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On October 27 2008 13:19 HeavOnEarth wrote:although you used basic.. F = ma then got the Ff .. i think im drawing the picture wrong  hmm alright now i get why you used sin for Fg and cos for Ff, made the triangle whoa Where'd Fpush and Fhor come from/ mean / represent ohhh hahaha Fhor means horizontal, I'm used to just writing Fn I'm still not sure what Fpush represents
Fpush is if you're pushing diagonally ( didn't realize you were pushing horizontally x_x)
I used F = ma because I'm still in B 
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On October 27 2008 14:32 HeavOnEarth wrote:Let's say, just for example, to anaylze the force there is A box weighing 630 N is pushed along a hor- izontal floor at constant velocity with a force of 250 N parallel to the floor. alright, you got a force going down, due to weight, and the force due to gravity you got a force, going across the floor, let's call this Fpush... of 250N did i miss anything? alright let's say it asks What is the coefficient of kinetic friction between the box and the floor? Ff = Fn x u ... so solve for coefficient ? please point out where my logic stopped, i don't want a numeric answer, i just want to understand physics  + Show Spoiler +
ma = Fpush - Ff
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United States24671 Posts
On October 27 2008 14:38 travis wrote:Show nested quote +On October 27 2008 14:36 micronesia wrote:On October 27 2008 14:21 HeavOnEarth wrote:thanks! now to um apply that with more examples ahh more hw problems D::: i ran out like this
Is it accelerating? Yes, since it's going in a circle. So, Fnet is Ff.
why? In order for an object to be accelerating, its velocity must be changing. You can change velocity by accelerating forwards (increasing speed), or accelerating backwards (deceleration; decreasing speed). However, if you go in a circle, your speed is constant, but the direction of the velocity is still changing... so you are accelerating. what if you go in a circle, while twisting on an axis do you go supernova?
Lol what? Serious question nestled in there?
Earth?
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It helps to draw all of the forces. One force is, as you mentioned, the force due to gravity BY the earth ON the box (630N). Another force (though it's not a real force, consider it that when drawing diagrams), is Force Normal BY the floor, ON the box (630N). That's why the box isn't accelerating straight downwards right? Because something is pushing it back up, which is force normal (one of newton's laws...an applied force must have an equal and opposite force or something like that). The obvious force is the 630 N, which is BY pusher, ON the box. The last force is the force of friction, which is BY the floor, ON the box. So now you have all the forces. Do this for every force problem, even if it's easy.
oh right, 3rd law, heh
Yeah i'll try and kill lots of trees next homework session
In the case of this problem, they're not really asking for a net force or anything like that, they're just asking for the coefficient. So in that case, knowing that Fn (or Force Normal) is the force of the surface pushing back on the object, you know that Fn is 630N. If you knew the force of friction you would just solve for u.
Sorry if I don't make sense.
oh it made perfect sense, Fn is the perpendicular force, so it's Fn, i had more problems identifying the Ff in that problem however.
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The mgcos(theta) is calculated because it is equal to the Fn, which is the force by the surface on the block. The mgsin(theta) is the force that it is sliding down the slope thanks to gravity.
edit: So the net force of the block sliding down is mgsin(theta)-Ff where Ff=Fn * mgcos(theta). So the force needed to keep it from moving is a force in the opposite direction with the same magnitude.
edit2: crap didn't read question, lemme read it again.
edit3: oh ok, so it's started to move, and you don't want it to accelerate. In that case, you just want the net force to be 0 down the slope so that there's no acceleration. So yea, you have the hypotenuse, which is the same as the net force of the block going down, you have the angle, so all you have to do is use trig to solve for the horizontal component.
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hehe ur adjacent side looks bigger than the hypotenuse
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On October 27 2008 14:32 HeavOnEarth wrote:
Let's say, just for example, to anaylze the force there is
A box weighing 630 N is pushed along a hor-
izontal floor at constant velocity with a force
of 250 N parallel to the floor.
Since it is not accelerating the sum of the force vectors must be zero. Have you done linear algebra? F(Gravity) is F(G)(0,0,-1), F(Normal) is F(N)*(Unit vector that is orthogonal against the surface), F(Push) is just a constant force in any direction(Usually given in the example), F friction is always parallel to the plane and always in the opposite direction of the sum of other forces.
Now assume that the plane is orthogonal to the force of gravity and that the pushing force is is parallel to the plane. We know F(G) is in the (0,0,-1) direction and that no other forces are acting in the z-axis except F(N), which since it is also orthogonal to the plane is in the direction (0,0,1). As such we can deduce that F(N)=F(G) and it is easy to see that the sum of them are 0.
To make it easy just assume that the pushing force is acting with a Force f along the positive x-axis, then we get f(1,0,0), since we only got a single force left friction must be f(-1,0,0), Now since the object is moving the friction force is always |friction|=µ|F(N)| which gives µ=|friction|/|F(N)|.
