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Forces in a wedge and a graphicall display of its forces


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#1 MatsNorway

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Posted 13 February 2013 - 12:26

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Is this the correct way to display the forces? Friction is not a factor.

I kinda struggle to understand my colleagues point about the wedge top force allways being bigger. to me this is just like a transmittion or a arm. you add a ratio(the angle) and you get different forces out.

It see so many ways to display the forces. there is one tangential to each side. Then there is Y and X. then there is along with the sides. in addition people mix up the counter forces so the display becomes all messed up.

With the forces being along the sides Fx becomes smaller when the wedge gets thinner. That does not seem right to me.

With the forces being tangential you don`t get a triangle at all. i haven`t looked into that so much but it should give me a bigger force on the sides than the input force.

Edited by MatsNorway, 13 February 2013 - 12:36.


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#2 Greg Locock

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Posted 13 February 2013 - 22:38

You need a force on the other side of the wedge for equilibrium, but yes you have the concept right, a thin wedge has greater 'amplification' than a fat one.

If that boggles your mind, turn the drawing on its side and you have a car on a slope, a gentle hill generates less resistance than a steep one.

#3 Kelpiecross

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Posted 14 February 2013 - 05:13


Mats - I think it is more appropriate if you draw the sideways wedging force at right angles to the applied downwards force - not as you have drawn it at a right angle to the inclined plane. Right angles to the applied force is the way a wedge lifts things or a screw clamps or a screw jack lifts. I see that Wiki has also got this slightly arse-up with their calculation (force divided by the tan of the wedging angle) giving the wedging force at a right angle to the applied force but their diagram shows the wedging force that is developed at right angles to the inclined face of the wedge. - which is the applied force divided by the sin of the wedging angle.

And I hope that I haven't got it a bit arse-up either.

#4 MatsNorway

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Posted 14 February 2013 - 07:48

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For me it makes perfect sense to think that a thinner shape pushes more to the side. Its distance 1 vs force 1 = distance 2 vs force 2. If d1 gets increased you get less d2 but the force go up.

They claimed that the highest force will allways be the one i put in.. small triangle above the wedge in the picture. to me it makes sence about energy but they continue to mix things up.

And i can`t say against them because i can`t explain this well enough.

#5 Greg Locock

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Posted 14 February 2013 - 10:31

They claimed that the highest force will allways be the one i put in.. small triangle above the wedge in the picture. to me it makes sence about energy but they continue to mix things up.

And i can`t say against them because i can`t explain this well enough.


If you can get back to an energy equation then generally things are easier.

In this case

Impact force* dy=2*Normal_force_on_wedge*(theta/2)*dy

for narrow wedges, with a tip angle of theta (there is a potentially large second order term I'm ignoring)



#6 MatsNorway

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Posted 14 February 2013 - 14:02

What is dy? distance in y direction?

Formulas are actually not what im after but more the graphicall way to find it. vectors?

Since there is no friction in this teorietical question surely the shape of the tip is irrelevant. once the angle goes to zero the force (Fx) goes to infinity. (dx) Distance traveled in x direction = 0

There is no impact if it matters for the formula. Its a push.

Edited by MatsNorway, 14 February 2013 - 14:32.


#7 Kelpiecross

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Posted 15 February 2013 - 04:35

What is dy? distance in y direction?

Formulas are actually not what im after but more the graphicall way to find it. vectors?

Since there is no friction in this teorietical question surely the shape of the tip is irrelevant. once the angle goes to zero the force (Fx) goes to infinity. (dx) Distance traveled in x direction = 0

There is no impact if it matters for the formula. Its a push.


Mats - the way I tend to do this type of problem is to calculate (by trigonometry) how much an object on the inclined face would be moved (in whatever chosen direction) relative to a certain distance moved by the applied force. This trigonometrical method amounts to a graphical method. The ratio of the distance moved by the applied force to the resulting movement of the object represents the mechanical advantage (or disadvantage) of the system and hence the multiplication of the applied force. As you say - the "pointier" the wedge the greater the multiplication of the original applied force - the straight line being infinitely an "pointy" wedge and thus producing an infinite sideways force.

You are right - the people telling you something different are wrong. Are they engineers at your factory?

#8 gruntguru

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Posted 15 February 2013 - 04:39

Are they engineers at your factory?

Ah doan wanna ride in no train they built.

#9 MatsNorway

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Posted 15 February 2013 - 10:07

http://dl.dropbox.co...90/1289_001.pdf

http://dl.dropbox.co...90/1290_001.pdf

I thought i would get the same force (Fx) as the angle was the same in total..

