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Friberg
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Prisms (some math)   prisma.png
« on: Aug 29th, 2003, 1:27pm »
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The shaded area is the prism (the picture shows a cut through it). The crane has two "arms" which shall hold these prisms by pressing them from each side. These forces are the red ones in the sketch and they work right-angled to the side-faces of the prism(s). Task 1:
Depending on the static friction value , the prisms the crane is able to carry are limited to those which have certain angles . Get an equation which shows the relation between and the possible 's; solved to . Something like this: ">2".
 
Task two:
Which force must the crane excert on a certain prism to be able to carry it? Derive a nice formula which contains all the variables la "F=m+g̦-45"
 
Check the attached image for the picture of the prisms..  
When you're done, send me a pm or something Smiley Thanks...
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Re: Prisms (some math)  
« Reply #1 on: Aug 30th, 2003, 6:18am »
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"" ,  ""  , ">2" and " la "F=m+g̦-45"
all look like giberish over here..
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Re: Prisms (some math)  
« Reply #2 on: Aug 30th, 2003, 3:34pm »
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Oh im sorry, it seems this forum cannot use those.. but ill make a file so you can see what i see.. Smiley
 
heres the file: http://thgf.mine.nu:82/task.jpg
 
Now it might be abit easier to solve =))))))))
Bye..
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Re: Prisms (some math)  
« Reply #3 on: Aug 31st, 2003, 12:30am »
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as of 12:30 AM 8/31/2003 PST, the image link http://thgf.mine.nu:82/task.jpg doesn't work; if you have it on your computer you could upload it to this server using the file attachment feature
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Friberg
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Re: Prisms (some math)   task.jpg
« Reply #4 on: Aug 31st, 2003, 1:49am »
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on Aug 31st, 2003, 12:30am, william wu wrote:
as of 12:30 AM 8/31/2003 PST, the image link http://thgf.mine.nu:82/task.jpg doesn't work; if you have it on your computer you could upload it to this server using the file attachment feature

 
Im so sorry... i had to change my port on the server... anyhow ill attach it to this one Smiley Else you can find it on : http://thgf.mine.nu:1337/task.jpg
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snifit
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Re: Prisms (some math)  
« Reply #5 on: Sep 2nd, 2003, 2:30am »
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Err...I worked on this for a while yesterday, but since then my scratch sheet has been thrown out...heh. Anyway, this is my answer for Task One (assuming that I haven't made a gross error somewhere, which is probably a big assumption):
 
F'sin(90 - [alpha]) + Fsin(90 - [alpha]') >= mg[mu]
 
at which point I got stuck trying to isolate the [alpha]...
 
Can it be assumed that the applied forces are equal? This seems reasonable to me.
Furthermore, can it be assumed that [alpha] and [alpha]' are equal? The only way they would be equal is if the prism formed an isosceles triangle, which I don't think can be assumed given the diagram.
 
Note: I derived the above formula using an equation for kinetic friction rather than static friction. Since the coefficient for kinetic friction is always lower than that of static friction, I didn't think it would cause any problems, other than not giving me an optimal solution.
 
Edit: It just occured to me that the equation for static friction would be the same as the equation for kinetic friction, but with a different coefficient.
 
Further Edit: Gravity isn't the cause of the friction, so the above equation's wrong. I guess I wasn't thinking very clearly when I first started working on this.  Smiley
« Last Edit: Sep 4th, 2003, 8:17pm by snifit » IP Logged
James Fingas
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Re: Prisms (some math)  
« Reply #6 on: Sep 2nd, 2003, 1:37pm »
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I think it will become clear that finding the condition on [mu] is independent of the mass of the object. It also seems to me that you'll get a condition on the locations of the forces on the sides of the prism. Just to guess at the answer for the first part, you can pick up the prism if:
 
[mu] > tan(([alpha] + [alpha]')/2)
 
For the second part, the minimum force will be related to the balancing of the prism and the locations of the forces. If the perpendicular bisector of the prism's base is vertical, as drawn, then the forces will be slightly lower than if the prism tilts a little so that the center of gravity is more directly between the crane arms. The most extreme case of this is when the prism "vertical" is pointing horizontally (or higher than horizontal) so that the prism is balancing exactly on one of the crane arms. Then one force is mg and the other is zero.  
 
