Need Math Help?

[Elyssa Xey Hexen]

Let's assume the block moves in the positive X direction.

Okay, the block accelerates overall with a net acceleration of 6.00m/s^2

The force applied to the block from friction will work against the force pushing the block. So F_push - F_friction = F_net

Since we know the net acceleration and the mass of the block, we get a force net of
4 kg * 6 m/s^2 = 24 N

The force of friction will be modeled by F_friction = nu where u is the coefficent of friction and "n" is the normal force. The F_push is the force applied to the block only along the axis at which the block is moving. We are told the block is pushed against the ceiling at an angle of 55 degrees from horizontal at a magnitude 85N. Splitting this into X and Y components of this force we get
F_x = cos(55) * 85N = 48.75 N
F_y = sin(55) * 85N = 69.63 N

Now, the F_push I mentioned before is this F_x force.

So,
F_push - F_friction = F_net

(48.75 N) - (nu) = 24 N

Now, the normal force is merely the force perpendicular to the direction of motion. This force is made up of the force pushing the block up, and the force pushing the block down. We already calculated one of the forces (F_y), but the force due to gravity will be F_gravity = (9.8 m/s^2) * (4kg) = 39.2 N So the total force pushing the block up is F_y - F_gravity = 30.43 N

Now with the normal force calculated, you can solve for u!

(48.75) - (30.43 * u) = 24
And merely solve for u,
u = (24 - 48.75)/ (-30.43)
u = .8133

 

[Kason Birdman]

oh thanks a lot man. helped greatly :D !

I've got one more if anyone wants to tackle this....

A 3.00 kg block is held on a 60 degrees incline by a horizontal force. If the coefficient of static friction between the block and the incline is 0.350, determine
a) the minimun value of the horizontal force and
b) the normal force exerted by the incline on the block

the answer for a is supposed to be 25.3 N... but me and all my friends keep getting 10.157 N and 20.314 N... for my normal force I got 14.7... is atleast that right? lol.

 

[Elyssa Xey Hexen]

Here is one way to think about it.

The block is not moving, so all forces acting on it must be balanced. The horizontal force will want to push the block against the incline and up it. But the force of gravity is also at work to push the block against the incline and down it. If neither of these forces can exceed the force of friction, the block won't move up or down the incline.

So, at minimum, to keep the block from moving,
F_push + F_gravity = F_friction

If we look at this problem from the blocks perspective by putting the x-axis along the incline, it can make it easier to look at this problem. This is because the normal force pushing the block against the incline will always equal the force of the incline pushing against the block. Newton's 2nd law. "For every action there is an equal and opposing reaction" So we only really need to care about the forces pushing the block up or down the plane.

F_friction is easy enough to calculate because it is F_friction = u * N
Where u is the coefficient of friction and N is the normal force.

The F_gravity along the incline will be sin(60 degrees) = F_gravity / 29.4 newtons
I got this by making a triangle between full force of gravity (29.4 newtons) and trying to find the magnitude of gravity in the direction of the incline. And this magnitude is 25.46 newtons.

Now, the F_push in the direction of the incline can be calculated similarly, but we do not know the full magnitude of the horizontal force yet. cos(60 degrees) = F_push / F_horizontal. So moving the equation around. F_horizontal * Cos(60 degrees) = F_push

Now, putting all these together in the form
F_push + F_gravity = F_friction
We get,
{F_horizontal * Cos(60 degrees)} + (25.46 Newtons) = u * N

So far we still need to calculated N since it is unknown. The normal force will be equal to the force of gravity and the horizontal force pushing the block against the incline. This can be calculated from the same two triangle from before, but instead the focus is the magnitude of the forces pushing the block into the incline.

For gravity it is, cos( 60 degrees) = F_gravity,normal / 29.4 newtons, and this normal force is 14.7 newtons. For the normal force on the block due to the horizontal force is sin(60 degrees) = F_normal / F_horizontal which moving it around will be sin(60 degrees) * F_horizontal = F_normal

Add these two together will get the total normal force so {sin(60 degrees) * F_horizontal} + 14.7 N

The full equation from the beginning is now,
{F_horizontal * Cos(60 degrees)} + (25.46 Newtons) = u ({sin(60 degrees) * F_horizontal} + 14.7 newtons)

The rest is inputting the number for coefficient of friction and algebra to find the value of the horizontal force.