We do maintenance not train production, hence the lack of calculation skills.

There is a lot of things wrong in how this firm is organised. Thats my main blame reason.

Edited by MatsNorway, 15 February 2013 - 10:09.


#10 Kelpiecross

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Posted 15 February 2013 - 12:12

[quote name='MatsNorway' date='Feb 15 2013, 21:07' post='6128739']
http://dl.dropbox.co...90/1289_001.pdf

I agree with your answers.

#11 MatsNorway

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Posted 15 February 2013 - 12:24

The angle on both of the wedges is the same but the total is different. What is it i am missing?

(im talkin pdf vs pdf)

#12 Kelpiecross

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Posted 15 February 2013 - 13:03

The angle on both of the wedges is the same but the total is different. What is it i am missing?

(im talkin pdf vs pdf)


I am just looking at the second diagram now - it is essentially the same wedge but the answers are quite different. The first wedge in the vertical position would appear to have a much greater splitting power than the second wedge - even though it is the same shape of wedge. I will have to have a bit of a think about this. I don't know about the relevance to the situation of Fx2 and F3 either.

I suspect it is because in the second case the force F is still vertical - so you are trying to bash the wedge in with it canted over slightly sideways to the left. Imagine tipping it over even more to the left with the same applied vertical force and the wedge would very quickly become impossible to move.

But as always - I could be wrong.

The ultimate situation would become when the wedge would be lying completely on its side with its flat surface parallel to the surface of the wood and no amount of an applied vertical force would have any effect.

Edited by Kelpiecross, 15 February 2013 - 13:08.


#13 GSpeedR

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Posted 15 February 2013 - 14:04

The angle on both of the wedges is the same but the total is different. What is it i am missing?

(im talkin pdf vs pdf)


In the top pdf you've divided the applied force in 2, where you haven't in the second pdf. Also the angles aren't the same (20 deg vs 10 deg) unless I'm misunderstanding your last post). Don't split the wedge in two, just do a single Free Body Diagram.

#14 MatsNorway

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Posted 15 February 2013 - 14:09

I calculated on half of it. half the force goes to one side. Then i added the forces as left/right should be the same.

I need a example on the free body diagram. not sure what it translates to for me.

Edited by MatsNorway, 15 February 2013 - 14:13.


#15 GSpeedR

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Posted 15 February 2013 - 15:04

I calculated on half of it. half the force goes to one side. Then i added the forces as left/right should be the same.

I need a example on the free body diagram. not sure what it translates to for me.


It's basically what you have done in the 2nd pdf just a more rigorous summing of forces in each direction.

One wedge is 20deg total included angle (both sides) and the other is 40deg, so the numbers will not be the same. [Unless I'm crazy and confused about your question. Entirely possible.]

Edited by GSpeedR, 15 February 2013 - 15:06.


#16 MatsNorway

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Posted 15 February 2013 - 15:19

It's basically what you have done in the 2nd pdf just a more rigorous summing of forces in each direction.

One wedge is 20deg total included angle (both sides) and the other is 40deg, so the numbers will not be the same. [Unless I'm crazy and confused about your question. Entirely possible.]


The symmetrical wedge is 20 too hence the usage of 10degrees in the calculations.

#17 manolis

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Posted 15 February 2013 - 15:41

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Manolis Pattakos

#18 MatsNorway

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Posted 16 February 2013 - 18:22

Ok manolis. you got another way that can be considered a way to simplify the model. what does the calculations say.
http://en.wikipedia.org/wiki/Friction

If an object is on a level surface and the force tending to cause it to slide is horizontal, the normal force N\, between the object and the surface is just its weight, which is equal to its mass multiplied by the acceleration due to earth's gravity, g. If the object is on a tilted surface such as an inclined plane, the normal force is less, because less of the force of gravity is perpendicular to the face of the plane. Therefore, the normal force, and ultimately the frictional force, is determined using vector analysis, usually via a free body diagram. Depending on the situation, the calculation of the normal force may include forces other than gravity

I struggle to get this but i got it sendt by one of my collegues.

Edited by MatsNorway, 16 February 2013 - 18:26.


#19 MatsNorway

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Posted 16 February 2013 - 18:43

Found it.
http://en.wikipedia....hanical_device)

But no graphical display of it.

Im assuming my calculations on the symmetrical wedge is correct. But im still confused on the asymmetrical one. But i have an idea.

Edited by MatsNorway, 16 February 2013 - 18:49.


