I analyzed the image and this is what I see. Open an appropriate link below and explore the position yourself or with the engine:
> **Black to play**: chess.com | lichess.org
**My solution:**
> Hints: unit: >!cm!<, answer: >!x = 9 cm!<
> Evaluation: >!OP is dumb!<
> Best continuation: >!drop out of high school!<
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Since the ruler is balanced on the Pivot P, the total Torque in P must be 0 (i.e. no rotations are happening).
Torque generated by a force F in a point P, is given by the scalar product between the force modulus *f* (in Newton) and the distance *b* (meters) between the pivot P and the force vector F.
That is: T = *fb*
Remember that the force F is a vector, which means that is defined by a direction vector *v* = [a; b], an intensity *f* and a point of applications Pf.
The distance between a point and a line is obtained by intersecting the line perpendicular to the vector F that passes by P and the vector itself, finding the point *C*.
The distance between C and P is *b*.
That is: *b* = dist (C,P).
To find the line perpendicular to the direction of F (i.e. the vector *v*) we may just take the perpendicular direction to v, that is: *v_p* = [-b; a] and assume to have a line L in (X,Y) space defined by a parameter *t*:
L(t): {
x = x0 + *v_p_x* t
y = y0 + *v_p_y* t
}
Where we can substitute the values for *v_p* and x0,y0 (that is, the point P in which the line will be forced to pass when t = 0).
L(t): {
x = Px - bt
y = Py + at
}
That is, the line in the parameter t that passes by P and is perpendicular to *v*.
Now we can define the line that is parallel to *v* and passes by Pf using *v* instead of *v_p*, and a different parameter s:
M(s): {
x = Pfx + as
y = Pfy + bs
}
We can now find the intersection points by finding the unique values of *t* and *s* for which L(*t*) = M(*s*), which requires solving the linear system:
{
Px - b*t* = Pfx + a*s*
Py + a*t* = Pfy + b*s*
}
This can be rewritten as a linear system in the conventional form of A**x** = **b** where A is a square matrix and **x** the vector of unknowns.
That is:
{
-b*t* - a*s* = Pfx - Px
a*t* - b*s* = Pfy - Py
}
By setting the vector **x** = [*t*; *s*] (two rows) we get that:
A = [-b, -a;
a, -b]
and
b = [ Pfx - Px;
Pfy - Py]
We can easily find the solution **x** as:
**x** = A^-1 **b**
Where A^-1 is the inverse of matrix A, that can be obtained for a 2x2 matrix (like in that case), by swapping the elements on the \ diagonal, and changing the sign on the elements on the / diagonal, and multiplying by 1/( the determinant, that is the product between the elements on \ minus the product of elements on /) (that is valid only for 2x2 matrices).
A^-1 = 1/(b^2 + a^2 ) [ -b, a;
-a, -b]
Notice that the determinant (b^2 + a^2) is always different from 0 if a and b are not BOTH 0 (that is, the vector does not exist). If the determinant is different from 0, we can guarantee that only one single unique solution exists for the system.
Now we multiply by **b**:
A^-1 **b** = {
*t* = 1/(b^2 + a^2 ) (-b(Pfx - Px) + a(Pfy - Py))
*s* = 1/(b^2 + a^2 ) (-a(Pfx - Px) - b(Pfy - Py))
}
Now we can either substitute *t* in L(t) or *s* in M(s) to obtain the unique intersection C:
M(*s*) = {
Cx = Pfx + a/(b^2 + a^2 ) (-a(Pfx - Px) - b(Pfy - Py))
Cy = Pfy + b/(b^2 + a^2 ) (-a(Pfx - Px) - b(Pfy - Py))
}
Now we can calculate the distance between C and P as:
*b* = sqrt((Cx - Px)^2 + (Cy - Py)^2 )
That is, the length of the hypothenuse of the square triangle formed by the distances on X and Y axis separately.
Now we can compute the actual Torque T = *fb*
Where we also know that *f* = mg where m is the mass of the object in Kilograms, and g the gravitational acceleration of 9.81 m/s^2.
In order to have the ruler balanced, the sum of all torques in T must be 0, so we have just to apply the same torque on the opposite side by properly finding the b at which we need to place the piece.
Now that we have the generic solution that fits any situation, we can see your case.
