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fluorescence
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Physics Exam (Question)
I'm sorry if I posted that into the wrong section.
But I didn't know where to put it.
I asked that question a couple of people today an none could help me.
This was the question in a physics exam at my university.
As drawn on the picture I have a box on an inclined plane with a mass m.
The plane has an angle of let's say 30 degrees. No I want to push that box up but the Box has the friction µ = 0,4.
And as I push up I don't push it horizontally to the plane but in an angle 15 degrees into the plane as shown as a red arrow.
The question is what Force do I need push the box up that way.
I would say
F = [ mg (sin(30) + µ cos (30) ) ]+ mg µ cos(15)
So the first part comes from the the normal inclined plane with friction. I need the power to move it against the downhill force + an additional
amount of power to move against the friction caused by the normal force.
So I have to add these two. The second part is the question how much power I need to put into this extra since I go for it in an angle of 15 °.
So I push the box even harder into the ground meaning I need an extra term with the friction and my angle in it.
So I thought of it like a second inclined plane and just added another normal force with the second angle. The problem is the following.
If I'd push it horizontally without any angle the cos becomes 1 and I need xyz N more force although there is no second actiont that would cause it.
So these xyz N musn't apear. So it's more likely to be a sin (15) but I can't see that mathematically. Can someone help me with the correct equation
please ?
Thank you in advance,
Greetings,
Martin
[Edited on 14-8-2015 by fluorescence]
[Edited on 14-8-2015 by fluorescence]
[Edited on 14-8-2015 by fluorescence]
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blogfast25
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Your second angle on the schematic is ill-defined. I'll assume the 15 degrees refers to the angle between the force in red and the horizontal.
1) Decompose the weight of the mass m into a component that's vertical to the plane, I'll call it F<sub>N,1</sub>, and one parallel to the
plane.
2) Do the same with the red force. The normal force is F<sub>N,2</sub>.
3) The total friction is now F = μ(F<sub>N,1</sub> + F<sub>N,2</sub>
4) Now balance the forces in the plane to work out what size the red force should be to get the block to move.
We don't provide ready-made solutions to homework on this site.
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fluorescence
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Okay thank yout but it's not a homework that's from my last physics exam that I unfortunately failed and now have to
repeat. I know they'll ask again for something like that and so I want to be prepared.
I didn't fully understand your explanation but I thought again about the angles if the box was on a
flat ground with 0° and I tried to push it with an angle (b).
In that case I confused them. The new force comes from a horizontal and a vertical force.
And the horizontal one has the cosine and thus no µ in it. And the vertical one is
another part of the force which has a sine in it and the friction µ. But since gravity is also
pointing downwards those two shoud add ( or do they multiply ? )
so it would be something like
F = µgm / [ sin(b)µ - cos(b) ] ?
So I'm pretty sure about the force I need to push the box up that is:
F = mg(sin(a) + cos(a)µ) And That force is now somehow linked to the formula above, but how.
I mean actually that would be the horizontal part of the force.
So it should be something like:
(mg + sin(b) F)µ - Fcos(b) and now F has to be [mg(sin(a) + µcos(a)]
But then I don't know how to rearrange the formula.
Did I get this correctely ? The force I need to push up is [mg(sin(a) + µcos(a)].
And that is now divided into an horizontal part and a vertical part
which follows the equations above ?
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blogfast25
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Call the angle between the plane and the horizon α, and the angle between the red force and the horizon β. The angle between the red force and the
plane is thus α + β.
Two Normal forces push the mass down to the plane, one from the weight, that is mg cosα, one from the red force F, that is F sin(α +
β). Together they provide a friction force μ[mg cosα + F sin(α + β)].
Now construct the balance of forces in line of the plane's surface:
F cos(α + β) > mg sinα + μ[mg cosα + F sin(α + β)], 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.
I could provide a much more formal treatment of this problem but I'm not sure the OP would be well served by it.
[Edited on 15-8-2015 by blogfast25]
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aga
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It might help if you get two bricks, and a plank of wood.
Try with the plank flat, and push a brick along it.
Then use the second brick to make the plank inclined at 3 different angles.
(advanced brick/plank users can create a variety of angles)
Try to push the first brick again for each different incline of the plank and feel the difference.