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#20 Greg Locock

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Posted 17 February 2013 - 00:12


An assymetrical wedge, say one with a vertical face and an angled face, without friction is just a right angled triangle of forces. All the forces are perpendicular to the surface they act on (that's what no friction implies). The horizontal force on the vertical side of the wedge = the horizontal force on the sloping side. Vertically the driving force = the vertical component of the force on the sloping side.



#21 manolis

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Posted 17 February 2013 - 04:45

MatsNorway, think it simply.

Posted Image

Given the F1 force (i.e. its amplitude and the point of the wedge wherein it is applied), given that the friction is zero (i.e. all forces are normal to the surfaces they apply), given the point of the wedge wherein the F2 is applied and given that the wedge is not moving (equilibrium), you have a single solution wherein F1/F2=a/b (i.e. F2= F1*b/a) and F3=SQRT(F2^2-F1^2).

Manolis Pattakos

#22 Kelpiecross

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Posted 17 February 2013 - 05:29

After much thought I think there is a fairly simple "real-life" way of explaining why the apparently same-shaped wedge generates different horizontal forces (that is; in the Fx direction at a right angle to the applied force F).

If you attack this problem from a "mechanical advantage" type of view: - in the first case (symmetrical wedge) if you calculate (by trigonometry) how far two points (I like to picture them as being like V8 roller bearing followers) are forced apart in a horizontal direction by 10 (for example) units of movement of the applied force F - and then use this distance to calculate the mechanical advantage - you get an MA of 2.836 - this gives a horizontal wedging force (Fx) of 28.36N if the applied force F is 10N.

In the second (non-symmetrical) case you find that per unit of movement of the applied force F the "followers" are being forced apart in a horizontal direction at a slightly higher rate - thus the distance they have moved after 10 units of movement of the applied force F is slightly more and the MA works out to be slightly less at 2.747 - resulting in a wedging force of 27.47N.

In the second case (non-symmetrical wedge) you find the the horizontal line of action of the "followers" lies across the wedge in a slightly "diagonal" manner an is thus a slightly longer distance compared to a horizontal line across the first (symmetrical) wedge.

Thus the more the wedge leans towards the left the longer the "diagonal" is, the lower the MA and the lower the wedging force is.

Also in the first example Mats calculated - I can't see that there is any Fxtotal of 56.71N that could be measured anywhere in the system - the force is 28.36N in both directions. Even if one "follower" is fixed and the other allowed to move - it is still 28.36 in both directions. Even with the non-symmetrical wedge it is still 27.47N in opposite directions.

Edited by Kelpiecross, 18 February 2013 - 04:46.


#23 Kelpiecross

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Posted 18 February 2013 - 05:16


Or if you are doing this problem by graphical/vector methods - the difference in the Fx wedging forces comes down to the fact that 2 times tan10 doesn't equal tan 20. But to me this doesn't "explain" the difference in real-world terms.


#24 MatsNorway

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Posted 19 February 2013 - 15:11

I think its related to a narrower interaction with the sides. both sides are closer to being alongside the force than the single 20 degree side is on the other.

This is what i consider to be fairly correct setups for the variations.

Upper figure just displays where the forces comes from and goes to.
https://dl.dropbox.c...90/1299_001.pdf

In my mind there is many ways to display it. I like this one to achieve equal amount of forces. if we where nit picking they should attack along the same lines. and the 90 degree bend should be where the upper corner is on the triangle. It doesn`t matter in this example but it could if you had limited contact areas the wedge where transfering the forces into. giving torques and so on.
https://dl.dropbox.c...90/1300_001.pdf

One of my collegues that i got into this understood this over the night and he could even explain me how the torque check.. stuff would work. and so on.

Im happy now. Ive proven a couple of civ engs and engineers wrong. Sometimes you gotta thrust your intuition.

Also it seems that getting the hangups on something isn`t a wise idea. Gregs comment about the missing force on one side did not make sence because i imagined that everything needed to be able to be drawn into triangles or something. Wich is silly..

Does it make sense to have the forces going in a straigh line there? yes.. well draw it straight there then..

Edited by MatsNorway, 19 February 2013 - 15:14.


#25 Kelpiecross

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Posted 19 February 2013 - 22:30

[quote name='MatsNorway' date='Feb 20 2013, 02:11' post='6132915']
I think its related to a narrower interaction with the sides. both sides are closer to being alongside the force than the single 20 degree side is on the other.

Another way to say this is that with one side vertical the wedge is effectively a bit thicker than when the wedge is symmetrical about a vertical line.