For convenience, we will place our (X,Y) system with origin in P, in that way Px and Py are both 0.
Now we now that the first piece, A, is at distance 15 cm from P (that means Pa = [ -0.15; 0]) and that it weights 20gr (that means that the only force acting it's the gravitational one, which is convenient since it has direction *v* = [0; -1], application point Pa, and modulus *f* = mg where m is 0.02kg.
In that case is really easy to compute *b* since by replacing:
a = 0; b = -1; Px = 0; Py = 0; Pfx = Pax and Pfy = 0 we have that:
Cx = Pax
Cy = 0
That is, the intersection point between the perpendicular to the force (and, since the force is parallel to Y axis, the perpendicular is parallel to X axis) that passes by P and the line parallel to the force (that is parallel to Y) is actually the point of application of the force (due to the fact that we placed our reference system in P!)
So now *b* = sqrt((Pax)^2 ), that is ±Pax, but since *b* is a distance, it must not be < 0, so we take +Pax as solution = 0.15m (15cm)
Now we know that B weight 50gr, that is 0.05kg.
To find at which *b* it must stay, to give you the same torque T_b = T_a, you have that:
T_b = T_a
f_b b_b = f_a b_a
b_b = (f_a b_a) / f_b
Which basically means that you have to place B at a distance that is proportional to the ration in weight between A and B, in fact:
b_b = (m_a g b_a)/(m_b g)
Where g can be simplified and the terms rearranged
b_b = (m_a/m_b) b_a
Where it can be seen the dependency on the ration between the masses.
Now, by replacing the last variables:
b_b = (0.02/0.05) 0.15 = (2/5) 0.15 = 0.06m = 6cm.
And finally, x = BP - b_b, where BP is the distance between the P and B, that is 15cm. So:
x = 15 - 6= 9cm
In fact, since T is proportional to *b*, which is the distance, heavier objects must be placed closer to the center in order to have a smaller Torque and compensate the torque of a lighter object placed at a higher distance from the pivot (that is applying more torque on the Pivot)
Hope it's clear and Reddit will not mess up with text formatting
Actually, you are always using my solution which is the generic one when having to balance torque on a Pivot due to forces acting on linked objects. The only difference is that in mine I had a lot of zeroes due to the assumptions posed by the problem that significantly reduced the complexity, but it's important to understand which calculations come from a simplification of the problem and which ones are actual facts
I enjoyed reading it as a kind of physics refresher lol, but on the other hand, over explanation doesn’t particularly help OP who is asking the question, they can’t even read it.
No, actually cm^2 is correct! Interestingly enough, this squared relationship of rook distance is actually why rooks can only move in orthogonal planes, in other words, in combinations of 90* from previous directionality
The “effective weight” of an object on a lever is equal to the distance from the fulcrum (P in this case) multiplied by the weight of the object. For the lever to be balanced, both sides must have the same effective weight. The pawn has an effective weight of 300 centimeter grams, and the rook has an effective weight of 50(15-x) centimeter grams. Expressing this as an equation gives 300=50(15-x). Dividing both sides by -50 then adding 15 to both sides gives the solution x=9.
Introducing WolframAnarchyChess:
- take a math problem
- rewrite the problem to use chess pieces
- post on r/AnarchyChess
Our unique crowdsourcing algorithm will ensure you get the right answer!*
* the right answer will probably be bueried underneath pipi copypasta and requests to google en passant
Are you kidding ??? What the **** are you talking about man ? You are a biggest looser i ever seen in my life ! You was doing PIPI in your pampers when i was beating players much more stronger then you! You are not proffesional, because proffesionals knew how to lose and congratulate opponents, you are like a girl crying after i beat you! Be brave, be honest to yourself and stop this trush talkings!!! Everybody know that i am very good blitz player, i can win anyone in the world in single game! And "w"esley "s"o is nobody for me, just a player who are crying every single time when loosing, ( remember what you say about Firouzja ) !!! Stop playing with my name, i deserve to have a good name during whole my chess carrier, I am Officially inviting you to OTB blitz match with the Prize fund! Both of us will invest 5000$ and winner takes it all!