Simple experimentation can sometimes help you get a grip on what the maths are trying to describe.
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blogfast25
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Quote: Originally posted by aga  |
Simple experimentation can sometimes help you get a grip on what the maths are trying to describe. |
That and a decent force diagram of the forces acting along the plane and perpendicular to it! 
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aga
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Not a lot of use to understand the Forces at play if you cannot Apply them to some physical Reality.
Do not mention friction.
Wickes now sell frictionless planks for exactly this experiment.
Edit
A variety of Bricks are also available, but the Planck is constant, regardless of it's inclination.
[Edited on 14-8-2015 by aga]
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blogfast25
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| Quote: Originally posted by aga | Not a lot of use to understand the Forces at play if you cannot Apply them to some physical Reality.
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Understanding can also come from writing down the governing equations. That is of course also the only way to answer the OPs question.
You could rig a system (corresponding to the OPs problem) with goniometers and force gauges but that would not provide a general
expression for F, only one that is particular to the parameters you've physically imposed.
[Edited on 15-8-2015 by blogfast25]
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blogfast25
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Ok, no need to go further off topic, agreed. So much gibberish in that last post I wouldn't know where to start refuting/clarifying it anyway. 
Let the OP have his say now.
Also, I've just heard the bell ring, so it's time for you to attend your QM/WM class!
[Edited on 15-8-2015 by blogfast25]
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fluorescence
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Ok thank you both for your replies. I'm already on the run now so I'll check that formula later
when I'm back. Just a question to your Formula. I thought if you replaced a > or < by an =
you'll have to add like a constant to it or am I wrong there ?
If I didn't swap anything that should be then:
F = [µmg cos(a) + mgsin(a)] / [cos(a+b) - µsin(a+b)]
That somehow looks like my formula but here both angles are combined. Okay I have to draw that
on a piece of paper. Thank you so much, I'll reply later when I'm back.
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gatosgr
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That's a high school problem which university used this for exams ? Anyway here is a FBD which you should always prioritize , then use ΣFx>0 and
solve for F seriously I doubt this is from exams it's too easy.
[Edited on 15-8-2015 by gatosgr]
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gatosgr
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Here is the full answer you should also draw the friction force on the opposite of x
-mgsin30+Fsin15-μ(mgcos30+Fcos15)>0 <=> Fsin15-μFcos15>mgsin30+μmgcos30 <=> F(sin15-μcos15)>mg(sin30+μcos30) <=>
F>mg(sin30+μcos30)/(sin15-μcos15)
if the angle is 15 degrees parallel to the x axis use 90-15=75 degrees instead for my solution
[Edited on 15-8-2015 by gatosgr]
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fluorescence
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I'm a chemistry student not a physics student. The questions aren't that hard. Still actually noone managed to correctly solve this question which is
just the a) Part of a whole series of questions to that topic. They had to set the points down till you only needed 30 of 100 points to make even some
students pass that exam. None of us had Physics in their highschool finals since we all chose chemistry and other stuff. And I don't remember that we
had this topic in school.
But thank you for your help .
So I calculated that with some imaginary numbers and came up
with about 85 N to even push that box up.
So if I do that with another angle that even pushes the box further
into the street I'll end up with a force that has to be bigger than 85 N.
I calculated that with both formulas given, one ended up with
200 N and your's with -660 N.
So did I do a mistake in solving that upper equation, given by blogfast25
for F ?
Oh and with 75 I'd get about 100 N
Edit:
Oh now I see it blogfast25 made the angle b to the horizon and not to the plane. so his (a+b) is actually my (b), or did I read this wrong ?
So I calculated both again and if I use 75 in your equation I get the same result as 15 in blogfast25's equation. So that should be it. Now I just
have to understand it.
PS:
I dunno why but I have some problems with classical mechanics ( I hope thats what it's called in English). I already had the exam on theoretical
chemistry which was full of quantum mechanics, wave functions, hamiltonians and other calculation and I already had physical chemistry which was about
thermodynamics, kinetics, electrodynamics/chemistry, ...
