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The math help thread - Corn - 2009-11-02
Gezus, I'm posting in here a lot more often.
I need help converting x - y = 2 into a polar equation
.
The math help thread - Hazzy - 2009-11-02
ClawofBeta Wrote:Gezus, I'm posting in here a lot more often.
I need help converting x - y = 2 into a polar equation
http://www.analyzemath.com/polarcoordinates/polar_to_rectangular_eq.html
I'm certainly no expert, but undoing what they did...
x=cos t. y=sin t
So...
rcost t - rsin t = 2
Edit: fak reverse ninjas.
The math help thread - KajitiSouls - 2009-11-02
ClawofBeta Wrote:Gezus, I'm posting in here a lot more often.Basic rectangular to polar conversion:
I need help converting x - y = 2 into a polar equation
x = r*cos(θ
y = r*sin(θ
r^2 = x^2 + y^2 <-- Pythagorean Theorem much?
In your case...
x - y = 2
r(cos(θ
- sin(θ) = 2That is, if there's no stupid hidden trig identity in there...
EDIT: Terrible ninja is terrible.
The math help thread - Hazzy - 2009-11-03
Two blocks are sliding towards each other, and hit with one far side hitting the mother in the middle.
A Picture.
Spoiler
Using math(what ever variables you'd need), how would you predict the movement of the blocks after impact? Include friction, and assume the blocks are identical in all characteristics that would affect the situation, and that equal forces are pushing the blocks together.
![[Image: 9iegwi.png]](http://i38.tinypic.com/9iegwi.png)
The math help thread - KajitiSouls - 2009-11-03
New problem that for some reason I can't get the right answer to.
Find the product of inertia of the triangle bounded by the x-axis, -b, and y = -(h/b)*x.
Product of inertia = ∫xy*dA
The math help thread - Veneni - 2009-11-06
I was going to make a whole new thread about this here, but then I saw this thread so I suppose it goes in here (even though it has nothing to do with homework).
It's about clean slate scrolls.
I was thinking about what scroll would have the highest possibility of getting you a slot back. I tried to solving the problem but didn't knew where to start.So, the problem is as follows: Say you have an infinite amount of 1,5 and 20% clean slate scrolls, what is the chance to success? Which has the highest chance of success?
If I should guess, it would be best to use the 20% scroll, because, suppose there are other clean slate scrolls at x% with x in (0,0.5], so 0.5 having a 100% destroy rate at fail. Then the scroll with 50% has 50% chance of success. Which is definately more than the 1% scroll. (You have success after on average 100 tries, while it gets destroyed after ~50.5 tries. If it were also 50% chance, the chances would be the same.)
It looks like I just answered my own question but what I'm really interested in is the exact chance of the 1,5 and 20% scrolls.
GL
The math help thread - Noah - 2009-11-06
Hazzy Wrote:Two blocks are sliding towards each other, and hit with one far side hitting the mother in the middle.
A Picture.
SpoilerUsing math(what ever variables you'd need), how would you predict the movement of the blocks after impact? Include friction, and assume the blocks are identical in all characteristics that would affect the situation, and that equal forces are pushing the blocks together.
Use Newton's 3. law:
![[Image: y9tkdqz.png]](http://mathurl.com/y9tkdqz.png)
By that, divide the blocks into two.
![[Image: 25tba85.png]](http://i37.tinypic.com/25tba85.png)
Furthermore, we now assume that the left part of the black block has zero mass, and that the right part of the red block has zero mass. What would happen then? They would simply have the same effect depending if the collision's elastic, inelastic or plastic.
That's the very basic part which you will go and use. But what should be assumed? I need more data. Does the blocks have the exactly same mass at any given point? What kind of collision-type is it? (Mostly this.)
KajitiSouls Wrote:New problem that for some reason I can't get the right answer to.
Find the product of inertia of the triangle bounded by the x-axis, -b, and y = -(h/b)*x.
Product of inertia = ∫xy*dA
Are you speaking of the moment of inertia? I assume that. However, there's no possibility to calculate the moment of inertia if you only have that data. At any given x, how heigh is the triangle at that position? You need a function for the triangle. y = ??
Veneni Wrote:I was going to make a whole new thread about this here, but then I saw this thread so I suppose it goes in here (even though it has nothing to do with homework).
It's about clean slate scrolls.
So, the problem is as follows: Say you have an infinite amount of 1,5 and 20% clean slate scrolls, what is the chance to success? Which has the highest chance of success?
If I should guess, it would be best to use the 20% scroll, because, suppose there are other clean slate scrolls at x% with x in (0,0.5], so 0.5 having a 100% destroy rate at fail. Then the scroll with 50% has 50% chance of success. Which is definately more than the 1% scroll. (You have success after on average 100 tries, while it gets destroyed after ~50.5 tries. If it were also 50% chance, the chances would be the same.)
