| Pages:
1
2
3
4 |
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
No, he wouldn't. If the OP had defined β unambiguously there would have been no need to clarify it. If anything I'd have earned points for doing
that.
β could be defined anyway you want as long as you have a second line defining it. How you define β affects the formulas but not the actual solution.
Sorry you can't see that.
Angles are angles, there are no "normal" angles. And "normal" people is just an irrelevant value judgement here.
[Edited on 15-8-2015 by blogfast25]
|
|
|
fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
You have seen that I actually redefined b correctely on page 1 at the bottom ? Just asking if your solution aims for that new defined angle.
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Quote: Originally posted by fluorescence  | | You have seen that I actually redefined b correctely on page 1 at the bottom ? Just asking if your solution aims for that new defined angle.
|
In my definition the angle β is simply the angle between the red force F and the horizontal. Pure and simple. As defined high up by me. No change on
my part.
Want to define it differently? Fine, just adjust the x and y components of the red force F accordingly. Not a problem at all.
The angles explained (as already shown above):
[Edited on 15-8-2015 by blogfast25]
|
|
|
fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
Okay that should be what I have. 15° to the horizon and another 30 from the plane which equals to 45.
Makes sense. So I probably just take you equation, it seems logic. And if the force pointed upwards I'd have to
substract them which equals 15 again ?
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
If you prefer, define β as the angle between the red force F and the x-axis (x-axis as defined in the schematic). In that case you can simply replace
the sum (α + β) by β, in the formulas that we worked out.
What matters is the ACTUAL value of the angle, in ACCORDANCE with the definition one uses (this seems to be the part 'gatogr' doesn't comprehend or
wants to make a HUGE deal of).
[Edited on 15-8-2015 by blogfast25]
|
|
|
aga
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
Despite blogfast25's patiently explained and correct method, it would be a good idea to ask another student on the course.
It may be that they were awake during that 1 lecture, and were taught in a different way.
Passing exams involves more than just getting the right answers.
Showing that you remember the Method you were Taught can also gain points.
Edit.
Likely that gatosgr was taught to approach this a different way, which is what is causing all this friction 
[Edited on 15-8-2015 by aga]
|
|
|
Fulmen
International Hazard
Posts: 1726
Registered: 24-9-2005
Member Is Offline
Mood: Bored
|
|
I agree with aga, start by checking with another student. Statics requires a bit "logistics", getting all your variables sorted out and organized in a
logical manner. While the physics is the same the methodology can vary quite a bit. Ideally your chosen method shouldn't affect your grades as long as
it's coherent and produces the right answer, but some censors are more picky than others.
A year ago I'd probably solve this in a heartbeat, but I really haven't done much statics since then so I'm a bit rusty. But perhaps I will give it a
go anyways just to see that I haven't forgotten everything.
We're not banging rocks together here. We know how to put a man back together.
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
It seems likely to me he had already done that or wasn't able to for some reason. Why else go to the trouble of becoming a member and posting the
problem here?
|
|
|
aga
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
Agreeing with aga is never likely to be a good choice.
It could be that the OP tried, didn't get it, came here ?
We all bend the truth a little when in need.
In any event, with what you;ve explained, even i get it, so the OP's problem should be well and truly sorted.
|
|
|
Fulmen
International Hazard
Posts: 1726
Registered: 24-9-2005
Member Is Offline
Mood: Bored
|
|
Good point, aga 
This one has peaked my curiosity, but presenting a complete solution isn't trivial. I usually solve these on paper, and then there is the whole
language barrier. I just realized that my vocabulary doesn't really cover all aspects of statics...
We're not banging rocks together here. We know how to put a man back together.
|
|
|
fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
Good Morning,
Why I didn't ask anyone else ? Well I did. I asked everyone I knew and had on skype, I even asked
a friend who studies physics. And they all gave me different solution. That from the physicist was the
worst where I needed like half of the force to push it up on an angle than whithout an angle and you
clearely saw that it had to be wrong.
The problem is that physics goes over 2 semesters meaning that it can only be written once in a year.
There is one exam for people who didn't write it after half a year but that one is usually harder than
the annual one ( retry exams are usually harder than the first ones ).