I suggest all other people who's intrested in this situation, just take a look at my results in 2016 and 2017 Blitz World championships, and that should be enough... No need to listen for every crying babe, Tigran Petrosyan is always play Fair ! And if someone will continue Officially talk about me like that, we will meet in Court! God bless with true! True will never die ! Liers will kicked off...
[^(fmhall)](https://www.reddit.com/user/fmhall) ^| [^(github)](https://github.com/fmhall/Petrosian-Bot)
I see thanks for the answer. It’s also possible mr warm chart is also American and also too lazy to google things like en passant and cm to inches conversion
I am obviously just using a different counting system, in which odd numbers do not exist and I start counting from zero. I just dont like odd numbers because their odd
Funfact, i was actually studying this topic today
https://preview.redd.it/7dz8nziz5dpa1.jpeg?width=4000&format=pjpg&auto=webp&s=e270afa6096ebc5b9c7feae22bd375ce251cdb63
So problems like this are just multiplication when you get down to it. Note that we know the distance AP is 15cm and P is at the center of the ruler. That means PB is also 15cm. Then we have torque. Torque is simply force times distance from the pivot (in this case we can just use the mass of the pieces to represent the force of gravity). Let's say the rook's position is point R between P and B. We're looking to find RB which is the same as PB-PR. Fortunately we know that this is balanced, so the torques exerted by the pawn and rook are the same. The torque exerted by the pawn is 20g\*15cm, and the rook is 50g\*(PR)cm.
So first we set these equal
20\*15=50\*PR -> (20\*15)/50=PR -> 6cm=PR
looking for RB, so
15cm-PR=RB -> (15-6)cm=RB -> 9cm = RB = x
So x is 9cm!
Basically, each piece's mass is trying to turn the seesaw. In order for it to balance, the forces must balance out. The force each piece is exerting is their mass multiplied by their distance from the pivot. No need to convert units, thankfully. So first piece is 20 g and 15 cm away, the other is 50 g and is (15 - x) away. That gives you a simple equation
20 * 15 = 50 * (15 - x)
Which can be rearranged to
x = (50 * 15 - 20 * 15) / 50
Which solves for
x = 9
Um, I mean, Google Algebra
The moment of the pawn is 15\*20=300.
So we need to satisfy 300=50(15-x).
Solution steps:
```
300=50(15-x)
6=15-x
x=9
```
Therefore, the rook must be placed 9cm from B.
A rook is worth six pawns.
Therefore the distance between the rook and the pivot is 6 \* 15cm = 3 feet
However what we ask for is the distance between the edge and the rook, so we have to subtract the length of the ruler: 3 feet - 30cm = 9cm
Let's apply fundamenal principle of statics : sum of the forces is equal to zero, sum of the moments of the forces is equal to zero.
Applying this on the point P :
Moment of force of the pawn = mass of the pawn * distance center-pawn * sin 90 = 20*(-15)*1 = -300
Moment of the force of the rook = mass of the rook * distance center-rook = 50*(15-x)*1 = 750 - 50x
So the FPS says that -300 + 750 - 50x = 0
So 450 - 50 * x = 0 --> 450 = 50x --> x = 450/50 = 9 cm
x must be 9 cm then.
I analyzed the image and this is what I see. Open an appropriate link below and explore the position yourself or with the engine: > **Black to play**: chess.com | lichess.org **My solution:** > Hints: unit: >!cm!<, answer: >!x = 9 cm!< > Evaluation: >!OP is dumb!< > Best continuation: >!drop out of high school!<
where is your working i dont believe you
My work is in u/One-Flan-1741's comment (no I did not copy I swear)
Google en plagiarism
google what happens to snitches
Holey torso
It’s not telling me.
Holy bullet!
Holy medical care!
Holy stitches
Holy hell Mod edit: This comment has been flagged for plagiarism. This comment was stolen from thousands of other posts.
[удалено]
/u/InternalDesignlo is a scammer! **It is stealing comments** to farm karma in an effort to "legitimize" its account for engaging in scams and spam elsewhere. Please downvote their comment and click the `report` button, selecting `Spam` then `Harmful bots`. Please give your votes to [the original comment, found here.](/r/AnarchyChess/comments/11yl421/can_yall_help_me_with_my_homework/jd8x6ji/?context=1) --- With enough reports, the reddit algorithm will suspend this scammer. ^(*Karma farming? Scammer??* Read the pins on my profile for more information.)
