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Mighty Hiker
climber
Vancouver, B.C.
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My orange fur ball mostly uses vegetation to sharpen her claws. Sometimes she eats the grassy variety, and then horks it up, accompanied by distressing sound effects. When you're a ferocious carnivore, the only thing that matters is prey.
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sac
Trad climber
Sun Coast B.C.
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a friend and I did some trundling today...
aaand ripped some vegetation from the granite...
Then, at the belay, we tiptoed around the arbutus,
trying not to disturb it,it's bark, or the "vancouver ground cone"(it's parasitic dontcha know) growing beneath.
We gently threaded our rope through, trying not to break, even the dead, branches. And then rapped, and did more ripping of veg from the crack
... go figure...
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Mighty Hiker
climber
Vancouver, B.C.
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Of course, a bazillion Hartounis doesn't even equal one Stannard, when it comes to energy output.
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Mighty Hiker
climber
Vancouver, B.C.
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Well, I'll probably see both of them at the FaceLift. The one about being caught between two irresistible forces comes to mind.
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Banquo
climber
Morgan Hill, CA (Mo' Hill)
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Ed Hartouni-
Better check your work.
Statics is a class engineers take after a first course in Newtonian mechanics. Engineering being applied science, the course teaches how to apply the equations learned in physics. As implied by the name of the course, the objects analyzed are static meaning that there is no change in velocity which is to say that all of the acceleration terms are equal to zero. There is a companion course called Dynamics which covers accelerating objects.
http://en.wikipedia.org/wiki/Statics
In physics F=ma but in statics, a=0 and all the forces acting on an object must result in a net zero force. The object is said to be in static equilibrium as opposed to dynamic equilibrium. The same is true for rotation, all of the torques or moments on the body must add up to zero. In a three dimensional space there are six independent equations that can be solved for 6 unknowns. In a two dimensional case like ours there are only 3 equations:
Sum of the forces x equal zero ΣFx=0
Sum of the forces y equal zero ΣFy=0
Sum of the moments z equal zero ΣMz=0
In my picture (reposted below), I drew a diagram showing all of the forces acting on the object (called a free body diagram). There are 3 unknown forces, P and the reaction force represented by Rx and Ry. I have three equations and can solve for all three but only solved for P. I used ΣMz=0 taken about the reaction force to get an equation in one unknown. There are only two terms in the equation, the overturning moment Ph and the resisting moment Wb/2.
This can also be easily solved using vector addition since the three vectors must add up to zero.
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Ed Hartouni
Trad climber
Livermore, CA
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I believe the line for your diagram should run through the center of gravity connecting one corner to the next, not to the half-way point of the upper surface...
when I push on that upper left corner, I push on the "lever" that runs from that corner to the contact point, the lower right corner.
The center of gravity sits on this line, so I can just concentrate all of the mass at that point. In my diagram, I resolved the vertical force of the mass of the block in the Earth's gravity, into two forces, one parallel to the line, and one perpendicular to the line.
Being smart,I would push along the perpendicular, which minimizes the force "wasted" either pushing into the contact corner, or in trying to lift the block... my first push, to rotate the block, must equal the force perpendicular to that line. The applied force gets less and less from there, until I get the center-of-gravity over the contact point, then a very small force tips the block.
I don't understand why you have drawn the line that you draw... note that all my forces are balanced, as are all my torques...
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Banquo
climber
Morgan Hill, CA (Mo' Hill)
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Ed-
The half way point at the top is the point of concurrency for all three forces. If there are only three forces acting on a static body, all three forces have to be concurrent or the sum of moments (torque) cannot be zero and angular acceleration results. If the weight and the reaction force are concurrent at the CG and the reaction force acts at the corner, the three forces have to be zero to be static. Calcs below.
Solving for the perpendicular force will find a smaller magnitude but since the block is so tall and skinny, the difference is very small. I didn't do this initially since it made the solution more complex and results in an insignificant difference. Calcs below.
The lever used in calculating torque or moment is measured perpendicular to the force. Pushing on the end of a wrench directly towards the bolt will not turn the bolt even though the force is applied the length of the wrench away from the bolt.
The only assumption I am making is that the acceleration is small enough so that the inertial force is negligible.
There is an exception to three forces acting on a static body must be concurrent, they can be parallel.
I've been teaching this stuff at SJSU since 1984.
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Ed Hartouni
Trad climber
Livermore, CA
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there are lots of ways to skin a cat....
in my way, I forgot that the line connecting the corners is a lever, and that the weight is at the mid point of that lever, giving me a 2:1 advantage...
so the force to move it is half the perpendicular force... that agrees with your calculation using torques, etc...
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Mighty Hiker
climber
Vancouver, B.C.
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It's not true that I weigh two tonnes, that I'm unbalanced, or that I can be easily tipped. I resemble all such remarks, not to mention that one about skinning cats. The very idea!
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Karl Baba
Trad climber
Yosemite, Ca
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I did some trundinttg last month and it only took one red alien to move a ton block
I think it would have been better not to be right under the block when it falls
peace
karl
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