And I passed them quite well. But with classical mechanics, that was the part that made me fail that exam. Together with some other stuff. But I guess
that is part of how we had our exercises. There was a whole semester on stuff like capacitors, coils and anything related to electricity with some
small bits of waves at the very end and all the rest of mechanics, thermodynamics, rotation and so on was the other semester, so too short to really
fully get how to construct diagrams like you showed above.
[Edited on 15-8-2015 by fluorescence]
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blogfast25
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| Quote: Originally posted by fluorescence |
Oh now I see it blogfast25 made the angle b to the horizon and not to the plane. so his (a+b) is actually my (b), or did I read this wrong ?
So I calculated both again and if I use 75 in your equation I get the same result as 15 in blogfast25's equation. So that should be it. Now I just
have to understand it.
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Clarifying the angles:
The angle between the red force and the slope is alpha + beta.
Then decompose each force into x- and y-components. The force balance in the x-axis tells you whether or not the mass moves along the x-axis or not.
Net x-force = 0 means no movement.
[Edited on 15-8-2015 by blogfast25]
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fluorescence
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Ok thank you so it's really (a+b). I see that from the picture now.
So is gatosgr's solution wrong or did I miscalculate something ?
a+b would mean that I have to use 45° since I push 15° into the box but the box and the ground
are coming into my direction with an angle of 30°C and so it's really 45°. Mhm...makes sense
I just seperated the angles as if they weren't influencing each other.
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blogfast25
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| Quote: Originally posted by fluorescence | Ok thank you so it's really (a+b). I see that from the picture now.
So is gatosgr's solution wrong or did I miscalculate something ?
a+b would mean that I have to use 45° since I push 15° into the box but the box and the ground
are coming into my direction with an angle of 30°C and so it's really 45°. Mhm...makes sense
I just seperated the angles as if they weren't influencing each other.
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As I indicated from the start: your angle β was poorly defined. So I interpreted it as shown in my schematic. An angle is ALWAYS between two straight
lines, just writing a number somewhere as you did in your first schematic means nothing and is open to interpretation. For instance,
if the asker states: 'the angle of the red force is 15 degrees' then that is meaningless: angle of 15 degrees between what and
what???
Poor problem definition is the cause of many, if not most, failures to solve simple problems like this one.
If you prefer you can define β differently, then work out the forces in x and y directions according to that particular definition. The formula will
be different, yet the solution will be the same.
[Edited on 15-8-2015 by blogfast25]
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fluorescence
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I think you interpreted it correctely. The angle b is between the top side of the box and the red force. So between the slope and the red force. It
should be like in your picture.
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gatosgr
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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.
anyway if you're confused with the sins replace sin15 with cos15 and forget a and b...... the reason you get the same result is sin90-x=cosx
and cos90-x=sinx here are the trigonometric identities http://www.mathwords.com/t/trig_identities.htm
-mgsin30+Fcos15-μ(mgcos30+Fsin15)>0 <=> Fcos15-μFsin15>mgsin30+μmgcos30 <=> F(cos15-μsin15)>mg(sin30+μcos30) <=>
F>mg(sin30+μcos30)/(cos15-μsin15)
[Edited on 15-8-2015 by gatosgr]
[Edited on 15-8-2015 by gatosgr]
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blogfast25
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| Quote: Originally posted by fluorescence | | I think you interpreted it correctely. The angle b is between the top side of the box and the red force. So between the slope and the red force. It
should be like in your picture. |
Sigh... NO! In my definition the angle between the slope and the red force is α + β. You need to be CLEAR, fluorescence!
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gatosgr
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that's a bad fbd blogfast it will confuse anyone who reads it
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fluorescence
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You have a lot of subjects if you study chemistry Physics, Biochemistry, Mathematics, ... actually those are the ones that kick most students out
since we mostly didn't learn these and focused on chemistry instead in school.
Here is the angle defined:
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blogfast25
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| 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]
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gatosgr
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You should learn how to draw a clean free body diagram OP it's all it takes for these kind of problems.
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fluorescence
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The problem is that I could, if I fail that exam again get exmatriculated although I've already passed all other exams. That would be 4 Semesters for
nothing just because of physics. I mean it's important no question but why do they have to make grades for these exams if it's not part of chemistry.
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gatosgr
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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, blame the guys that built your course, I blame mine.
[Edited on 15-8-2015 by gatosgr]
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