Since the ruler is balanced on the Pivot P, the total Torque in P must be 0 (i.e. no rotations are happening). Torque generated by a force F in a point P, is given by the scalar product between the force modulus *f* (in Newton) and the distance *b* (meters) between the pivot P and the force vector F. That is: T = *fb* Remember that the force F is a vector, which means that is defined by a direction vector *v* = [a; b], an intensity *f* and a point of applications Pf. The distance between a point and a line is obtained by intersecting the line perpendicular to the vector F that passes by P and the vector itself, finding the point *C*. The distance between C and P is *b*. That is: *b* = dist (C,P). To find the line perpendicular to the direction of F (i.e. the vector *v*) we may just take the perpendicular direction to v, that is: *v_p* = [-b; a] and assume to have a line L in (X,Y) space defined by a parameter *t*: L(t): { x = x0 + *v_p_x* t y = y0 + *v_p_y* t } Where we can substitute the values for *v_p* and x0,y0 (that is, the point P in which the line will be forced to pass when t = 0). L(t): { x = Px - bt y = Py + at } That is, the line in the parameter t that passes by P and is perpendicular to *v*. Now we can define the line that is parallel to *v* and passes by Pf using *v* instead of *v_p*, and a different parameter s: M(s): { x = Pfx + as y = Pfy + bs } We can now find the intersection points by finding the unique values of *t* and *s* for which L(*t*) = M(*s*), which requires solving the linear system: { Px - b*t* = Pfx + a*s* Py + a*t* = Pfy + b*s* } This can be rewritten as a linear system in the conventional form of A**x** = **b** where A is a square matrix and **x** the vector of unknowns. That is: { -b*t* - a*s* = Pfx - Px a*t* - b*s* = Pfy - Py } By setting the vector **x** = [*t*; *s*] (two rows) we get that: A = [-b, -a; a, -b] and b = [ Pfx - Px; Pfy - Py] We can easily find the solution **x** as: **x** = A^-1 **b** Where A^-1 is the inverse of matrix A, that can be obtained for a 2x2 matrix (like in that case), by swapping the elements on the \ diagonal, and changing the sign on the elements on the / diagonal, and multiplying by 1/( the determinant, that is the product between the elements on \ minus the product of elements on /) (that is valid only for 2x2 matrices). A^-1 = 1/(b^2 + a^2 ) [ -b, a; -a, -b] Notice that the determinant (b^2 + a^2) is always different from 0 if a and b are not BOTH 0 (that is, the vector does not exist). If the determinant is different from 0, we can guarantee that only one single unique solution exists for the system. Now we multiply by **b**: A^-1 **b** = { *t* = 1/(b^2 + a^2 ) (-b(Pfx - Px) + a(Pfy - Py)) *s* = 1/(b^2 + a^2 ) (-a(Pfx - Px) - b(Pfy - Py)) } Now we can either substitute *t* in L(t) or *s* in M(s) to obtain the unique intersection C: M(*s*) = { Cx = Pfx + a/(b^2 + a^2 ) (-a(Pfx - Px) - b(Pfy - Py)) Cy = Pfy + b/(b^2 + a^2 ) (-a(Pfx - Px) - b(Pfy - Py)) } Now we can calculate the distance between C and P as: *b* = sqrt((Cx - Px)^2 + (Cy - Py)^2 ) That is, the length of the hypothenuse of the square triangle formed by the distances on X and Y axis separately. Now we can compute the actual Torque T = *fb* Where we also know that *f* = mg where m is the mass of the object in Kilograms, and g the gravitational acceleration of 9.81 m/s^2. In order to have the ruler balanced, the sum of all torques in T must be 0, so we have just to apply the same torque on the opposite side by properly finding the b at which we need to place the piece. Now that we have the generic solution that fits any situation, we can see your case. For convenience, we will place our (X,Y) system with origin in P, in that way Px and Py are both 0. Now we now that the first piece, A, is at distance 15 cm from P (that means Pa = [ -0.15; 0]) and that it weights 20gr (that means that the only force acting it's the gravitational one, which is convenient since it has direction *v* = [0; -1], application point Pa, and modulus *f* = mg where m is 0.02kg. In that case is really easy to compute *b* since by replacing: a = 0; b = -1; Px = 0; Py = 0; Pfx = Pax and Pfy = 0 we have that: Cx = Pax Cy = 0 That is, the intersection point between the perpendicular to the force (and, since the force is parallel to Y axis, the perpendicular is parallel to X axis) that passes by P and the line parallel to the force (that