Edit:
After a bit of rearranging I found that I got the positive and negatives mixed up. The force of gravity along the incline is greater than the horizontal force. So block will want to move down the slope, but the static friction will oppose the block going down this direction. So the actual equation from before should have been

F_gravity = F_friction + F_push

So, if you replace these quantities with the stuff we derived before, and solve for Horizontal force, you'll get 25.3 newtons as you mentioned the answer should be.

The answer to part B is simple enough. The normal force pushing the block against incline will be the same magnitude as the incline pushing against the block. The normal force pushing the block against the incline we already derived to be
{sin(60 degrees) * F_horizontal} + 14.7 N.

Now, with the value of the horizontal force, you can directly calculate the normal force. Which it should be 36.6 newtons.

 

[Kason Birdman]

you are a man upon men. the whole st. marys smash crew thanks you

 

[Elyssa Xey Hexen]

On of the few times I've been able to put my semester of Newtonian mechanics to good use lol
Just hope I made everything clear enough. I always think I never explain stuff thoroughly.

 

[Kason Birdman]

now it was all pretty good actually.
and newtonian mechnics? is that a joke or is that actually what its called? haha

 

[Elyssa Xey Hexen]

Pretty much, his laws and other derivations and applications from those laws fit under the term of Newtonian mechanics. I think you could think about it more specifically as Newton made these laws that work well, and from them we created a branch of physics called dynamic, kinematics, etc.

 

[:mad:]

This might sound weird, but I'm stumped.

 

[Corpsecreate]

This is an easy one:

First you should you should figure out what the gradient is, then you should figure out what the y-intercept is. Then you should convert all the 6 equations into standard form ( y = mx + c) and see which ones match the equation of that line.

And just as a memory refresher, the gradient = (y2-y1) / (x2-x1)

 

[:mad:]

Thank you kindly! It's making more and more sense now.

 

[Elyssa Xey Hexen]

lol I forgot slope of a line was also called the gradient. After doing vectors, I keep associating it with only vectors.

 

[Corpsecreate]

What are gradients in the context of vectors? O.o o.O

 

[Elyssa Xey Hexen]

Say you have some vector function f(x,y,z), the gradient will be the partial derivative of f with respect to x in the i-hat direction plus the partial derivative of f with respect to y in the j-hat direction plus the partial derivative of f with respect to z in the k-hat direction.

Although, this gradient vector can be of any number of variables, you just add on more partial derivatives of the function with respect to a variable in the direction that variable represents.

 

[Kason Birdman]

lol I have never heard the term gradient before (well, in that context, obviously) I have only ever heard it called slope.

 

[Deleted member]

ugh vector calculus. boy am I glad I never have to do that stuff again.

 

[Corpsecreate]

The question is...do you still remember how to do it?

 

[kevo]

Hey guys. This may seem a little silly, but I'm stuck on a real world math problem pertaining to running a smash tournament. Can I get some input?

I know that the number of games in a double elimination tournament is at most 2n-1, with n being the number of entrants. My question is: at what point n (# of contestants) does it become more timely to do pools before bracket? Assume that doing pools eliminates exactly half the number of entrants.

As I began working this problem out, I found that how pools are split, number of setups, etc all affect the answer and the way this problem can be approached. Am I thinking about this wrong? What's a more effective way to model this situation?

 

[Elyssa Xey Hexen]

I just want to clarify. You want to know at what number of entries (contestants) does it take more time to go through pools than it does to actually go through bracket?

Going off of that, I would say you can really only measure that in the number of sets that have to be played. As you already mentioned, the number of set-ups can determine how quickly you can get through a number of matches including how pools are split.

If the number of sets played in all pools exceeds that of how many sets would have to be played in bracket, then you know that your configuration for pools might take longer than the bracket itself.

The number of sets, B, in pool size, A, is:
A=1; B=0
A=2; B=1
A=3; B=3
A=4; B=6
A=5; B=10
A=6; B=15
A=7; B=21
A=8; B=28
It follows a pattern, but I'm not sure how to write it out well. Say you had 9 people in a pool, the total sets played would be the total sets played in 8 sets and the number of people in the pool. 28 + 9 = 37 sets played. That's the pattern.

