
#1
Feb1813, 03:19 AM

P: 27

Hi
I have the last days had lots of discussion about wedges on work. All of my collegues claim that the higest force your able to get out of a wedge is the one you put in on the top. While i claim that the wedge can push more to the side the thinner it is. To simplify matters we say no friction on any surface. My issue is that i can`t get a good source on how to do the vectors on the wedge. Initial thread on my regular forum. http://forums.autosport.com/index.php?showtopic=181135 My only decent source. http://en.wikipedia.org/wiki/Wedge_%...ical_device%29 



#2
Feb1813, 03:33 AM

P: 27

http://hyperphysics.phyastr.gsu.edu...s/incline.html
So the hypotenus of the forces goes 90degrees from the long side(s)? Marked L called incline. My english is not perfect. thank you. 



#3
Feb1813, 05:34 AM

P: 27

Does it even matter if the wedge is moving?
The Force in x direction will allways be bigger surely.. Unless the angle is greater than 45 degrees or there is friction in play. I found a ton of articles on it. http://hyperphysics.phyastr.gsu.edu...s/incline.html http://www.ehow.com/how_6544271_calc...agewedge.html http://www.accessscience.com/abstrac...%3fid%3d743200 http://www.accessscience.com/loadBin...3200FG0010.gif http://www.crsep.org/PerplexingPairs...rt5.Aprl01.pdf “A knife cutting butter functions in the same way. You push downward on the top of the butter with a knife. The butter is not crushed under the edge of the knife, it is pushed apart into two pieces as the knife moves through it.” http://weirdrichard.com/wedge.htm Every article talks about ratios.. I think there should be some about the graphical way of doing it. Its not like all wedges are perfectly straight. or a simple wedge shape. 



#4
Feb1813, 09:07 AM

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PF Gold
P: 4,453

Wedges and their abilities
Assume that the wedge is symmetric about its center plane and the included angle of the wedge is 2θ. Assume that the wedge has penetrated an arbitrary distance into the work, and that the work is in contact with the wedge over the length of penetration. Do a force balance on the wedge. Since you are assuming no friction, the contact forces of the work on the wedge are perpendicular to the wedge surface. Resolve these forces into components in the vertical and horizontal directions. Set the sum of these components equal to the applied vertical force. See what you get.




#5
Feb1813, 09:26 AM

P: 27

Symmetrical wedge. The 2θ i don`t get/understand what is.
The way i read you makes me think its this one. (its a wedge with a stripe in the middle) i calculated only one side. with half the total angle of the wedge. 20degrees total. http://dl.dropbox.com/u/72823890/1289_001.pdf 



#6
Feb1813, 10:22 AM

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P: 4,453

Yes. That's what I had in mind.
Chet 



#7
Feb1813, 11:57 AM

P: 27

You claim my setup on that wedge is correct? everything?
Is there any situasjon the loads would split in a different manner? movement etc? 



#8
Feb1813, 12:51 PM

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MatsNorway: Your solution in the attached file in post 5 currently appears correct.
The only minor "mistake" is, you should reverse the direction of the arrowheads on your F2, F3, Fx1, and Fx2 force vectors. Why? Because your free body currently appears to be the wedge. 



#9
Feb1813, 01:28 PM

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P: 4,453





#10
Feb1813, 02:58 PM

P: 27

Counter forces should be separate i think. But thank you 



#11
Feb1813, 06:39 PM

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MatsNorway: You chose the wedge to be your free body, in your freebody diagram. In a freebody diagram (FBD), one shows the forces acting on the free body. When you do this, if the object is in equilibrium, then the summation of horizontal forces will equal zero, the summation of vertical forces will equal zero, and the summation of moment will equal zero.




#12
Feb1913, 02:53 AM

P: 27

One of my collegues just came to me and said he had been thinking about it. And he agrees now so im very happy. He also displayed how you go about getting the moment to zero. Pick a point and work around that basically. Im going to throw this in everyones faces now.
Thank you for you help everyone. I think these two are close to how they should be. Main input force and the forces needed to keep it in steady state. http://dl.dropbox.com/u/72823890/1299_001.pdf http://dl.dropbox.com/u/72823890/1300_001.pdf 



#13
Feb1913, 09:27 AM

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MatsNorway: Nice work. Your solutions in both attached files in post 12 currently appear correct. I usually like to also write out the fundamental equilibrium equations. Below is the summation of vertical (V) forces, and the summation of horizontal (H) forces.
∑ V = 0 = F1 + F2. Therefore, F2 = F1. But F2 = Fh*sin(θ). Therefore, Fh = F1/sin(θ). ∑ H = 0 = Fxv  Fxh. Therefore, Fxv = Fxh. But Fxh = Fh*cos(θ). Therefore, Fxh = Fh*cos(θ) = [F1/sin(θ)]*cos(θ) = F1/tan(θ) = (10 N)/tan(20 deg) = 27.5 N. 



#14
Feb2013, 01:39 AM

P: 27

Hehe. My collegue did something similar. I can`t really follow you. Whats behind the cosinus and tangens stuff i don`t know. i just use them. I have to look in books to know what goes where.
The discusion about wedges is still ongoing at the motorsport forum. http://forums.autosport.com/index.ph...&#entry6133775 



#15
Mar2713, 04:55 AM

P: 27

http://dl.dropbox.com/u/72823890/1363_001.pdf
So now im curious about friction. I have reasoned that the friction angle is a handy thing. I think it should be doable to subtract its angle from one of the sides of the triangle of forces. As suggested in the pdf above. the dotted lines are the resultant force with the friction added. the bigger the friction the smaller the force to the side. And if the friction is small enough. you get no side forces. 


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