Sciencemadness Discussion Board - Physics Exam (Question)

Physics Exam (Question)

fluorescence - 14-8-2015 at 07:45

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]

blogfast25 - 14-8-2015 at 09:07

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 FN,1, and one parallel to the plane.

2) Do the same with the red force. The normal force is FN,2.

3) The total friction is now F = μ(FN,1 + FN,2

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.

fluorescence - 14-8-2015 at 13:24

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 ?

blogfast25 - 14-8-2015 at 14:20

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]

aga - 14-8-2015 at 14:38

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.

blogfast25 - 14-8-2015 at 14:55

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!

aga - 14-8-2015 at 15:02

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]

blogfast25 - 14-8-2015 at 15:11

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.

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]

blogfast25 - 14-8-2015 at 16:04

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]

fluorescence - 14-8-2015 at 23:40

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.

gatosgr - 15-8-2015 at 01:32

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]

gatosgr - 15-8-2015 at 01:43

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]

fluorescence - 15-8-2015 at 02:29

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]

blogfast25 - 15-8-2015 at 06:36

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.

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]

fluorescence - 15-8-2015 at 06:56

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.

blogfast25 - 15-8-2015 at 07:05

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.

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]

fluorescence - 15-8-2015 at 07:11

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.

gatosgr - 15-8-2015 at 07:17

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]

blogfast25 - 15-8-2015 at 07:18

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!

gatosgr - 15-8-2015 at 07:19

that's a bad fbd blogfast it will confuse anyone who reads it

fluorescence - 15-8-2015 at 07:22

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:

blogfast25 - 15-8-2015 at 07:23

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.

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]

gatosgr - 15-8-2015 at 07:24

You should learn how to draw a clean free body diagram OP it's all it takes for these kind of problems.

fluorescence - 15-8-2015 at 07:27

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.

gatosgr - 15-8-2015 at 07:28

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]

gatosgr - 15-8-2015 at 07:33

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.

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.

blogfast25 - 15-8-2015 at 07:33

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.

gatosgr - 15-8-2015 at 07:36

Read the answer above your post, I'm done with this. OP use my solution and learn how to draw diagrams, bye.

fluorescence - 15-8-2015 at 07:38

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.

blogfast25 - 15-8-2015 at 07:38

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]

blogfast25 - 15-8-2015 at 07:50

@fluoresence:

If you're struggling with these problems, try and go back to 'basics'. Revisit thoroughly the material you were initially taught.

fluorescence - 15-8-2015 at 07:57

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.

blogfast25 - 15-8-2015 at 08:04

Quote: Originally posted by fluorescence

So should I take your formula. It seems logic to take both angles into consideration and
use a+b instead of only b.

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]

fluorescence - 15-8-2015 at 08:13

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.

blogfast25 - 15-8-2015 at 08:20

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]

fluorescence - 15-8-2015 at 08:23

Oh that's what you meant, sorry.

So the y should be

for F: f* sin(a+b)

for mg: mgcos(a)

blogfast25 - 15-8-2015 at 08:28

Now determine y components.

fluorescence - 15-8-2015 at 08:31

Didn't I just do that ? Or do you mean now with friction ?

blogfast25 - 15-8-2015 at 08:34

Quote: Originally posted by fluorescence

Didn't I just do that ? Or do you mean now with friction ?

No, we've just determine the x components, not the y components. You need the latter because they affect the friction.

gatosgr - 15-8-2015 at 08:36

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]

fluorescence - 15-8-2015 at 08:37

?

We had x:

Fcos(a+b) and mgsin(a)

And y:

Fsin(a+b) and mgcos(a)

What did I miss ?

gatosgr - 15-8-2015 at 08:39

read my solution or fail

[Edited on 15-8-2015 by gatosgr]

blogfast25 - 15-8-2015 at 08:41

Quote: Originally posted by gatosgr

Blogfast did you pass your physics exam?

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]

blogfast25 - 15-8-2015 at 08:42

Quote: Originally posted by fluorescence

?

We had x:

Fcos(a+b) and mgsin(a)

And y:

Fsin(a+b) and mgcos(a)

That's correct.

Now determine the friction force, obviously in the x direction.

fluorescence - 15-8-2015 at 08:45

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.

blogfast25 - 15-8-2015 at 08:48

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.

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?

fluorescence - 15-8-2015 at 08:50

µ[Fsin(a+b) + mgcos(a)]

blogfast25 - 15-8-2015 at 08:53

Quote: Originally posted by fluorescence

µ[Fsin(a+b) + mgcos(a)]

Exactly.