If the forces are not in the right locations, you'll get rotational forces on the prism, making it tilt to one side, increasing the friction force on one side and decreasing it on the other. This will make the prism harder to hold.
 
Your drawing and description are not completely accurate. When you say the forces are perpendicular, that can't be completely true, because F and F' pointing downwards plus mg pointing downwards adds to a net downwards force. With friction, the forces will not point downwards any more. In fact, a good way of solving the problem might be to show that friction gives the cotangent of the minimum angle that the force can make with the surface. If the force can be slightly pointing upwards (allowing it to counteract gravity), then the prism can be held.
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Re: Prisms (some math)  
« Reply #7 on: Sep 3rd, 2003, 10:51am »
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on Sep 2nd, 2003, 1:37pm, James Fingas wrote:
I think it will become clear that finding the condition on [mu] is independent of the mass of the object. .

 
I think this is only true assuming that the 'vertical' of the prism is indeed vertical - not slanted in anyway. If the 'vertical' were anything but, then the mass of one side of the prism would contribute to friction, while the other side's mass would not. (I hadn't thought of this until you brought it up).
 
on Sep 2nd, 2003, 1:37pm, James Fingas wrote:
For the second part, the minimum force will be related to the balancing of the prism and the locations of the forces. If the perpendicular bisector of the prism's base is vertical, as drawn, then the forces will be slightly lower than if the prism tilts a little so that the center of gravity is more directly between the crane arms.

 
I'm not sure what you mean here. If the prism was tilted a little, then gravity would add to the force of friction, lowering the force needed to lift the prism, not raising it.
 
on Sep 2nd, 2003, 1:37pm, James Fingas wrote:

Your drawing and description are not completely accurate. When you say the forces are perpendicular, that can't be completely true, because F and F' pointing downwards plus mg pointing downwards adds to a net downwards force. With friction, the forces will not point downwards any more.

 
Again, i'm not sure what you mean here. Yes, we get a net downward force, but friction acts to cancel out that force so that we can lift the prism.
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Re: Prisms (some math)  
« Reply #8 on: Sep 3rd, 2003, 9:05pm »
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The problem statement here is unclear. There is no mention of gravity, but it seems to be intended. It is possible for the red forces to act normal to the faces: only if they pull the wedges apart with gravity downward, or squeeze them together if gravity acts in an upward direction. Only the second possibility could be stable.
 
Are the forces really intended to be normal to the faces, or should it be assumed there is friction between the crane grips and the prisms? Is that friction coefficient, [mu], for between the crane and the blocks or between the two prisms, or should slip at all three contacting locations be considered? Is gravity assumed to act downward? ...
 
I have a feeling the question is meant to be, "A crane picks up the pair of contacting prisms of the same density by applying forces to the two faces as shown in the figure. The prisms are held in a position with the bottom edge horizontal. Find conditions for friction coefficient (which has the same value at all contact points) such that no slip occurs between the prisms or between the crane and the prisms." Figure should show an Mg arrow acting downward.  Or maybe the intent is that the crane has a firm grip on the wedges and cannot slip (so the forces do not act normal to the faces).
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Re: Prisms (some math)  
« Reply #9 on: Sep 4th, 2003, 9:49am »
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I. a=a'
II. solve Part one to a
III. The prism is hold absolutely symetrically. So you can (and have to Wink ) derive an equation for the force which is the minimum that is needed to hold a certain prism.
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Re: Prisms (some math)  
« Reply #10 on: Sep 4th, 2003, 5:35pm »
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That does not help clarify much.  If [alpha]=[alpha]', by symmetry the friction load at the interface must be zero. If the red forces act normal to the faces, the friction load there is zero too. This makes [mu] irrelevant in the problem.
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Re: Prisms (some math)  
« Reply #11 on: Sep 4th, 2003, 8:13pm »
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I don't think so. The horizontal components of F and F' (which we are assuming to be equal?) will add to give a force of friction at the interface of the two prisms where
 