It looks like I just answered my own question but what I'm really interested in is the exact chance of the 1,5 and 20% scrolls.
GL
The exact chance? Do you mean the scroll with the highest chance of succeeding in recovering a slot, given you have unlimited amounts of the scroll?
For 1, 3 and 5% slates, there's twice the chance of destroying the item. For the 20% one, there's 50% chance of destroying the item.
Assume p is the chance of working the clean slate, and 2p is the chance of destroying the item. Then, the chance of no effect is (1 - 3p). We have then that
![[Image: y9t36s8.png]](http://mathurl.com/y9t36s8.png)
This is the chance for 1, 3 and 5% slates.
For 20% one, we have something else:
The chance of working a clean slate is still p, but the chance of destroying the item is increased to 2.5 instead of 2 only. Then, the chance of no effect is (1 - 3.5p). We have that
![[Image: y9cuq6w.png]](http://mathurl.com/y9cuq6w.png)
And yeah, 2/7 is a lower chance than 1/3, so it follows that the slates with the highest probability of working, assuming you have infinite of them, are the 1, 3 and 5% slates.
Noah
The math help thread - KajitiSouls - 2009-11-06
Noah Wrote:Are you speaking of the moment of inertia? I assume that. However, there's no possibility to calculate the moment of inertia if you only have that data. At any given x, how heigh is the triangle at that position? You need a function for the triangle. y = ??
Noah
Nope, I'm speaking about the product of inertia. I suggest you look closer as I gave you what y is. (looking back at the post one of the equations is incomplete, oops)
Triangle is defined by:
y = 0
x = -b
y = -(h/b)*x
The math help thread - HooKarez - 2009-11-07
I don't really understand how to do these two Physics problems:
Top: Calculate the force of the pin joint and the tension in the rope.
Bottom: Determine the mass needed to balance the beam in the figure.
![[Image: Phyzzzz2.jpg]](http://i201.photobucket.com/albums/aa232/crazyswimmer22/Phyzzzz2.jpg)
The math help thread - Hazzy - 2009-11-07
Noah Wrote:Use Newton's 3. law:
By that, divide the blocks into two.
Furthermore, we now assume that the left part of the black block has zero mass, and that the right part of the red block has zero mass. What would happen then? They would simply have the same effect depending if the collision's elastic, inelastic or plastic.
That's the very basic part which you will go and use. But what should be assumed? I need more data. Does the blocks have the exactly same mass at any given point? What kind of collision-type is it? (Mostly this.)
What kind of collisions are there...?
For simplicity, assume the blocks have a uniform density.
The math help thread - Russt - 2009-11-07
HooKarez Wrote:I don't really understand how to do these two Physics problems:For the bottom one, I think the force*distance on both sides needs to be equal. Since we're talking gravity, that's proportional to mass*distance.
Top: Calculate the force of the pin joint and the tension in the rope.
Bottom: Determine the mass needed to balance the beam in the figure.
Spoiler
Since it's a meterstick, the distance between each of those dots is 10cm (though it actually doesn't matter). The 40cm of meterstick on either side of the balance point cancel out, leaving the 20cm on the right, which is 20% of the meterstick.
That part of the meterstick weighs 10g and its midpoint is 50cm to the right of the balance point, so its mass*distance is 500g*cm.
The mass you're trying to find is 20cm left of the balance point, so it's 500g*cm/20cm = 25g
Disclaimer: I haven't taken any Physics beyond 8th grade "Physical Science."
The math help thread - Noah - 2009-11-07
KajitiSouls Wrote:Nope, I'm speaking about the product of inertia. I suggest you look closer as I gave you what y is. (looking back at the post one of the equations is incomplete, oops)
Triangle is defined by:
y = 0
x = -b
y = -(h/b)*x
I'd assume y = kx where k = -h/b is the formula for the triangle then. If that's the case, what length is the triangle? Is that given, or is that a variable? (I'm very confused.)
HooKarez Wrote:I don't really understand how to do these two Physics problems:
Top: Calculate the force of the pin joint and the tension in the rope.
Bottom: Determine the mass needed to balance the beam in the figure.
Man, I just found a nice hack to use for mathurl.
![[Image: yl39p4l.png]](http://mathurl.com/yl39p4l.png)
Edit: Meh, it's not per cm, but per 10 cm. Though, the result doesn't matter.
Hazzy Wrote:What kind of collisions are there...?
For simplicity, assume the blocks have a uniform density.
http://en.wikipedia.org/wiki/Collision
The math help thread - KajitiSouls - 2009-11-07
Noah Wrote:I'd assume y = kx where k = -h/b is the formula for the triangle then. If that's the case, what length is the triangle? Is that given, or is that a variable? (I'm very confused.)
Man you must live in a different mathematical universe or something o.O
Triangle is b wide and h high. Rise over run. Simple as that. Remember that the triangle is on the left side of the y-axis.
