@WillHunting so ... we are both abnormal :)

yes

Oui

ja

Sim

Is it possible to write the solution of $x(t)=A(t)x(t)$ as $x(t)=X(t,t_0)x_0+\int_{t_0}^{t}X(t,s)A(s)ds$ where "X(t,t_0)" is the principal matrix solution ?

@WillHunting ...and retire, right?

@WillHunting Stupid little details! (I'm a perfectionist!) That particular OP, I think, doesn't like me...a while back, given a comment I made. Oh well, I won't waste more time on him/her!

@WillHunting hehehe...I've started do nibble (I nibble, rarely feast!) How about you?

@WillHunting Where are you?

@WillHunting Really! Wow...good morning to you, then!

@WillHunting I'm at 5:10 p.m.!

@WillHunting hehehe...Yes indeed!

must get some sleep - maybe tomorrow I will have a kitchen again ...

goodnight Will- and all

@OldJohn Good night, John!

@Charlie Bye for now

@WillHunting Can I make a stupid question?

@WillHunting If a function $f$ is uniformly continuous in $[0,\infty)$ is it uniformly continuous in $[0,a)$ ?

@WillHunting, you're welcome!

it is, isn't it?

:_:

I don't know what to study.

@WillHunting, should I attempt to kill some of my Calculus book?

@WillHunting Thanks.

@WillHunting, Calculus for Dummies ;)

@WillHunting, why?]

I rather like them

@WillHunting I don't like these Dummies' books either :)

@WillHunting no just went to my upstairs computer which is still logged-in :)

@WillHunting Of course I did! I can't remember, did we have icecream too?

and wondered if anyone would notice the subtle apostrophe in my last comment ...

@WillHunting, Schaum's Outline of Calculus gets hardcore.

Triple integrals toward the very end

Schaum's complex analysis is pretty good

@WillHunting Yes, I need to eat better (and a bit more, too)!

@WillHunting, for now I do. Off and on.

I think I am going to go get some studying done.

Ciao.

darn - nobody noticed the difference between "Dummies Books" and "Dummies' Books" :(

Sup guys

Sups

@Link sup

Nothing much

can you help me with a physics problem?

@Link sure

Maybes.

@WillHunting :)

Must sleep - g'night again

Night

Night

I have a ramp of angle $\theta$, the ramp is mostly frictionless, except for a small piece in the middle. A block $m$ is going up the ramp, when it hits the friction area, it has acceleration of $1.13g$ down the ramp. The block eventually stops and then goes down the ramp, when it hits the friction area it has accl of $0.27g$ up the ramp. What is $\theta$?

$g = 10N/kg$

Any ideas?

I drew a FBD, but stuck :C

I am using mint

@Link I'm sorry, i did not understand, the block is going up, hit the friction pint and then goes down?

No

it goes up

past the friction bit

eventually stops

and goes back down

oh! got it

when going over the friction bit on the way up, accelration is downwards, when going down the ramp, accelration is upwards

okay

hmm....

thinking

wait..loading....

note that friction only depends on the normal force, friction coefficient, and direction. so the friction force on the way up is equal and opposite of the friction force on the way down. call this force $f$ ...

on the way up, the acceleration is $1.13 = g \sin(\theta) + f/m$ and on the way down the acceleration is $0.27 = g \sin(\theta) - f/m$. so add those two equations together, and you get $1.13 + 0.27 = 2 g \sin(\theta)$

wait, can you explain that a bit more?

Especially the second part of both equations?

like the g sin(0) +- f/m?

@Link you understand why the force of friction is equal and opposite on the trip up vs down?

Yes, since it is opposite the relative motion

right. now on the way up, gravity has magnitude $g \sin(\theta)$ and is pulling it down the ramp. friction has magnitude $f/m$ and is also pulling it down the ramp. they're in the same direction, so we add

okay

but what is f/m?

m = mass

f is?

on the way down, gravity and friction have the same magnitude, and gravity is still pulling it down the ramp, but friction is pulling up the ramp so we subtract

Okay, i get that, but what is F?

cool. so the point is that $f$ depends on $m$, which you don't know ... but it ends up canceling out, as you can see in the equations above

$f$ also depends on $\mu$, which you also don't know

$f$ denotes the force of friction, if that's what you're asking

oh

that makes sense then

but why are you dividing by m?

cool cool cool

i mean by m?

well, because $f$ is the force and $F = ma$, so the acceleration is $f/m$ ... you could just as well call $f$ the acceleration due to friction and not divide by $m$

okay

so f is friction, and since f = ma, a = f/m?

but in the end it doesn't matter as it cancels out..

precisely!

awesome

thanks

no problem

I hate not being able to solve this on my own :C, i can draw diagrams and stuff, and now what to do, just can't get the equations down

Also, why do you set the equations equal to .27 and 1.13?

is it because that is the netforce at that time?

whoops, should have been $.27g$ and $1.13g$ ... just because we were calculating the acceleration of the block, and those were the accelerations given in the problem

...

okay

oh

i get it now!