Now write the balance of ALL forces in the x-direction, friction force included.

gatosgr - 15-8-2015 at 08:54

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]

fluorescence - 15-8-2015 at 09:01

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.

blogfast25 - 15-8-2015 at 09:02

Quote: Originally posted by gatosgr

Any physics professor would substract points for this mess of yours.

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 - 15-8-2015 at 09:06

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 - 15-8-2015 at 09:10

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 - 15-8-2015 at 09:36

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 - 15-8-2015 at 09:49

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 - 15-8-2015 at 12:04

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 - 15-8-2015 at 12:21

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.

blogfast25 - 15-8-2015 at 12:25

Quote: Originally posted by Fulmen

I agree with aga, start by checking with another student.

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 - 15-8-2015 at 13:09

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 - 15-8-2015 at 15:26

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...

fluorescence - 15-8-2015 at 23:32

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 - 16-8-2015 at 05:43

Quote: Originally posted by fluorescence

blogfast did a really good job there, and I thank him very much and of course thank you
to all of you here, too

Thanks and good luck with your exams.

fluorescence - 16-8-2015 at 07:33

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 - 16-8-2015 at 08:06

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 - 16-8-2015 at 08:17

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 Fx,red. 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 = Fx,red 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 - 16-8-2015 at 11:22

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]

blogfast25 - 16-8-2015 at 11:53

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 - 16-8-2015 at 12:12

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

blogfast25 - 16-8-2015 at 12:55

Quote: Originally posted by Eddygp

6.626x10^-34, actually

6.626x10-34 J.s, actually.

fluorescence - 16-8-2015 at 13:20

@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 - 16-8-2015 at 13:45

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 - 16-8-2015 at 14:09

Blogfast: I'm having a little trouble following the sketches. Please give me the values of the angles α and ß.

blogfast25 - 16-8-2015 at 14:19

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 - 16-8-2015 at 14:41

For µ = 0.4 that gives F=0.9815mg. Is that correct?

blogfast25 - 16-8-2015 at 15:30

Quote: Originally posted by Magpie

For µ = 0.4 that gives F=0.9815mg. Is that correct?

Yes, that's correct.

F = mg (sinα + μcosα)/(cosβ - μsinβ)

[Edited on 16-8-2015 by blogfast25]

Magpie - 16-8-2015 at 16:12

I got the same answer with my own equations.

As you can see there was a lot of erasing.

[Edited on 17-8-2015 by Magpie]

blogfast25 - 16-8-2015 at 16:16

Quote: Originally posted by Magpie

I got the same answer with my own equations.

Your equations are different? If so I'd like to see them.

[Edited on 17-8-2015 by blogfast25]

Magpie - 16-8-2015 at 16:31

I edited my post to include a photo of my calculations. If you can't read it I can provide a better edition.

[Edited on 17-8-2015 by Magpie]

blogfast25 - 16-8-2015 at 17:06

They're the same, except you filled in the numerical values of the angles right from the start. No problem with that, of course. There really is only one solution to this kind of problem.

And I honestly can't see the erasing...

[Edited on 17-8-2015 by blogfast25]

gatosgr - 16-8-2015 at 22:58

bad diagrams people stop confusing OP

[Edited on 17-8-2015 by gatosgr]

Magpie - 17-8-2015 at 02:41

Quote: Originally posted by blogfast25

There really is only one solution to this kind of problem.

You're right, of course. The only difference might be in the use of different trigonometric identities, eg:

Sinß instead of Cos (90°-ß)

gatosgr: I'll see if I can make a better diagram.

[Edited on 17-8-2015 by Magpie]

[Edited on 17-8-2015 by Magpie]

MrHomeScientist - 17-8-2015 at 07:06

gatosgr, didn't you say you were leaving the thread way back on page 2?

Very interesting physics problem, by the way. I always enjoyed drawing free body diagrams!

blogfast25 - 17-8-2015 at 07:17

Quote: Originally posted by MrHomeScientist

gatosgr, didn't you say you were leaving the thread way back on page 2?

Promises, promises, MrHS... promises, promises.

Strange also how someone who considers free body statics problems 'not physics' (or not 'worthy' of physics or whatever Mr no need to puncuate gatosgr actually meant) keeps coming back to this thread...

I think he should concentrate on preparing HCl his favourite way:

Quote: Originally posted by gatosgr

Are drain cleaners allowed or should we bubble chlorine gas through water to make HCL [sic]? Electrolysis and such... although drain cleaners are very common everywhere.

[Edited on 17-8-2015 by blogfast25]

Magpie - 17-8-2015 at 12:00

Here's my sketch in "paint." I hate drawing with a computer - it's so slow and tedious. I might add the calculations later.

friction sketch.bmp - 703kB

blogfast25 - 17-8-2015 at 12:33

Quote: Originally posted by Magpie

I hate drawing with a computer - it's so slow and tedious.