HooKarez Wrote:I don't really understand how to do these two Physics problems:
Top: Calculate the force of the pin joint and the tension in the rope.
Bottom: Determine the mass needed to balance the beam in the figure.
Freebody diagrams are fun! Hooray!
Top:
In order to do this problem correctly, we'd need a few more assumptions. For both of our sake, I'm assuming the system is at rest. This means that the following equations are true:
ΣForce = 0
ΣTorque = 0
I'm also going to assume that the mass can swing around the pin, or else the pole/wall comes into play, making the problem even more complicated.
Next, we find out all the forces involved. Red arrows are total forces, and green arrows are component (x,y) forces.
Free body diagram
![[Image: hookarezmath.jpg]](http://img195.imageshack.us/img195/4905/hookarezmath.jpg)
Since ΣF = 0, ΣT = 0, and the component forces in one direction have no bearing on forces in another direction, set up equations for the system and solve. For the last equation, we're taking the torque around the pin.
Code:
Fp = pin force
Fr = rope force
Fm = mass force = 10kg * 9.8m/sec^2
98 = Fr*sin(45) + Fp*sin(θ)
Fr*cos(45) = Fp*cos(θ)
0 = 98*1 + Fr*sin(45)*2The math help thread - HooKarez - 2009-11-09
The top problem above still confuses me. I understand your assumptions with the fact that F and T both have a net of zero. However, with all that said can I assume that Fpy is zero? And your last equation is puzzling. I just don't really understand why the 98*1 is there. I dunno, that last past just instilled mass confusion. I think my values are right though going from Noah's values above. 98.1N for the Pin Joint and 137.7N for the Tension in the Rope.
The math help thread - KajitiSouls - 2009-11-09
HooKarez Wrote:The top problem above still confuses me. I understand your assumptions with the fact that F and T both have a net of zero. However, with all that said can I assume that Fpy is zero? And your last equation is puzzling. I just don't really understand why the 98*1 is there. I dunno, that last past just instilled mass confusion. I think my values are right though going from Noah's values above. 98.1N for the Pin Joint and 137.7N for the Tension in the Rope.
No you may not assume Fpy is zero because it is a pin that holds the mass on an axis (going through the paper). Fpy is zero only if the element at that point can move, such as a rolling wheel. And the last equation is a torque equation centered around the pin. 98*1 is simply the force of the mass multiplied by the distance from the pin.
With problems as simple as this one, there are two things that should be true:
1.) There should be a complete force triangle. If you took all the force vectors and laid them head to tail with each other, you should get a complete triangle. If not, the system is in motion. Failure to satisfy this condition is a pretty big indicator that you interpreted the system wrong.
2.) The forces involved (when not parallel) all should intersect at a single point when you extend their vectors (line of action). I have no idea how true this is when you have four or more forces...
EDIT: Looks like I screwed up the last equation. It should really be 0 = 98*1 - Fr*sin(45)*2 xD
The math help thread - HooKarez - 2009-11-09
How are we certain that the torque of the mass is 98*1? Since the entire mass is from 0 to 2m. It is just because 1 is the average distance?
The math help thread - KajitiSouls - 2009-11-09
HooKarez Wrote:How are we certain that the torque of the mass is 98*1? Since the entire mass is from 0 to 2m. It is just because 1 is the average distance?
Yes.
To be more technical about it, 1 is the location of the mass' center of gravity (centroid), where you can represent the object as a point mass for theoretical purposes. The same thing applied to the meter stick in the other problem you presented, no?
EDIT: If your textbook does the problem assuming Fpy is 0 or something to that effect, you should probably do it that way. But as a future reminder, that is not how reality works.
The math help thread - Noah - 2009-11-09
KajitiSouls Wrote:No you may not assume Fpy is zero because it is a pin that holds the mass on an axis (going through the paper).
![[Image: main-chemistry.gif]](http://scienceworld.wolfram.com/chemistry/images/main-chemistry.gif)
So is my interpretation of a pin joint wrong? I'm assuming the pin joint is the thingy D in the figure. Because of its way of working, it's impossible for that thing to be affected by any vertical force from the pole. It will be cancelled by the wall on the other side. Assume no friction: That means it will fall with the force g towards the ground because no other forces affect it.
KajitiSouls Wrote:EDIT: If your textbook does the problem assuming Fpy is 0 or something to that effect, you should probably do it that way. But as a future reminder, that is not how reality works.Keep in mind that physics always use models, not reality.
Also, for your question, I understood your triangle. I don't understand why that took me so long to understand it however, maybe I was just lazy. As I don't really know what the product of inertia is, I can't really answer your problem. I'm sorry about that.
Noah
The math help thread - KajitiSouls - 2009-11-09
The math help thread - HooKarez - 2009-11-09
All right. I did everything and got two answers.. but they're both equal. I guess that's possible...? I got answers of 69.37 N for both Fr and Fp. -shrug-