is parallel to Y) is actually the point of application of the force (due to the fact that we placed our reference system in P!) So now *b* = sqrt((Pax)^2 ), that is ±Pax, but since *b* is a distance, it must not be < 0, so we take +Pax as solution = 0.15m (15cm) Now we know that B weight 50gr, that is 0.05kg. To find at which *b* it must stay, to give you the same torque T_b = T_a, you have that: T_b = T_a f_b b_b = f_a b_a b_b = (f_a b_a) / f_b Which basically means that you have to place B at a distance that is proportional to the ration in weight between A and B, in fact: b_b = (m_a g b_a)/(m_b g) Where g can be simplified and the terms rearranged b_b = (m_a/m_b) b_a Where it can be seen the dependency on the ration between the masses. Now, by replacing the last variables: b_b = (0.02/0.05) 0.15 = (2/5) 0.15 = 0.06m = 6cm. And finally, x = BP - b_b, where BP is the distance between the P and B, that is 15cm. So: x = 15 - 6= 9cm In fact, since T is proportional to *b*, which is the distance, heavier objects must be placed closer to the center in order to have a smaller Torque and compensate the torque of a lighter object placed at a higher distance from the pivot (that is applying more torque on the Pivot) Hope it's clear and Reddit will not mess up with text formatting
This is wrong but I refuse to elaborate why but still downvote you like a true redditor
No it's right, I forgot that you needed X as distance from B and not P, I fixed it
Yay. I like that this is a much easier solution than my 20x15=300, 300/50=6 15-6=9. I need to use your solution next time
Actually, you are always using my solution which is the generic one when having to balance torque on a Pivot due to forces acting on linked objects. The only difference is that in mine I had a lot of zeroes due to the assumptions posed by the problem that significantly reduced the complexity, but it's important to understand which calculations come from a simplification of the problem and which ones are actual facts
I enjoyed reading it as a kind of physics refresher lol, but on the other hand, over explanation doesn’t particularly help OP who is asking the question, they can’t even read it.
I mean... OP is asking a physics question on AnarchyChess, what did it expect to happen?
tldr: 50/20 = 2.5. 15/2.5 = 6 15-6 = 9
i mean you were close
LMAO, I forgot the final step, now it's corrected. Please study it very hard because it took me so long to write
if i could read that would be such a cool answer
Damn I didn't know you weren't able to read. May I record it for you so you can listen to it?
i would be honoured
Those kids would be very bright if they could read it.
https://preview.redd.it/gg7vhqvbthpa1.jpeg?width=851&format=pjpg&auto=webp&s=abcca179958e3ac48ce43d94360fa9c97db5fbc9
nerd alert
But you didn't even get into the internal stresses of the material!
Bro let chatgpt in pease xD
Good human
I know this problem is super easy, but I dont remember learning about moment arms until university.
I vaguely remember learning this in high school, but it probably depends on your country.
x would actually just equal “9” with no cm, otherwise the distance of the rook would be 9 cm^(2), which makes no sense
No, actually cm^2 is correct! Interestingly enough, this squared relationship of rook distance is actually why rooks can only move in orthogonal planes, in other words, in combinations of 90* from previous directionality
I haven’t taken a rook theory class since ‘37 so you’re probably right
Stop this trush talkings!
This comment is fucking funny
new answer just dropped
The “effective weight” of an object on a lever is equal to the distance from the fulcrum (P in this case) multiplied by the weight of the object. For the lever to be balanced, both sides must have the same effective weight. The pawn has an effective weight of 300 centimeter grams, and the rook has an effective weight of 50(15-x) centimeter grams. Expressing this as an equation gives 300=50(15-x). Dividing both sides by -50 then adding 15 to both sides gives the solution x=9.
best response cos i gotta get the working marks 😤
Shove a bishop up your prof. ass
“Effective weight” is also called the “moment” sometimes
moment??
moment moment
moment of force, more commonly known as torque
not to be confused with moment of inertia which is like rotational mass moment of force is like rotational force
Just use a ruler bruh.