A lot faster than Etch A Sketch, though!

blogfast25 - 17-8-2015 at 16:00

Here’s an interesting mechanics ‘free body’ problem but not a Statics one.

A mass m is dropped from a sufficiently high platform, while attached to a sufficiently strong elastic rope of unstretched length l (the other end of the elastic rope is attached to the platform). The elastic rope behaves like a Hookean spring.

When the mass m is merely suspended from the platform by the elastic rope, the stretched (equilibrium) length of the rope is twice the unstretched length.

Prove that when the mass is dropped from the platform while attached to the elastic rope it falls to a depth of:

d = (2 + √3) l

Ask for any clarifications, if you need them. Enjoy!

[Edited on 18-8-2015 by blogfast25]

Oscilllator - 17-8-2015 at 17:53

I get d = (3 + 2√2)l

blogfast25 - 17-8-2015 at 18:16

Quote: Originally posted by Oscilllator

I get d = (3 + 2√2)l

I haven't checked your solution completely but I can see one simple error: at equilibrium mg = kl, not 2kl. For a Hookean spring, extending force = spring constant (k) times spring extension (NOT spring length). Here spring length at equilibrium is 2l, so spring extension is 2l - l = l, not 2l.

Solving this as an energy conservation problem is of course the right way of doing it.

[Edited on 18-8-2015 by blogfast25]

Oscilllator - 17-8-2015 at 18:47

That's probably it. In my defence though, I had a lecture in 15 minutes at the time I started the problem.

blogfast25 - 17-8-2015 at 18:51

Try that correction and it should work, I think. Haven't had time to verify it myself.

fluorescence - 18-8-2015 at 03:35

I found another one in this years exercises.

There is that sqared ladder frame that moves with a speed v = a*t
into a magnetic field that changes it's strengh over time.
So this is a combination of two known problems. The question is for
the induced Potential/ Current (U), don't know how to correctly call that in English. The even gave a formula for the magnetic field ( I marked that orange).

So I went through some books and found the following lines (marked blue) since it's hard to find that much on that topic.

This is for a different shape that a simple linear conductor. So like the shape I have. It's said that the change of Φ with time can be made into a sum of two Φ.

Φ1 equals the induced potential for the change of the magnetic field where Φ2 is for a movement of the frame.

Now I have my problems integrating that. Isn't the given formula (orange box) already the integral for the equation Φ1 ? Since I dunno how to integrate that. What do I integrate that for ? I have a dependecy on t so shouldn't there be a dt somewhere ? Do I have to integrate it for dA since this is the only thing that is mentioned in that equation.

And then there is this ring integral Φ2. I dunno how to solve these. I know there is a formula for a straight conductor U = BlV. But that has no area in it. I'm not sure how to solve that one ? Can I use that given formula as solution for Φ1 already ?

Edit:

I thought about that Φ2 again.

U for a moving linear conductor in a magnetic field should be

U = n * ΔΦ/Δt where n is equal to 1.

And Φ is defined as Φ = B * A

No I just asume B to be constant for a second and just deal with A.
A is like the only thing that changes now where
ΔA = l*v*Δt

So U = n * ΔΦ/Δt = B * ΔA = B*l*v

So the area is not important for that step ? Do I have to add up all four sides of the frame to get the lenght ?

And then question is for the Φ1. Do I just take that formula and define B here as non constant and just use the information in the box. Or do I do it as said in the blue box and calculate two Φ's and then add them up. But if so, how do I have to treat the B in Φ2, constant or not constant ?

[Edited on 18-8-2015 by fluorescence]

fluorescence - 22-8-2015 at 09:54

Okay so I tried to figure out another
solution. Perhaps this could be better.
At least I think I'm close to the solution now
although there are some unknown letters
left.

Could this perhaps be anything close to an
answer to the question ?

fluorescence - 7-9-2015 at 06:28

So today was my physics exam. It was quite hard, nearly nothing on mechanics,
thermodynamics or waves, most was on electromagnetism and there was only
one inclined plane which was quite cool and took me a moment to understand.

Now I don't want to hear a solution since I just hope mine was correct and I don't
want that to ruin my holidays now but just to keep you updated on how it acutally was.

There was an inclined plane with a mass (with a charge ) on it that slides down ( no rolling ). A
Magnetic Field is ponting into the background and the question was when so at
what time the obejct lifts off which was quite cool and most of the students I asked
didn't came up with the solution.

I just wanted to say thank you to all of you for the support !

blogfast25 - 7-9-2015 at 08:19

All the best!