Ffriction = 2Fhorizontal[mu]
 
In order for the crane to be able to lift the prisms,  
 
2Fhorizontal[mu] >= 2Fvertical - mg where Fvertical is the downwards force exerted by one of the crane arms.
 
I'm pretty sure I'm right so far, but from here I can't see how to get an answer for the first part of the question - that pesky mg.
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Re: Prisms (some math)  
« Reply #12 on: Sep 5th, 2003, 7:48pm »
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That is incorrect, Snifit. If the friction force is upward on the right prism it must act downward on the left prism. If downward on right prism it must act upward on left prism. Either way is not symmetrical so friction force must be zero.
 
If one object is in the process of sliding over another then the friction force is [mu]k*Fnormal. The prisms are not in the process of sliding here.
 
If objects are in static contact, the maximum friction force that may be supported before sliding begins is [mu]s*Fnormal. While in static contact, the force of friction may be anything from zero up to this maximum.
 
By your logic, if you take two blocks of wood and tightly wrap a wire around them to squeeze together and cause a high contact force, the wire could be wrapped so tighlty that the friction between the blocks would lift them off the ground and cause them to levitate.  Shocked
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Re: Prisms (some math)  
« Reply #13 on: Sep 5th, 2003, 9:18pm »
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Anti-gravity! You've figured out the trick!!.  Shocked
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Re: Prisms (some math)  
« Reply #14 on: Sep 5th, 2003, 10:34pm »
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You're right - there shouldn't have been an inequality. I do know that friction only opposes motion, I just didn't think the inequality would allow friction to cause motion (I meant to allow for the maximum static friction force to be higher than the downwards force - whoops).
 
And you're right again about the zero-friction interface. Trust logic to come in there and make me a fool. Thanks for setting me straight. Smiley
 
So is this solvable with the current information?
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Re: Prisms (some math)  
« Reply #15 on: Sep 15th, 2003, 8:58am »
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I still havn't figured it out Sad
Need help Smiley Thanks...
 
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Re: Prisms (some math)  
« Reply #16 on: Sep 15th, 2003, 9:52am »
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supposing the crane wants to keep the prism in one place, then we don't need to equate [alpha]1 and [alpha]2, since the forces in the horizontal direction need to cancel out.
And of course the forces in vertical direction need to be equal as well.
« Last Edit: Sep 15th, 2003, 9:55am by towr » IP Logged

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Re: Prisms (some math)  
« Reply #17 on: Sep 15th, 2003, 10:27am »
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The problem is not solvable given only the information shown above. We also need to know the orientation of the prism (is the base actually horizontal?) and the location of the crane arms on the prism sides (this could have a very large effect on allowable values of [mu]).
 
Without this information, we're just making assumptions. One set of assumptions we could make is that the prism will slip so that the base is not necessarily horizontal, and so the crane arms are directly opposed to one another, so that the necessary [mu] is minimized. This leads to the solution I gave earlier.
 
Besides this, we're not exactly clear on whether there is one prism or two, or where the friction force applies.
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Re: Prisms (some math)  
« Reply #18 on: Sep 15th, 2003, 10:41am »
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Well, I think it's clear enough there is only one prism.
And assuming a steady state, the forces can only be perpendicular to the sides of the prism if the forces are equal, and the angles [alpha] are the same (easy to check).
Titlting the prism would change the (direction of the) crane-forces.
After that it's just a matter of observing that the friction force need to counteract both the downward forces of the crane, and gravity.
 
