Focus?
Does anyone know how to calculate the center of gravity of this figure?
Do you generally like comic characters?
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ByLucia598
My task is: Investigate the following question: After what time would the ICE reach a speed of 300km/h using the distance-time law s(t) = 0.7t^2? The thing is, though, I haven't specified either acceleration or distance… Do I need to plug two formulas into each other? Thanks to everyone in advance
That’s a little about it…
Do you mean the area center of the area marked red in the following image?
Do you mean the line center in the following image blue marked line?
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I used to draw the relevant emphasis on the pictures of my answer. More or less random, the focus is on the same place.
But before I arrive on possible computing paths, I would like to know which one is meant.
So keep the main focus
However, there is no indication of the density how difficult which parts are. This is only a 2-dimensional drawing, without any indication that indicates a mass.
And even if one starts from a homogeneous distribution (all the same density), i.e. the center of gravity is equal to the geometric center of gravity, it is still not clear…
Then we’d be back on the question… Is that what I mean in picture 1 or what I mean in picture 2? [As described in my answer, it is rather coincidence that in the example the same focus comes out. It is therefore important to clarify this so that it may also fit in other situations.]
So that should be meant as a sheet
Then my mistake
Well, assumed the density is constantly distributed.
Then you can calculate the center of gravity once per square.
Then you can add the centers of gravity weighted by their mass and catch the center of gravity of the structure.
How to add them?
I get from the coordinates of the five focal points of the sub-areas as the focus of the overall figure:
(but please calculate)
Addendum by comment: Obviously, the total spelling is not common, therefore written here:
“Ai =35/8”
the A stands for the surface, right?
but what does the
is wrong, and I didn’t write that either.
In the end, I am the weighted average of the two coordinates and accordingly Ai the area of the i-th rectangle, just as xi the x.coordinates of the center of gravity of the Si the i-th rectangle.
The “i” only numbers the individual rectangles from 1 to 5 so that one knows what belongs to it.
In which class you learn
I don’t know – my school time is almost half a century back.
Oh, that’s a bit her😅
Thank you.
The calculation of the center of gravity is achieved by the sum of the individual focal points. It is assumed from the view that each surface element leaves a torque on the coordinate axis via a lever arm x_n. The sum of all torques divided by the sum of all surfaces then gives a “effective” lever arm for all surfaces. This then represents the center of gravity line for a dimension. In the following picture, the exemplary for the x-direction has been carried out. A subsequent passage for the y-direction then leads to the other center of gravity line. The intersection is the focus. In this context, it is irrelevant where one places the origin of the coordinates.
The following shall apply:
Specific
The same consideration for the y-direction is carried out analogously. In this particular case, even a simple symmetry consideration would make this computation step unnecessary.
I can’t calculate it, but I’d guess.
In the upper square between the 3rd and 4th row and column.
Reasons:
The figure would be the 4th. Part and the connections there would be the center focus.
At the top right and at the bottom left, you can move the point towards the top left.
I’d tap the middle of the line.
As I said, it’s only guessed.