As an example, say only 15 people show up for tournament and you want to do a double elimination bracket of 8 people. You could split pools into say 3 pools of 4 and one pool of 3. You could split it other ways, but I think you already know that it would be hard to find a cut off limit for pools that way. Anyway, a pool of 4 would need to complete 6 sets and a pool of 3 would take 3 sets to complete. Overall, 6*3 + 3 = 21 sets played for the entirety of pools. As you said the number of sets was 2n-1 for sets played in double elimination bracket. That is 15 sets played. Pools requires more sets played than the bracket itself.

But then you have to consider how much time do you have to play all those sets, the number of set-ups with you, the talent there, and are people entering multiple events (this can really delay a tournament). I think the issue is too complicated to really come up with some final answer to your question other than look at it in the number of sets and weigh the situation at the tournament and make a final decision from there.

 

[Corpsecreate]

This problem is a complicated one. The calculations involve logs and estimated factorials of decimal numbers. I figured it out a while ago and made a program out of it.

http://www.mediafire.com/?m2pq31vcvt7d5zu

You will need java installed to run it. If you want to see how long a pool would take, go to the Round Robin tab and change the number of pools to 1.

The really tough calculation part is figuring "how many sets can be played 'at once'". In my program I labelled this number as "Consecutive Matches" though I should have named it Simultaneous Sets as that makes more sense. What I mean by this is lets say you have a single elimination with 16 entrants. The number of sets from first round to grand finals is:

8
4
2
1

Assuming each set goes for exactly T minutes, then if you have 8 setups the tournament will go for:

8/8 = T
4/8 = 0.5 --> round up --> T
2/8 = 0.25 --> round up --> T
1/8 = 0.125 --> round up --> T

The tournament will take 4T. This problem gets especially difficult when you have say 3 TV's and even more difficult when you have say 19 entrants and even MORE difficult when you apply it to double elimination instead of single. My program does its best to estimate the "Simultaneous Sets" which in my above example is 4. The total time taken = (Time Per Match)*(Number of Matches)/(Simultaneous Sets).

I just dug up some of my code and the calculation for the ConsecMatches is:

ConsecMatches = Math.ceil(2*entrants/tvs + 2*Math.log(tvs)/Math.log(2)-1);

 

[Kason Birdman]

haha thats awesome.. Imagine if questions in highschool were actually like that.

cant you just use inequalitys with one side being an equation for how many pools sets there would be and the other side being (2n-1)/2, then solve for n? ofcourse, that just figures out how many sets. maybe thats wrong. actually yeah nvm it is wrong lol.

 

[M.K]

Blah, I'm dying on the inside -_- This problem is giving me a headache.

Paramount Electronics has an annual profit given by

P = -100,000 + 5000q - 0.25q^2
where q is the number of laptop computers it sells each year. The number of laptop computers it can make and sell each year depends on the number n of electrical engineers it employs, according to the equation

q = 30 n + .01n^2

The average profit PP per computer is given by dividing the total profit P by q:

PP = (-100,000/q )+ 5000 - 0.25q

Determine the marginal average product (dPP/dn) at an employee level of 10 engineers and interpret the results.

Please help I'm awful at math

 

[Corpsecreate]

if I'm understanding this correctly, it just wants you to find dPP/dn and evaluate that result at n = 10 and interpret the results.

PP = -0.25q + 5000 - 100,000/q
q = 0.01n^2 + 30n

Sub the equation for q into pp:

PP = -0.25(0.01n^2 + 30n) + 5000 - 100,000/(0.01n^2+30n)
PP = -n^2/400 - 15n/2 + 5000 - 100,000/(0.01n^2+30n)

dPP/dn = something, get out your pen/paper

 

[Elyssa Xey Hexen]

An electric field is a vector at every point in space caused by some arrangement of charges like electrons or protons. If some other charge is placed at some point in this electric field, it will experience a force pushing it in some direction along with a certain magnitude.

Now, the electric field generated by a single charge has magnitude E = (k *q) / d,
where E is the magnitude of the electric field at a particular point, k is a constant, q is the charge creating the electric field, and d is the distance from the charge to a point in space. The direction of the electric field at a particular point points away from positive charges, but towards negative charges.

Say that you have an infinitely long line of positive charge. Does the magnitude of the electric field
A) increase as you get further away from the line of charge.
B) decrease as you get further away form the line of charge.
C) stay the same.
D) Not enough information.

Explain your answer.

A little help on this one.