#26 Kelpiecross

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Posted 19 February 2013 - 23:08


In the symmetrical wedge case you were able to say the the applied force F would be split equally in opposite directions (5N). In the asymmetrical case you were able to say that the whole of F (10N) was to one side.

If the wedge were positioned 5 degrees to the right and 15 degrees to the left - how would you divide F into forces in opposite directions so you could use your method of working out the answer?


#27 GSpeedR

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Posted 19 February 2013 - 23:29

In the symmetrical wedge case you were able to say the the applied force F would be split equally in opposite directions (5N). In the asymmetrical case you were able to say that the whole of F (10N) was to one side.

If the wedge were positioned 5 degrees to the right and 15 degrees to the left - how would you divide F into forces in opposite directions so you could use your method of working out the answer?


There is a reaction normal to every surface in static equilibrium, so the asymmetrical case doesn't have 'the whole of F to one side'. The forces have to sum to 0 in all directions. Manolis has provided the diagrams (force F3 in both cases).

Edited by GSpeedR, 19 February 2013 - 23:31.


#28 Kelpiecross

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Posted 20 February 2013 - 04:26

There is a reaction normal to every surface in static equilibrium, so the asymmetrical case doesn't have 'the whole of F to one side'. The forces have to sum to 0 in all directions. Manolis has provided the diagrams (force F3 in both cases).


Show a method to work out an answer to the 5/15 degree orientation (and an actual numerical answer for an applied force of 10N).

#29 gruntguru

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Posted 20 February 2013 - 05:51

Show a method to work out an answer to the 5/15 degree orientation (and an actual numerical answer for an applied force of 10N).

The vector triangle has angles of 75, 85 and 20 degrees and the input force is 10N and opposite the 20 degree angle. A bit of trig (A/sin a = B/sin b = C/sin c) gives the other 2 sides (force) magnitudes at 29.12 and 28.24 N. The largest force acts on the longest side.

Note. If all three forces are perpendicular to the surface of application (ie 3 frictionless faces) the force is proportional to the length of the side.

Pressure analogy. Assume the wedge is prismatic. immerse the wedge in a pressurised fluid in a weightless environment. The wedge must be in equilibrium. The force on each face is proportional to the area of that face since P=F/A.

This allows you to solve any frictionless wedge problem the same way if you first angle the small face to be perpendicular to the applied force.

Edited by gruntguru, 20 February 2013 - 22:58.


#30 Kelpiecross

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Posted 20 February 2013 - 07:17

The vector triangle has angles of 75, 85 and 20 degrees and the input force is 10N and opposite the 30 degree angle. A bit of trig (A/sin a = B/sin b = C/sin c) gives the other 2 sides (force) magnitudes at 29.12 and 28.24 N. The largest force acts on the longest side.

Note. If all three forces are perpendicular to the surface of application (ie 3 frictionless faces) the force is proportional to the length of the side.

Pressure analogy. Assume the wedge is prismatic. immerse the wedge in a pressurised fluid in a weightless environment. The wedge must be in equilibrium. The force on each face is proportional to the area of that face since P=F/A.

This allows you to solve any frictionless wedge problem the same way if you first angle the small face to be perpendicular to the applied force.



Without a sketch I am not too sure just what you mean - but these are the answers I get as well. I did my calculation by getting the forces normal to the applied force by the mechanical advantage method (28.14 in each direction) then using vectors etc. to work out the forces normal to the faces.

The point I was trying to make previously was it seems to me that Mat's method only works in special cases - that is; when you know how the applied force is split up - 5N in the first case and 10N in the second case. For "in-between" cases (such as the 5/15 example) you wouldn't know the applied force is split up.

#31 MatsNorway

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Posted 20 February 2013 - 07:29

Another way to say this is that with one side vertical the wedge is effectively a bit thicker than when the wedge is symmetrical about a vertical line.


Perhaps.. i feel you are kinda giving the wedge different caracteristics, its just the same wedge with a job that can`t be done as efficient due to the direction of the force in relation to the angle of the working side/sides.

I feel that ratio is inferior to angle when your working out the loads it can give. Or at minimum it gives you something extra to be vary of.

http://dl.dropbox.co...3890/Wedges.pdf

#32 Kelpiecross

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Posted 20 February 2013 - 12:08


I think everybody sees their own particular method of calculating the forces as "intuitive". I still prefer my method as it just suits me - the various steps in the calculation seem logical and I can picture in real-life just what seems to happening. I am inclined to make mistakes in the arithmetic and methods like the one I favour tend to show up mistakes as the calculation is being done. I try and avoid calculations which just involve "plugging" numbers into a formula without really knowing the logic behind the formula - you then have little idea if the answer seems reasonable or not. And the mechanical advantage method I use can be applied to any situation - as does GG's method.

