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gatosgr
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Quote: Originally posted by blogfast25  | | Quote: Originally posted by gatosgr | why do you do physics if you are a chemistry student? physical chemistry has nothing to do with physics actually and physicists or engineers don't do
it the same way as chemists.
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What the HELL are you talking about???? That chemistry students should not have to attend physics courses??? Or only physical chemistry courses?
Do you realise just how self-limiting that would be??? 
Don't 'do it the same way'? Don't do what the same way? Rumpy pumpy? Nooky? Making whoopy? Dear G-d... ROFLOL!
[Edited on 15-8-2015 by blogfast25] |
Attend some thermodynamics physics classes or statistical thermodynamics engineering classes you'll sure learn the difference to what you're taught in
chemistry instead of making comments.
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blogfast25
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| Quote: Originally posted by gatosgr | That problem is not physics, it's high school physics learn to differentiate the two.. I told you it's stupid and pointless it's not real physics
either you get almost nothing from this.
[Edited on 15-8-2015 by gatosgr] |
Firstly, you might learn to use punctuation.
Secondly, problems like the OPs are REAL physics, whether you like it or not. It's Mechanics, specifically Statics. It's what's taught before
Kinematics and Dynamics. You're being ABSURD.
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gatosgr
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Read the answer above your post, I'm done with this. OP use my solution and learn how to draw diagrams, bye.
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fluorescence
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The problem is that I need pass it.
Everyone says that question is easy to answer and yet I get different equations.
I searched through the net for something like this. I went through all of our exercises
I even looked through the 500 page exercise book that our professor recommended
and there is questions like this one where I have an inclined plane + another angle.
This isn't the only question I couldn't answer correctly but it's the one that gave me zero points
and I have to go through a whole list of these questions I remember from the first exam to make
them better this time ( they won't be asked again but I have to know how to solve them).
I'd need an example where I already have a solution and then learn how to construct it.
As you see I have no idea how to combine these two problems plane+ angle.
I know how to solve them seperately but I don't know how to combine them.
Your answeres where quite clear till now and showed how to construct a force diagram.
And if you look for blogfasts explanation he ordered them quite well so the components of the equation
are like the would be on the inlclined plain.
The problem is just what angles do I use now. I understood why the formula looks like this now,
I see the different parts of it now the question is just for the angles (a+b) or (b) ?
I can't anwer it, if I could I wouldn't ask that question. I need that answer once so I can
construct similar problems from that scheme.
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blogfast25
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| Quote: Originally posted by gatosgr |
Attend some thermodynamics physics classes or statistical thermodynamics engineering classes you'll sure learn the difference to what you're taught in
chemistry instead of making comments. |
Listen you twit, I've got a degree in chemical engineering and am very well versed most parts of physics too, including statistical thermodynamics.
Your whole approach here makes no sense at all: the physics of Static bodies is an integral part of physics and an important one. That I have
to defend that notion on a science forum is truly remarkable.
You're attitude is astonishing: you seem to think you can define what is physics and what is not based on what you value as
worthwhile and what not. NONSENSE!
A very decent representation of physics paradigms, here:
http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html
[Edited on 15-8-2015 by blogfast25]
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blogfast25
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@fluoresence:
If you're struggling with these problems, try and go back to 'basics'. Revisit thoroughly the material you were initially taught.
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fluorescence
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I already did that. That was the only problem where I didn't have an answer for.
I already worked through all the stuff we did but we never did anything like this
and then they ask for it.
So should I take your formula. It seems logic to take both angles into consideration and
use a+b instead of only b. I uploaded that inclined plane again where the angle is better
defined.
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blogfast25
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Yes but only if you understand it. For example, do you know how to decompose the red force F into its x- and y-components, using
basic trigonometry? Similarly for the force mg?
[Edited on 15-8-2015 by blogfast25]
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fluorescence
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The x-part should be with a cosine, so F * cos(a) and since it has no friction it's just that
the y-part has friction since it's like the normal force and thus should be F*µsin(a).