So we mostly chose to wright it after a year again where it would be easier.
So those who didn't make it the first time weren't really able to solve it.
And those who did did it more than a year ago. And we had two semesters
full of exams, practical works in the lab (6 of them to be honest) and lot's of other stuff.
Noone remembers what he wrote back then.
Then I mailed the guy that made excercises but I couldn't reach him and since they ususally do that
during there final years it might be that he already left the university or has a new mail.
Then I looked through the books and the net and couldn't find anything and so
I went here in the hope that someone had a clue. Cause if you think about it.
You know what could possibly be in an exam and you have no chance to solve it.
I know have the time to find someone who explains it to me and there was noone who did.
blogfast did a really good job there, and I thank him very much and of course thank you
to all of you here, too  .
I recieved a mail from a fellow student yesterday whom I asked for his physics summary
and never go anything and he only asked me how you would photopolymerize 2-methyloxirane
with Ar2JPF6. I sent him a long explanation for how the reaction would be and that was it.
They just come and ask for stuff and if you want something it's like "ewww...physics, no stop that"
[Edited on 16-8-2015 by fluorescence]
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Thanks and good luck with your exams.
|
|
|
fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
Thank you .
Another question. If I'm asked for the work that is needed to push it for a certain distance.
The formula is W = F*s but with vectors there is also a cosine in that formula.
I guess that is just for the directions of the forces since work equals the area in a
force to distance diagram. And the work is defined for any direct line between two points
in space. So if there is an inclined plane it shouldn't change anything but the force needed.
But just to be sure. That angle has nothing to do with my plane, the formula will shorten
to W = F*s, doesn't it ?
|
|
|
gatosgr
Hazard to Others
Posts: 237
Registered: 7-4-2015
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by fluorescence | Thank you .
And the work is defined for any direct line between two pointsin space. So if there is an inclined plane it shouldn't change anything but the
force needed.
But just to be sure. That angle has nothing to do with my plane, the formula will shorten
to W = F*s, doesn't it ? |
Don't ever write that down it's not right but for your case you can hypothesize it, only when the convergence of the force field is zero the path
taken doesn't matter for the work done because the force field is conservative and still you need to further analyze. The mechanical work done is
W=Fcos15*s since this force vector is what pushed the box and the vertical vector produces friction aka thermal dissipation to the environment.
youtube has enough problem for you to practice
[Edited on 16-8-2015 by gatosgr]
[Edited on 16-8-2015 by gatosgr]
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by fluorescence |
The formula is W = F*s but with vectors there is also a cosine in that formula.
I guess that is just for the directions of the forces since work equals the area in a
force to distance diagram. And the work is defined for any direct line between two points
in space. So if there is an inclined plane it shouldn't change anything but the force needed.
But just to be sure. That angle has nothing to do with my plane, the formula will shorten
to W = F*s, doesn't it ? |
Hmmm. You need a better understanding of what precisely constitutes F in the W = Fs equation.
In the y-direction no useful work is being performed because all y-components cancel each other out.
Useful work is being performed ONLY by the x-component of the red force F, I'll call that
F<sub>x,red</sub>. We've already calculated it.
The work performed by this force when it drags the object along the x-axis for a length Δx is then:
W = F<sub>x,red</sub> Δx (the line of force and the line of displacement have to be the same).
Note that this is only true when that force is constant over Δx but that is what we generally assume in problems like this one.
[Edited on 16-8-2015 by blogfast25]
[Edited on 16-8-2015 by blogfast25]
|
|
|
Magpie
lab constructor
Posts: 5939
Registered: 1-11-2003
Location: USA
Member Is Offline
Mood: Chemistry: the subtle science.
|
|
So, what is the final answer? I want to compare it to my result.
I recommend that you don't memorize the methods for specific problems. Instead develop a good understanding of the principles involved, ie, ΣF = 0,
force due to gravity = mg, force due to friction = µ times the force normal to the sliding surface, etc.
Then it just becomes a problem in trigonometry.
Then practice, practice, practice.
[Edited on 16-8-2015 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Magpie:
fluorescence defines the angle β as the angle between the red force F and the plane as 15 degrees and the angle α as the angle between the plane and
the horizontal as 30 degrees.