I presume GG's method appears to be intuitive to him - it certainly doesn't suit my way of doing things. (The pressure analogy is very interesting though).

Nice to see that Mats is making good use of his company's drawing facilities.



#33 gruntguru

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Posted 20 February 2013 - 12:09

Without a sketch I am not too sure just what you mean - but these are the answers I get as well.

Look at Manolis Post#21. The right hand diagrams show the vector triangle(s). For a wedge with three frictionless faces, each force is normal to the face and the vector triangle has the same angles (and therefore the same ratio of forces) as the wedge itself.

#34 Kelpiecross

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Posted 20 February 2013 - 12:29

Look at Manolis Post#21. The right hand diagrams show the vector triangle(s). For a wedge with three frictionless faces, each force is normal to the face and the vector triangle has the same angles (and therefore the same ratio of forces) as the wedge itself.


Where are the 85, 75 and 20 degree angles? Opposite what 30 degree angle?

#35 manolis

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Posted 20 February 2013 - 15:31

Where are the 85, 75 and 20 degree angles? Opposite what 30 degree angle?


I hope this helps:

Posted Image

By mistake Gruntguru wrote 30, instead of the correct 20 degrees.

Manolis Pattakos

#36 MatsNorway

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Posted 20 February 2013 - 17:45

I think everybody sees their own particular method of calculating the forces as "intuitive". I still prefer my method as it just suits me - the various steps in the calculation seem logical and I can picture in real-life just what seems to happening. I am inclined to make mistakes in the arithmetic and methods like the one I favour tend to show up mistakes as the calculation is being done. I try and avoid calculations which just involve "plugging" numbers into a formula without really knowing the logic behind the formula - you then have little idea if the answer seems reasonable or not. And the mechanical advantage method I use can be applied to any situation - as does GG's method.

I presume GG's method appears to be intuitive to him - it certainly doesn't suit my way of doing things. (The pressure analogy is very interesting though).


Im not saying do it like person X. Im saying there is weaknesses to using ratio only. Be aware of them and if not sure double check with the other method.

Nice to see that Mats is making good use of his company's drawing facilities.


haha. I think they don`t mind me using some time to freshen up my knowledge. if i had more knowledge around calculations i could document my designs better and sell alot of them outside the workshop. Outsiders have raised wishes to buy some of them. Friction and angles has indeed been a challenge in the past.

The discussion at work today as one of the guys came back has now progressed towards friction and friction angle.

Manolis what program do you use to make the figures?

Edited by MatsNorway, 20 February 2013 - 17:46.


#37 gruntguru

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Posted 20 February 2013 - 23:00

Where are the 85, 75 and 20 degree angles? Opposite what 30 degree angle?

Whoops typo there is no 30 deg - should be 20. Post fixed.

#38 Kelpiecross

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Posted 21 February 2013 - 03:39


Mats - the "foreign order" is a fine old tradition in Oz and the UK (and apparently Norway). After all - that is half the reason one works for a company isn't it?

Manny (and GG) - thank you for the diagrams. I did have a vague idea that this was what the GG was getting at (I did wonder if the 30 degrees was a typo). If I used a method like this (which to me personally is non-intuitive/non-logical etc.) I would very quickly go wrong without a lot of practice at this type of calculation.

#39 Greg Locock

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Posted 21 February 2013 - 04:12

Free body diagrams are, in my experience, badly taught, badly learned and badly deployed. Certainly I managed to get into uni to study engineering without ever having specifically constructed a complex 2D one, never mind 3D. We sort of vaguely drifted past the topic at school. To some extent that doesn't matter, if you can do vector maths, or have that sort of brain, but I would say that the ability to punch out good FBDs is pretty essential in many engineering fields, and high school science teachers aren't quite equipped for it.

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#40 Kelpiecross

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Posted 21 February 2013 - 04:53

Free body diagrams are, in my experience, badly taught, badly learned and badly deployed. Certainly I managed to get into uni to study engineering without ever having specifically constructed a complex 2D one, never mind 3D. We sort of vaguely drifted past the topic at school. To some extent that doesn't matter, if you can do vector maths, or have that sort of brain, but I would say that the ability to punch out good FBDs is pretty essential in many engineering fields, and high school science teachers aren't quite equipped for it.


