you choose, the torques of gravity and the string on the rod will cancel as described.

>>8525030
just plug in 89° and experience the joys of experimentation.
Oh boy, I tell ya, 'tis gonna be fun.

>>8525030
If you decreased the mass of the block the rod would fall straight down and if you increased the mass of the block then the rod would rotate counter clockwise until the string segment it is connected to is parallel with the rod, in which case the system gets "stuck" because the tension of the string is now pulling parallel to the pivot point. In either case the string segment is not going straight down as the illustration clearly implies is occuring.

>>8524878
well, i get why that works.

not 100% were the thinking error is in my approach. i found minor mistakes, but the idea to use angular momentum should still theoretically work to use torgue=MoI*angular accel.

whatever. i stand corrected.

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Indeed, the angle doesn't play a part here. The counterweight's mass must be equal to a half of the leg's mass.

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>>8524919
>mfw you think calculus is required to solve this

>>8525094
Cross product (torque) is part of calculus.

>>8525104
Σmoments(fulcrum) = 0 = (4cosθ)(15g) - (8cosθ)(mg)

Where is the calculus in this expression?

>>8525112
Definition of moment/torque is [math]\bar{\tau}=\bar{r}\times\bar {F}[/math]. You just skipped steps.

>>8525104
Hilariously stupid. We are also assuming the isotropic nature of time which is part of relativity but no one with a brain would try to classify this problem as such.

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>>8524553

You tried

>>8524553
Holy shit what the fuck is this shit

>>8524874
>just an elementary statics problem
this^ not even difficult.

>>8524553
you gotta seek some help, really...

>>8525144
Your mother is stupid. I was only one to be able to present an exhaustive solution to the problem.

>>8525094
You need calculus for center of mass and moments. distributed load calculations. there isnt a correct answer in this thread.

>>8525542
(but that's because the problem is another fucking stupid mspaint problem)

>>8525542
>>8525485
Yeah so wheres your exhaustive solution? Clearly one makes the assumption that the bar is uniform otherwise the problem cannot be solved. To say that you need calculus is just nonsense, you don't.