Anyhow, the hard part comes when you have forces which are not orthogonal. For example if we have a tilted plane F(N) could go in the (0,-sina,cosa) direction were a is the tilt degree around the x-axis, and thus (If every other force is orthogonal to the G force, which is usually not the case) cosa*F(N)=F(G) as long as the object is not accelerating.
However if we have an acceleration along the plane we can add the negative of that force on the force equations.
Hmm, maybe I am going overboard...
Anyway doing problems like this is all about linear algebra. And F=ma just means that the force vector is equal to the acceleration vector divided by the mass. If the acceleration is orthogonal to the velocity no change of speed is made. For example if I go 100m/s north, and accelerates 10m/ss to the east in this instant my speed do not change. Speed is equal to the length of the velocity vector, which is sqrt(A^2+B^2+C^2) were A,B,C are the components of the velocity. Now if we take the a-derivative of that we get 2*A*(lots of shit), assuming that the other terms are not zero and that A is zero we get that a small change of A do not change the speed if the A component of the velocity vector is zero. That goes for all directions since you can always transform the components of each vector to a new coordinate system were that direction would be one of the unit vectors.
However the vector still changes, and thus the velocity, but the length do not.
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whoaa wall of confusing* text o,o gimme a sec
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By the way, if you want it in simpler terms just do one equation for every dimension of freedom like this:
Sum of forces in x-axis=(Acceleration in the x-axis)*mass
Sum of forces in y-axis=(Acceleration in the y-axis)*mass
Sum of forces in z-axis=(Acceleration in the z-axis)*mass
Are you doing momentum yet? Then you also need to check those equations to see how much its rotation is accelerating/deaccelerating.
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Since it is not accelerating the sum of the force vectors must be zero. Have you done linear algebra? F(Gravity) is F(G)(0,0,-1), F(Normal) is F(N)*(Unit vector that is orthogonal against the surface), F(Push) is just a constant force in any direction(Usually given in the example), F friction is always parallel to the plane and always in the opposite direction of the sum of other forces.
Fg ,down on the z axis
Fn, perpendicular = orthogonal ?
Fpush, kk noted
Ff, noted
Now assume that the plane is orthogonal to the force of gravity and that the pushing force is is parallel to the plane. We know F(G) is in the (0,0,-1) direction and that no other forces are acting in the z-axis except F(N), which since it is also orthogonal to the plane is in the direction (0,0,1). As such we can deduce that F(N)=F(G) and it is easy to see that the sum of them are 0.
alright, Fn is just pushing back on Fg so they're equal (in magnitude )?
To make it easy just assume that the pushing force is acting with a Force f along the positive x-axis, then we get f(1,0,0), since we only got a single force left friction must be f(-1,0,0), Now since the object is moving the friction force is always |friction|=µ|F(N)| which gives µ=|friction|/|F(N)|.
alright friction moves against Fpush , however, it's not nessecary equal , is it? unless it comes o a complete stop, even then friction might > force, unless then co-oridinate plane you're writing just represents the direction and not magnitude, then nevermind heh.
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On October 27 2008 15:17 Klockan3 wrote:
By the way, if you want it in simpler terms just do one equation for every dimension of freedom like this:
Sum of forces in x-axis=(Acceleration in the x-axis)*mass
Sum of forces in y-axis=(Acceleration in the y-axis)*mass
Sum of forces in z-axis=(Acceleration in the z-axis)*mass
Are you doing momentum yet? Then you also need to check those equations to see how much its rotation is accelerating/deaccelerating.
z - axis = Fg+Fn ,
x axis = Ff+ Fpush and
y axis = ?
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Personally, I think it's more confusing it learning to think about it in that way right away.
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On October 27 2008 15:23 HeavOnEarth wrote:Show nested quote +On October 27 2008 15:17 Klockan3 wrote:
By the way, if you want it in simpler terms just do one equation for every dimension of freedom like this:
Sum of forces in x-axis=(Acceleration in the x-axis)*mass
Sum of forces in y-axis=(Acceleration in the y-axis)*mass
Sum of forces in z-axis=(Acceleration in the z-axis)*mass
Are you doing momentum yet? Then you also need to check those equations to see how much its rotation is accelerating/deaccelerating. z - axis = Fg+Fn , x axis = Ff+ Fpush and y axis = ?
y axis was zero in my example.
And the important thing to note here is that if the object is not accelerating then all of these must equal zero, no exceptions at all.
On October 27 2008 15:27 nevake wrote:
Personally, I think it's more confusing it learning to think about it in that way right away.
He should already have learned doing it with only one dimension of freedom since these problems needs vectors.