And as you said there is also the mass so there is mg that is also pointing into y-direction.
so they have to be taken into consideration, too.
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blogfast25
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| Quote: Originally posted by fluorescence | The x-part should be with a cosine, so F * cos(a) and since it has no friction it's just that
the y-part has friction since it's like the normal force and thus should be F*µsin(a).
And as you said there is also the mass so there is mg that is also pointing into y-direction.
so they have to be taken into consideration, too. |
No.
The x component of the red force F equals F cos(α + β) . Don't concern yourself with friction just yet.
The x component of mg equals mg sinα.
Now determine the y-components of F and mg.
[Edited on 15-8-2015 by blogfast25]
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fluorescence
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Oh that's what you meant, sorry.
So the y should be
for F: f* sin(a+b)
for mg: mgcos(a)
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blogfast25
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Now determine y components.
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fluorescence
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Didn't I just do that ? Or do you mean now with friction ?
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blogfast25
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No, we've just determine the x components, not the y components. You need the latter because they affect the friction.
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gatosgr
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Blogfast did you pass your physics exam?
That's just fun to read but I'm sorry for OP that he has to deal with this.. you might get weird solutions sometimes and I think it's what's confusing
you because for some angles and friction coefficients there is no solution meaning the body will never start moving. Use the angles and coefficients
they gave you.
[Edited on 15-8-2015 by gatosgr]
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fluorescence
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?
We had x:
Fcos(a+b) and mgsin(a)
And y:
Fsin(a+b) and mgcos(a)
What did I miss ?
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gatosgr
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read my solution or fail
[Edited on 15-8-2015 by gatosgr]
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blogfast25
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Your being an obnoxious turd and an obstacle to any progress for the OP. READ what I wrote from the beginning. I defined the problem correctly and
solved it too.
Please FUCK OFF. You're now effectively trolling.
[Edited on 15-8-2015 by blogfast25]
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blogfast25
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That's correct.
Now determine the friction force, obviously in the x direction.
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fluorescence
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In the x-direction ?
There is friction as I move it in the x-direction, but isn't the friction caused by the y-components?
At least on the normal inclined plane you only use the friction for the normal force which is the y-direction.
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blogfast25
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| Quote: Originally posted by fluorescence | In the x-direction ?
There is friction as I move it in the x-direction, but isn't the friction caused by the y-components?
At least on the normal inclined plane you only use the friction for the normal force which is the y-direction.
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The friction is in the x direction and it is caused by the y components: it's the sum of the y components (aka 'the 'Normals') multiplied by the
friction coefficient μ. Can you manage that?
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fluorescence
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µ[Fsin(a+b) + mgcos(a)]
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blogfast25
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Exactly.
Now write the balance of ALL forces in the x-direction, friction force included.
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gatosgr
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There's a reason normal people use proper angles instead of mixing them all up, it makes for easier solutions, you might get away with this simple
problem but if you try it with more complex ones you're fkd. Bad solution indeed blogfast. Any physics professor would substract points for this mess
of yours.
[Edited on 15-8-2015 by gatosgr]
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fluorescence
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mgsin(a) + µ[Fsin(a+b) + mgcos(a)]
and later I have that Fcos(a+b) on the other side of the equation.
That's pretty much what you answered in your first posts and it even has the structure of the inclined plane.
F = downhill force + normal force
here downhill is mgsin(a) like in the original equation for the inclined plane.
the normal force has two components, the usual normal force from the inclined plane mgcos(a) and
the y-component from the red Force Fsin(a+b)
And F is now not the force to push up but rather the x-Compontent of the red force.
I understood where they come from. I wasn't sure about the angles that (a+b) thing. If they really add up
or not.
Like what happens if I push with 15° but not down but up, so mirror that red force on a horizontal line.
Wouldn't they add up to 45°, too ? Or is upwards then -15° and so I just get 15° for b ?
Thinking about this logically it has to be a different amount of force since the friction increases when I
push into the ground.
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