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#1
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Re: Help on calc 3 problem would be incredibly appreciated...
Hint 1: This is a one dimentional problem.
Hint 2: Normalize the three vectors. Hint 3: Ignore the x and y components Hint 4: Think Ratio |
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#2
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Re: Help on calc 3 problem would be incredibly appreciated...
Joe,
Do you know what form the expected answer should take? If it's simply looking for the ratio of the magnitude of OA to the projection, then that's pretty easy and Jack's advice applies. If they're expecting you to calculate an actual load on OA, then that's a little trickier. In that case it turns into a system of 3 equations. On that track, the idea is that the loads on OA, OB, and OC add up to your load, but in the opposite direction. So FOA + FOB + FOC = (0,0,1000). The trick there is, again, to nomalize (turn into a unit vector) OA, OB, OC. Then you can multiply those unit vectors by a magnitude and end up with a system of 3 equations. |
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#3
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Re: Help on calc 3 problem would be incredibly appreciated...
Quote:
The exact question from the book is in quotes in the original post... admittingly it is pretty inspecific. |
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#4
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Re: Help on calc 3 problem would be incredibly appreciated...
In that case, I suspect you're looking at my second paragraph of hints, then.
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#5
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Re: Help on calc 3 problem would be incredibly appreciated...
Well... here's what I got so far:
I pretty much focused on OA... I did normalize the vector equation to get [2/sqrt(21)] i + [1/sqrt(21)] j + [4/sqrt(21] k *This is after simplification, the magnitude of OA is 5sqrt(21)* Realizing this will create ratios of force, and that the magnitude of the normal equation is 1, I then put this through the projection formula: projv u = (u.v/||v||^2)v Letting u = OA and v = the 1000kg vector... the x and y components logically no longer matter and its projection comes out to be (4sqrt(21))/21 k Then finding the magnitude of that projection resulted in: 16/21 And that is how the cookie crumbled for me... does that seem right??? ![]() I do like to learn different routes of logic, and vector math in calc 3 has definitely introduced me to that... so I really do appreciate the hints and everything. Helps me out a lot. ![]() _joe Update: thinking about it... the 16/21 had to be its portion of the load... so the actual load bearing capacity i came up with was ~762... but i still dont know if its right... Last edited by JoeXIII'007 : 06-09-2007 at 09:18. Reason: putting down thoughts |
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#6
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Re: Help on calc 3 problem would be incredibly appreciated...
Joe,
Why don't you sanity check the answer you got there? Apply the same procedure to OB and OC and see if your answers add up to 1000. If they don't... well then you're barking up the wrong tree. |
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