Oddly enough I have a similar viewpoint. I liked physics in high school (nerd) and just continued to use these methods (in a self-modified sort of way) in first year physics and engineering at uni. The engineering methods especially seemed vaguely incomprehensible to me - I was able to work out the problems using the physics methods I was familiar with - Christ knows what the lecturers marking test papers thought - but I did quite well.

As an aside here - seeing there are so many engineers on this forum - I found Engineering 1 very disappointing. I had imagined that Eng 1 would involve at least some fiddling with nuts-and-bolts, engines, cutting things up with hacksaws etc. - but there was absolutely none of this. In my course I did Eng 1 as a "terminating" subject for non-engineering students - maybe the "real" engineering students did more interesting things.

#41 gruntguru

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Posted 21 February 2013 - 06:45

I had imagined that Eng 1 would involve at least some fiddling with nuts-and-bolts, engines, cutting things up with hacksaws etc. - but there was absolutely none of this.

Sounds like you should have studied Mech Eng. (I seem to remember you saying you were a Civil)

#42 MatsNorway

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Posted 21 February 2013 - 07:28

Mats - the "foreign order" is a fine old tradition in Oz and the UK (and apparently Norway). After all - that is half the reason one works for a company isn't it?


What?

#43 Kelpiecross

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Posted 21 February 2013 - 12:48

Sounds like you should have studied Mech Eng. (I seem to remember you saying you were a Civil)


No - science degree.

#44 Kelpiecross

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Posted 21 February 2013 - 12:50

What?


You mean - what is a "foreign order"? - doing your own work on the company's time.

#45 Tony Matthews

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Posted 21 February 2013 - 12:50

What?

I'm with Mats - what?


Ah, thanks!

Edited by Tony Matthews, 21 February 2013 - 12:50.


#46 MatsNorway

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Posted 21 February 2013 - 16:06

You mean - what is a "foreign order"? - doing your own work on the company's time.


It started as work related. And i guess i wasted a lot of company time on it. BUT its for the god of the company.

#47 Kelpiecross

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Posted 22 February 2013 - 05:35

It started as work related. And i guess i wasted a lot of company time on it. BUT its for the god of the company.


I should point out that I was not criticising you - I may not have been the world champion of "foreign orders" - but I was probably a contender for the title.

I worked in a research establishment in the mid-sixties (as a lowly trainee) where it seemed to me that 90% of the work done in the place was "foreign orders"

Doing it for the "god" of the company is a bit crawly-bumlick isn't it? (Attempted joke).

#48 manolis

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Posted 22 February 2013 - 08:33

Manolis what program do you use to make the figures?


AutoCAD (it is like using a Ferrari for delivering pizza downtown!).
These figures can be made using any design program, even the Microsoft Paint.

For the wedge forces:
The calculation of the forces in a body is not a difficult issue.
In case there is interest, we can analyze it: what are the laws, how they are applied, case of immovable body, case of moving body etc.

Thanks
Manolis Pattakos

#49 Rasputin

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Posted 31 March 2013 - 06:16

Free body diagrams are, in my experience, badly taught, badly learned and badly deployed. Certainly I managed to get into uni to study engineering without ever having specifically constructed a complex 2D one, never mind 3D. We sort of vaguely drifted past the topic at school. To some extent that doesn't matter, if you can do vector maths, or have that sort of brain, but I would say that the ability to punch out good FBDs is pretty essential in many engineering fields, and high school science teachers aren't quite equipped for it.


The ability to mouth your way around displaying your true engineering skills is quite obviously very thoroughly taught.

A question to Mats; What is the point of this exercise if you exclude the friction forces, which are at the focus of most wedge applications?

#50 MatsNorway

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Posted 01 April 2013 - 17:54

It started of as a discussion about the distribution of forces on a point in a FEA analysis and after some initial warming up (uuuu umm with confused discussions period) I realised how it was but to my surprice no one else where agreeing with me initially.

We eliminiated the friction to simplify the discussion topic.

So now friction angle comes into play. I read somewhere that the friction coefficient was to be tangens with a funny symbol.. if its inverse tangens i don`t know.. When i think about it inverse converts the number back to the angle right? sounds right.. and sounds simple enough.
http://dl.dropbox.co...90/1363_001.pdf

Figure shows With out friction as the main triangle. And as friction increases the Fn decreases in size to the point where there is no side forces.

Edited by MatsNorway, 01 April 2013 - 17:56.