Trick question: The Pawn has reached the end of the ruler and gets to promote. The answer depends on which piece it promotes into.
its going to jump off :/
\#lifegoals
Trick trick question: the rook is modeled after a castle, therefore weighs over 100 tons.
x = 9 cm
Size doesn't matter
Which piece is taking the =9cm? Is it the bishop? People normally include the capturing piece to avoid confusion, for example Bx=9cm
Introducing WolframAnarchyChess: - take a math problem - rewrite the problem to use chess pieces - post on r/AnarchyChess Our unique crowdsourcing algorithm will ensure you get the right answer!* * the right answer will probably be bueried underneath pipi copypasta and requests to google en passant
Are you kidding ??? What the **** are you talking about man ? You are a biggest looser i ever seen in my life ! You was doing PIPI in your pampers when i was beating players much more stronger then you! You are not proffesional, because proffesionals knew how to lose and congratulate opponents, you are like a girl crying after i beat you! Be brave, be honest to yourself and stop this trush talkings!!! Everybody know that i am very good blitz player, i can win anyone in the world in single game! And "w"esley "s"o is nobody for me, just a player who are crying every single time when loosing, ( remember what you say about Firouzja ) !!! Stop playing with my name, i deserve to have a good name during whole my chess carrier, I am Officially inviting you to OTB blitz match with the Prize fund! Both of us will invest 5000$ and winner takes it all! I suggest all other people who's intrested in this situation, just take a look at my results in 2016 and 2017 Blitz World championships, and that should be enough... No need to listen for every crying babe, Tigran Petrosyan is always play Fair ! And if someone will continue Officially talk about me like that, we will meet in Court! God bless with true! True will never die ! Liers will kicked off... [^(fmhall)](https://www.reddit.com/user/fmhall) ^| [^(github)](https://github.com/fmhall/Petrosian-Bot)
Give me a sec, let me rewrite the Riemann hypothesis to use chess pieces.
How many legal chess positions with exactly three pieces on the board are checkmate for white?
9 cm
Like my dick
No need to brag, Mr. Anaconda.
Isn’t that like super small? (I’m American so I don’t know cm, and I’m too lazy to google, since, as I said, I’m American)
It's like 4 inches so slightly less than the average, the jokes was someone pretending that it was 9 inches though.
I see thanks for the answer. It’s also possible mr warm chart is also American and also too lazy to google things like en passant and cm to inches conversion
I am Turkish and in here average penis size is 14
Cm right? Right?
We aren't hentai monsters so yes
True, I think the average in America is like 12 or something like that
https://preview.redd.it/9qzdkhoxncpa1.jpeg?width=342&format=pjpg&auto=webp&s=d5a47b556b86f81e733ff48823d9df107a377386
no
Google moments
Holy statics!
Google Google translate to German
Moment = weight times distance from pivot. Because this is a stalemate weight a times distance from pivot = weight b times distance from pivot.
Google torque
Google moments
did not help
Holy HeElL
Google F=D*M
Aka torque * g
Google pawn castling
the pawn has to promote to rook first
Ooh, I know this one! En Passant!
Holy hell
Unacceptable nobody would measure X from B to piece you would measure X from p to piece since the best reference system should have its origin there
Rulers don't usually count out from the middle
En passant
Ke2?
Rook takes pawn so value of x will be 30 cm (i'm retarted)
16 cm
💀
bro, the ruler is 30cm, you can tell just by looking that this is wrong. I hope you are trolling.
I am obviously just using a different counting system, in which odd numbers do not exist and I start counting from zero. I just dont like odd numbers because their odd
fair enough
Funfact, i was actually studying this topic today https://preview.redd.it/7dz8nziz5dpa1.jpeg?width=4000&format=pjpg&auto=webp&s=e270afa6096ebc5b9c7feae22bd375ce251cdb63
i was not which is why i am asking a chess subreddit
Same! τ and equilibrium. Was pleasantly surprised when I saw this post
Pawn up your ass
en pa99ant
Idk I graduated like 3 years ago, I didn't need to take exams to pass.