In that case the solution is:
F cosβ > mg sinα + μ[mg cosα + F sinβ], for any upward movement to happen.
Solve this equation for F. Set the 'larger than' sign to 'equal' if you're not familiar with inequalities.
[Edited on 16-8-2015 by blogfast25]
|
|
|
Eddygp
National Hazard
Posts: 858
Registered: 31-3-2012
Location: University of York, UK
Member Is Offline
Mood: Organometallic
|
|
| Quote: Originally posted by aga |
A variety of Bricks are also available, but the Planck is constant, regardless of it's inclination.
[Edited on 14-8-2015 by aga] |
6.626x10^-34, actually
there may be bugs in gfind
[ˌɛdidʒiˈpiː] IPA pronunciation for my Username |
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
6.626x10<sup>-34</sup> J.s, actually.
|
|
|
fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
@Magpie: So in my case it should actually be just b if I think about it again. Cause I just saw that blogfast defined
b between horizon and red force, but in my description b is the angle between the plane or the box and the red force so I have to use (15° instead of
45°).
Edit:
Okay I calculated this stuff through again now with correct angles.
If I define the angle b to be between the plane and the red force as I drew it in the diagram I need (with an example) about 98 N.
To move the box itself without these angles I'd need 84 N so it would fit. I need more Energy to push it with angle that increases the friction.
Now what happens if I push with the same angle but upwards, so the red force mirrored on a horizontal line ? If I use -15 as value I end up with 79 N.
So less then just moving the box. Do I have to use another angle here or is it okay ?
Logically I wouldn't increase the amount of friction as I don't push into the plane. I'd even lift it up a bit since I'm pushing it upwards so it
could even reduce the amount of friction created.
[Edited on 16-8-2015 by fluorescence]
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by fluorescence |
Okay I calculated this stuff through again now with correct angles.
If I define the angle b to be between the plane and the red force as I drew it in the diagram I need (with an example) about 98 N.
To move the box itself without these angles I'd need 84 N so it would fit. I need more Energy to push it with angle that increases the friction.
Now what happens if I push with the same angle but upwards, so the red force mirrored on a horizontal line ? If I use -15 as value I end up with 79 N.
So less then just moving the box. Do I have to use another angle here or is it okay ?
Logically I wouldn't increase the amount of friction as I don't push into the plane. I'd even lift it up a bit since I'm pushing it upwards so it
could even reduce the amount of friction created.
|
Yes, the angle β would indeed become -15 degrees, the way you described it. The y-component of red F would now point upwards, reducing friction and
reducing the value of red F needed to get any movement. Pointing red F somewhat downward as in the original problem is an inefficient way of pushing
anything up a plane!
[Edited on 16-8-2015 by blogfast25]
|
|
|
Magpie
lab constructor
Posts: 5939
Registered: 1-11-2003
Location: USA
Member Is Offline
Mood: Chemistry: the subtle science.
|
|
Blogfast: I'm having a little trouble following the sketches. Please give me the values of the angles α and ß.
The single most important condition for a successful synthesis is good mixing - Nicodem
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Magpie:
fluorescence defines the angle β as the angle between the red force F and the plane as 15 degrees and the angle α as the angle between the plane and
the horizontal as 30 degrees.
In that case the solution is:
F cosβ > mg sinα + μ[mg cosα + F sinβ], for any upward movement to happen.
Solve this equation for F. Set the 'larger than' sign to 'equal' if you're not familiar with inequalities.
|
|
|
Magpie
lab constructor
Posts: 5939
Registered: 1-11-2003
Location: USA
Member Is Offline
Mood: Chemistry: the subtle science.
|
|
For µ = 0.4 that gives F=0.9815mg. Is that correct?
The single most important condition for a successful synthesis is good mixing - Nicodem
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Yes, that's correct.
F = mg (sinα + μcosα)/(cosβ - μsinβ)
[Edited on 16-8-2015 by blogfast25]
|
|
|
| Pages:
1
2
3
4 |
![[*]](images/xpblue/default_icon.gif){kind=icon}