 

[Deleted member]

okay so what I'm guessing here is that at every whole unit along the Y-axis is one positive charge (all of the same power).
so at any point in the plane (there really is no difference between the 2D and 3D case here since the situation is symetrical around this axis of charges), you experience a force from all (ie infinte amount of) the charges.
this means you can make an infinite sum of these charges to calculate the total magnitude (which would always point away from the line at a right angle (why?))

now, if this sum converges to some real number, and you would position yourself twice as far away from the line, the sum should also decrease by some factor (probably half but I'm not entirely sure, but if will decrease).
however if that sum diverges, there isn't really a point in moving away als infinity/2 is still infinity.

now I don't really know how this infinte sum will look I hope this at least helps a bit.

 

[Rici]

I think because it's a line of positive charge , the electric field will be perpendicular to that line(because every particle causes the field lines of the others to go straight). So if you get further away, the distance d gets bigger, and because E=(k*q)/d, this means that the magnitude will get smaller.

@The Paprika Killer,

I see where you're getting from, and you would be indeed correct if it wasn't for the fact that electric field lines never cross each other.

 

[Deleted member]

well obv the field lines are all parallel to eahcother and perpendicular to the line of charges, but all charges still apply a positive sideways force to any point of the plane, their vertical force component just gets cancelled by all the other charges. thus at any point in the plane you will experience a force from all particles, thus you'd still need an infinite sum to calculate that sideways force, which is a sum that could diverge. If it does you would experience an infinite (sideways)force at any point except when you're actually on the charge line (since all the charge's horizontal force components are 0 there).

not saying the sum actually will diverge (since I don't know what comes out of it), just that it can.

 

[Elyssa Xey Hexen]

I got it figured out after my exam. Its as Paprika said. Assume the line of charge is vertical. The vertical components of all charges at a particular point in space will cancel out leaving only there horizontal components to add together.

Now, the closer the point is to the line of charge, the closest charges will contribute more to the strength of the field at that point than the charges further out. This is because the charges far away are almost vertical and do not contribute a whole lot to the overall electric field.

As a point gets further away from the line of charge, the closest charges will contribute less (since its further away), but charges further away are no longer acting like they are vertical and more charges start to contribute to the overall electric field at that point.

So, apparently given an infinite line of charge, the electric field is the same everywhere in space (except where the line of charge is). If I was to add up the contribution from every charge, it should come out to be independent of distance from the line of charge. I'm glad I was not expected to do an integral like that.

 

[Kason Birdman]

yo guys. i got more fun physics I donno if I'm doing it right or if im missing anything.

A small block is fired from a compressed spring up a ramp sloped at 25 degrees to the horizontal. The mass of the block is 0.0035 kg, the coefficient of friction is 0.43 and the spring force constant is 9.5 N/m. Calculate the compression of the spring in order for the block to slide a vertical height of 5.7 cm.

man.. I just realized.. we have a lot of questions involving small blocks.

 

[Elyssa Xey Hexen]

@Anyone: Does anyone have any slick methods when trying to reverse the area of integration when taking double integrals. The same for when converting area of integration restrictions to polar coordinate restrictions? The only way I have been doing it so far is to draw out the whole area of integration, then figure out the new bounds of integration from that. It's not difficult to do that when I have functions such as lines or simple curves, but when I do not know how to draw an equation like x = y - 2y^2, it takes up a lot of time to figure it out.

@Kason Birdman:

I think you need to model the compression of the spring as a conservation of energy.

The initial potential energy + kinetic energy will equal the kinetic energy and potential energy afterwards. When the spring is compressed, it will have some potential energy, but there is no kinetic energy. After it is released, the potential energy of the spring goes into the kinetic energy of the block. If the distance the block travels during compression is much less than the distance it travels up the slope, then I think it is safe to assume that energy was conserved.

U = K, where U = 0.5 * k * s^2; k is the spring constant and s is the distance compressed. And, K = 0.5 * m * v^2; m is the mass of the block and v is the velocity of the final velocity of the block.

The final velocity of the block will have the force of gravity and the force of friction acting against it. So, the blocks velocity needs to slow down to zero in the span of 5.7 cm (vertical height).