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Anyhow, the hard part comes when you have forces which are not orthogonal. For example if we have a tilted plane F(N) could go in the (0,-sina,cosa) direction were a is the tilt degree around the x-axis, and thus (If every other force is orthogonal to the G force, which is usually not the case) cosa*F(N)=F(G) as long as the object is not accelerating.
AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
damn, spamming caps didn't make it easier to understand
However if we have an acceleration along the plane we can add the negative of that force on the force equations.
Hmm, maybe I am going overboard...
Anyway doing problems like this is all about linear algebra. And F=ma just means that the force vector is equal to the acceleration vector divided by the mass. If the acceleration is orthogonal to the velocity no change of speed is made. For example if I go 100m/s north, and accelerates 10m/ss to the east in this instant my speed do not change.
not understanding that, why does it not change? actually , what does orthogonal mean heh
Speed is equal to the length of the velocity vector, which is sqrt(A^2+B^2+C^2) were A,B,C are the components of the velocity.
I learned this one a while back.
Now if we take the a-derivative of that we get 2*A*(lots of shit), assuming that the other terms are not zero and that A is zero we get that a small change of A do not change the speed if the A component of the velocity vector is zero. That goes for all directions since you can always transform the components of each vector to a new coordinate system were that direction would be one of the unit vectors.
However the vector still changes, and thus the velocity, but the length do not.
with... X -> V - > A deriv. / 2nd deriv. / intergals ?
i shall balance redox and half reactions and turn on my memorization hacks for chem solubility/oxidation rules as i waits :O chem is so easy compared to physics ^.^
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I take it as you have not learned linear algebra, everything gets much easier if you had done that.
http://en.wikipedia.org/wiki/Orthogonality
In other words, orthogonal is just a more general term for perpendicular.
Vector transformations are done by applying an orthogonal matrix to the whole field, but since you have not done linear algebra you wont understand that...
Sorry, but I thought that maybe you would have read such things when you did a physics question like this. I assume that you only work with forces that are perpendicular then, in that case all you would need are those 3 equations I mentioned earlier.
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I take it as you have not learned linear algebra, everything gets much easier if you had done that.
it's not that i didn't learn it at all, it's just our teacher just gives us the essentials to solve the problem , shortcuts to calculus, then proceeds in teaching us physic II
is this a good site to learn linear algebra from?
http://www.egwald.ca/linearalgebra/vectors.php
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On October 27 2008 15:44 HeavOnEarth wrote:Show nested quote +
I take it as you have not learned linear algebra, everything gets much easier if you had done that.
it's not that i didn't learn it at all, it's just our teacher just gives us the essentials to solve the problem , shortcuts to calculus, then proceeds in teaching us physic II is this a good site to learn linear algebra from? http://www.egwald.ca/linearalgebra/vectors.php
I need to go now, but I can help you when I get back. Since I live in sweden my timezone is a bit different :p
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good night or stuffs `
ill be here in about 15hours, right nows.. 2am so ill get home at around 5
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On October 27 2008 14:38 HeavOnEarth wrote:
Likewise,
A tired worker pushes with a horizontal force
of 400 N on a 100 kg crate initially resting on a
thick pile carpet. The coefficients of static and
kinetic friction are 0.6 and 0.2, respectively.
The acceleration of gravity is 9.81 m/s2 .
Find the frictional force exerted by the car-
pet on the crate. Answer in units of N.
Ok, i know Ff = u*Fn
I know F=ma
Fg = mg
and i can picture the forces 400->>>> horizontally, and 100kg, 9.81 downwards
, now the coefficients confuse me =/
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On October 27 2008 16:05 HeavOnEarth wrote:Show nested quote +On October 27 2008 14:38 HeavOnEarth wrote:
Likewise,
A tired worker pushes with a horizontal force
of 400 N on a 100 kg crate initially resting on a
thick pile carpet. The coefficients of static and
kinetic friction are 0.6 and 0.2, respectively.
The acceleration of gravity is 9.81 m/s2 .
Find the frictional force exerted by the car-
pet on the crate. Answer in units of N.
Ok, i know Ff = u*Fn
I know F=ma
Fg = mg
and i can picture the forces 400->>>> horizontally, and 100kg, 9.81 downwards
, now the coefficients confuse me =/
I would guess that 0.6 is the static friction and 0.2 is the kinetic friction.
Since the crate is initially resting and 400 N is not enough to go above the static friction the answer must be: ???????
Edit: And for what to read to learn the linear algebra needed, read roughly everything in this that is not about matrixes:
http://en.wikipedia.org/wiki/Dot_product
The dot product is usefull all the time in mechanics since you need to constantly project forces on to surfaces or split them up to get the equations straight etc.
Might be a bit abstract but if you plan on doing something that requires college level math then there is no reason not to learn about it now, it is a powerful tool at the same time as the definition of it is really simple.
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