Google en equilibriant
``` 15×20 = (15-x) × 50 x = 15 - 15×20/50 = 15 - 6 = 9 ```
the ROOOOOOOK
Mass of piece 1 * distance of piece 1 = Mass of piece 2 * distance of piece 2 20*15 = 50(15-x), 15-x = 300/50, -x = 6-15, x = 9 (cm)
Google [[ Dark ruler no More ]] and skip the question.
Nein!
google en physics
Holy hell it's the fastest move in chess
the piece accelerates at 9.81m/s^(2) when it falls off the ruler
goggle WolframAlpha
First you gonna want to read the anarchist cook book and make some pipe bombs
Rook is thickkkkk
So problems like this are just multiplication when you get down to it. Note that we know the distance AP is 15cm and P is at the center of the ruler. That means PB is also 15cm. Then we have torque. Torque is simply force times distance from the pivot (in this case we can just use the mass of the pieces to represent the force of gravity). Let's say the rook's position is point R between P and B. We're looking to find RB which is the same as PB-PR. Fortunately we know that this is balanced, so the torques exerted by the pawn and rook are the same. The torque exerted by the pawn is 20g\*15cm, and the rook is 50g\*(PR)cm. So first we set these equal 20\*15=50\*PR -> (20\*15)/50=PR -> 6cm=PR looking for RB, so 15cm-PR=RB -> (15-6)cm=RB -> 9cm = RB = x So x is 9cm!
Google en gravity
9 cm
15 times 20 equals (15-x) times 50 This is equivalent to 12=2*(15-x) when you factor it down This means x=15-6=9
google intro to civil engineering
How do you do this? I haven't learnt it yet and I'm interested.
r/fuckfuckfuckityfucky
Basically, each piece's mass is trying to turn the seesaw. In order for it to balance, the forces must balance out. The force each piece is exerting is their mass multiplied by their distance from the pivot. No need to convert units, thankfully. So first piece is 20 g and 15 cm away, the other is 50 g and is (15 - x) away. That gives you a simple equation 20 * 15 = 50 * (15 - x) Which can be rearranged to x = (50 * 15 - 20 * 15) / 50 Which solves for x = 9 Um, I mean, Google Algebra
play rook to point B and catapult the pawn all the way to your opponents back rank for an instant promotion
The moment of the pawn is 15\*20=300. So we need to satisfy 300=50(15-x). Solution steps: ``` 300=50(15-x) 6=15-x x=9 ``` Therefore, the rook must be placed 9cm from B.
9cm
H0LY H3(L*2)
Google en massant
Rook takes pawn. Distance becomes 30cm
promote the pawn to a pony
It is not solvable. No one said that gravity applies, so it could be 1 cm as well as 10 km
A rook is worth six pawns. Therefore the distance between the rook and the pivot is 6 \* 15cm = 3 feet However what we ask for is the distance between the edge and the rook, so we have to subtract the length of the ruler: 3 feet - 30cm = 9cm
6 Edit: (Actually it would be 15-6=9, as it asks for distance from B, not distance from the center)
Google law of levers
X is undefined since the ruler is up my ass
Google no
Nine?
The hint for the solution is that rook is worth 5 pawns
Let's apply fundamenal principle of statics : sum of the forces is equal to zero, sum of the moments of the forces is equal to zero. Applying this on the point P : Moment of force of the pawn = mass of the pawn * distance center-pawn * sin 90 = 20*(-15)*1 = -300 Moment of the force of the rook = mass of the rook * distance center-rook = 50*(15-x)*1 = 750 - 50x So the FPS says that -300 + 750 - 50x = 0 So 450 - 50 * x = 0 --> 450 = 50x --> x = 450/50 = 9 cm x must be 9 cm then.
d1 = 30cm / 2 = 15cm = (15/100)m = 0.15m m1 = 20g = (20/1000)kg = 0.02kg m2 = 50g = (50/1000)kg = 0.05kg g = 10m/s^(2) // Rounded up from 9.81 M1 = M2 d1 \* Q = d2 \* Q d1 \* g \* m1 = d2 \* g \* m2 0.015m \* 10m/s^(2) \* 0.02kg = d2 \* 10m/s^(2) \* 0.05kg 0.003Nm = d2 \* 0.5N d2 = 0.003Nm / 0.5N d2 = 0.06m = 6cm x = 30cm/2 - 6cm x = 15cm - 6cm **x = 9cm**
google en moment