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Re: Prisms (some math)  
« Reply #19 on: Sep 15th, 2003, 11:01am »
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The forces can never be perpendicular to the prism sides, because then there would be three forces pushing the prism down, and no forces pushing it up.
 
The force a crane arm applies to the prism is the sum of the normal force (perpendicular by definition) and friction. If this sum were normal to the surface, there would be no friction force, and the prism would fall.
 
Assuming you knew where the crane arms were, and what direction the base of the prism was pointing, you would solve the problem by simultaneously solving the equations for vertical, horizontal, and rotational force acting on the prism. There are four unknowns: two normal forces and two frictional forces. Even after solving these equations, however, there is an additional degree of freedom (corresponding to the force the crane arms apply in excess of what is needed to hold up the prism, in the direction pointing from one crane arm interface point to the other). Varying the magnitude of this force will let you uniquely determine F and [mu]eff = frictional force/normal force. Graph these to determine the minimal [mu] necessary and the minimal F necessary.  
 
Without knowing the points at which the crane arms touch the prism, we can't do the rotational balance of forces. Without knowing the orientation of the prism, we can't calculate the gravity force's location. There are currently too many unknowns.
« Last Edit: Sep 15th, 2003, 11:11am by James Fingas » IP Logged

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Re: Prisms (some math)  
« Reply #20 on: Sep 15th, 2003, 11:17am »
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on Sep 15th, 2003, 11:01am, James Fingas wrote:
The forces can never be perpendicular to the prism sides, because then there would be three forces pushing the prism down, and no forces pushing it up.
 
The force a crane arm applies to the prism is the sum of the normal force (perpendicular by definition) and friction. If this sum were normal to the surface, there would be no friction force, and the prism would fall.
I think that's only the case if you take into account all forces at the point, i.e. including gravity, which isn't the case here.
If you stand on the ground, there is only a normal force, but there is also friction. Which gets noticed when someone tries to drag you backwards (a force different from gravity).
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Re: Prisms (some math)  
« Reply #21 on: Sep 15th, 2003, 11:53am »
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on Sep 15th, 2003, 11:17am, towr wrote:

I think that's only the case if you take into account all forces at the point, i.e. including gravity, which isn't the case here.
 
 
Not sure what you mean. We're supposed to take gravity into account. It acts on the center of mass of the prism. Gravity must be counteracted by friction, or the prism will fall. If we ignore friction, the force of the crane arms on the prism is perpendicular--by definition. But this adds nothing to our understanding of the problem.
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Re: Prisms (some math)  
« Reply #22 on: Sep 15th, 2003, 12:22pm »
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The friction is a reaction-force in respose to a force in the opposite direction, but that needn't be a force cause by the crane.
Regardless perpendicular forces would still cause a downward force in the prism. And that force would have a component in the direction of friction, and thus result in a reactionary friction-force..
I think.. (therefore I am ((,) right (?)))
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Re: Prisms (some math)   prism.gif
« Reply #23 on: Sep 15th, 2003, 1:55pm »
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on Sep 15th, 2003, 12:22pm, towr wrote:
The friction is a reaction-force in respose to a force in the opposite direction, but that needn't be a force cause by the crane.

 
The friction force acts on both the crane and on the prism. I am attaching a diagram that I hope clears up what I am talking about. There are exactly three forces acting on the prism: gravity, F, and F'. F and F' can be split up into two components each. All we have to do to solve the problem is balance the forces for rotation and translation of the prism. However, we are missing information (i.e. the locations of the forces).
« Last Edit: Sep 15th, 2003, 1:55pm by James Fingas » IP Logged


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Re: Prisms (some math)  
« Reply #24 on: Sep 15th, 2003, 2:26pm »
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I see it a bit differently, but it comes down to the same thing.
Where you have 'normal' I just have the F, and where you have F I have an Fres.
(And as a result I'd draw them originating at different places, though that doesn't make any real difference either)
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