So with all of that, I think you should be able to figure out the amount of compression. Although, I am not sure if you should assume no friction or gravity acts on the block when it moves along that little bit of compressed distance. If it did, you wouldn't exactly have conserved energy. A way to test that might be to look at the distance it travels during compression to the distance it travels up the slope and comes to a stop. If the compression distance is much less than the distance the block travels, I think you can assume that energy was conserved since any distance the block traveled during compression is not a whole lot of time for it to be acted upon by the opposing forces.

 

[Kason Birdman]

ah ok thanks. and yeah the thing about the friction making it so not all energy is conserved is what I was having problems with.

 

[Elyssa Xey Hexen]

Then, I only have one idea. The springs potential energy gets transferred into the block, and where it stops is the point at which the block has zero potential energy. This change in potential energy should equal the work done by the block on the system. Or you could say the system does negative work on the block through conservative forces against the direction the block moves in.

So U_f - U_i = -W = - force * distance
U_f is defined to be zero, while you have U_i as the spring potential energy. The force is the net force of gravity and friction over a distance that equates to 5.7 cm in height. That's my best guess. And a quick calculation I did with that idea was 1.39 cm compressed from equilibrium. No idea if you have the answer, you did before hand.

 

[Kason Birdman]

Yeah I don't have the answer. but once I do I'll let you know. And yeah I see where you're going with that.. looks pretty solid I'de say.

 

[Kason Birdman]

I got another one.. I think I know how to do it but the way I did it seems too easy.

A ball is dropped frmo a height of 1.00 m onto a vertical spring which has a force constant of 500 N/m. If the maximum compression of the spring is 18.0 cm, calculate the mass of the ball.

I did it using the law of conservation of energy.

before = after
Ee + Ep + Ek = Ee' + Ep' + Ek'
Ep = Ee' (at the biginning all there is is potention enegery due to gravity, at the end all there is is the elastic potential enegery in the spring)
mgh = 1/2kx
m(9.8)(1) = 1/2(500)(0.18)
m = 500(0.18) / (2(9.8))
m = 4.59 kg

is this how you would do the question?

 

[Elyssa Xey Hexen]

Go with your method. Potential energy due to gravity goes into kinetic and kinetic will go into the potential energy of the spring since there is no energy loss.

But isn't potential energy of dependent on the square of the springs displacement? You do not have it squared.

 

[Kason Birdman]

yeah you're right, the x is squared

 

[Kason Birdman]

do you know anything about planetary mechanics?

I guess this isn't really the place.. its not THAT math related.. but just incase..

In 1620 you are applying to become an assistant to Mr. Kepler. In the interview he asks you to calculate how long it would take the earth to complete one full revolution around the sun if the distance from the earth to the sun was tripled (keep in mind that the earth-sun distance was not known at that time.) How would you answer the question and land the job?

Has something to do with the Kepler constant (3rd law of his i think...) I'm guessing? but how do you apply it to get an answer?

 

[Corpsecreate]

There is a 25 character password on a computer. The characters can be any letter and any digit and the password is not case sensitive. You are given a hint about the password, the hint is "No character can be repeated in a row, for example F5EE has 2 E's in a row. However T6EO4E is fine as the 2 E's are not in a row".

How many possible combinations are there?

If we discount the hint, the answer is 36^25. With the hint, I dont know how to do it. Any help?

 

[Rici]

Not totally sure, but it sounds right:
As you said the first character has 36 possibilities. The one after that cant be the same so then there are 35 possibilities. And that goes on till the end. So then you get:
36x35x35x35x... So it should be 36 x 35^24.

 

[Jim Morrison]

Welcome to high school maths, can anyone help me out!?

Given is the function f(x)= cos 2x - sin x + 1 in the domain of [0, 2π]
For which value of p does the line y=p have exactly 4 intersections with the function of f(x)?

I don't even know where to start.

While I'm here, another one;

Given is the function f(x) = cos 2x - 2sin x + 2 in the domain of [0,2π]
Calculate the coordinates of the tops A, B, C and D (the 4 extremes in the domain) of this function.

 

[Elyssa Xey Hexen]

2nd part you simply take the derivative of the equation. Once you do that, you'll use a double angle formula. Then, you can factor something out. Put that equation equal to zero and you can find the values of X that are critical points within the domain of 0 to 2pi

You could then take the 2nd derivative. Then put the x values you found and plug into the 2nd derivative. If its greater than zero, then you have a min at that X value. If it is less than zero, you have a max at that X value.

Not sure how to do the first one.