Note: Graph>New Parameter creates a new parameter in a sketch. You can name it and you can give it a value. Later you can double-click it and change the value or the name.
On page A1, there is a sketch with two points A1 and A2 and two parameters m1 and m2 representing masses at the points. Compute the sum M = m1 + m2 and calculate ratios m1/M and m2/M.
Mark an appropriate ratio and use dilation to construct the point P that is the solution of (m1 + m2) P = m1A1 + m2A2.
Make a tool that will both compute M and construct P. It should take as inputs the two points A1 and A2 and also the two parameters m1 and m2.
Check that P is in the right location by trying some simple values of This should continue to be correct if you change the values of the parameters m1 or m2.
Set the values of m1 and m2 the numbers 6 and Ð10 to find P that satisfies (-4)P = 6A1 + (-10)A2.
Note that this is the same as saying 10A2 = 6A1 + 4P, so the arrangement of points as for this equation with positive coefficients. But the construction should work for one or more negative masses (so long as the sum is not zero).
Page A2 shows another way to control the masses that determine the location of the center of mass P. The sliders M1 and M2 determine ratios displayed on the screen that can be used as input to the Center of 2 Masses tool. In this case the construction has already been done, but soon you will use your custom tool with sliders.
On page B3 is a triangle and 3 masses given by sliders. The instructions say to construct 3 centers of mass B1, B2, B3 of the 3 pairs of points. Then using the mass (m1 + m2) at B3 and m3 at A3, construct the center of mass P3 of A3 and B3. Also construct centers P1 and P2 for the other pairs A1, B1 and A2, B2, with appropriate masses in the same way. Then you should find that the points P1 = P2 = P3 = P, the center of mass of the system of 3 masses at the 3 points A1, A2, A3.
P satisfies the relation (m1 + m2 + m3) P = m1 A1 + m2 A2 + m3 A3.
This shows an important fact: The center of mass of 3 or more points can be computed two masses at a time, but replacing a pair of point masses with the sum of the two masses at the point that is the center of mass.
So in this case, (m1 + m2 + m3) P = m1 A1 + (m2 + m3) B1. You can check this by algebra.
From this make a 3Bary Tool that takes 3 points and 3 masses and constructs the center of mass. Hide some extra stuff before making the tool, so the construction is pretty neat.
On page B3, set the sliders so that as nearly as possible, all the masses are equal. Then the points B1, B2, B3 are midpoints and P is the centroid G of the triangle A1A2A3 satisfying 3G = 1A1 + 1 A2 + 1 A3.
On page C4 is a slider parameter t. Let s = 1-t. Then if s and t are taken as masses, the sum s + t = 1 always.
Suppose two points A and B are given and the center of mass P(t) is constructed with mass s at A and t at B, so 1P(t) = (1-t)A + tB. Then t is an affine coordinate on line AB, with t = 0 and t = 1 giving P(0) = A and P(1) = B, with values of t such as t = 2, t = 3, etc, marking a ruler along line AB.
Now we apply this idea with repetition to construct an interesting curve. Follow the directions on page C4.
To see the quadratic nature of the point C1, write the formulas using t and (1-t) for B1, B2 and C1.
In this section, we will mostly work with point masses where the mass = 1.
On page D5 are given 3 points A, B, C. Construct the centroid of these 3 points and make a Centroid Tool. The inputs to the tool should be the three A, B, C (no mass measures).
On page D6 are given 4 points A, B, C, D. Follow the directions on the page to construct segments connecting the midpoints of complementary pairs of segments such as AC and BD.
These segments all pass through M, the center of mass of the points A, B, C, D (with equal masses at each point).
A set of 4 points A, B, C, D in the plane can be visualized as the "picture" (projection) of a 3D figure with 4 vertices, namely a general tetrahedron. Then page D6 shows that the segments connecting midpoints of opposite edges are concurrent at the (3D) center of mass of the vertices of the tetrahedron.
On page D7, this idea is continued. Construct the centroid of each of the faces of the tetrahedron and connect the centroid to the opposite vertex by a segment. Each of these segments also passes through the center of mass of the tetrahedron.
Suppose a figure is made from a flat board or a flat piece of metal. The center of mass is computed to a certain degree of accuracy by thinking of many points spread uniformly across the 2D shape, with the approximate center of mass computed from this large number of points. Then imagining the number of points increasing and the size of the mass decreasing proportionately, we arrive at an exact theoretical center of mass for a 2D region (calculus comes into play in general).
On Page E8, we learn that the center of mass of a uniform triangular region is the centroid, the center of mass of the vertices (with mass = 1 at each vertex).
On page E9, we learn that the same is not the case for quadrilateral regions.
On page F10, as a preliminary, start with triangle ABC and sides a, b, c. Measure the lengths of a, b, c and construct the center of mass of points A, B, C, with masses being the lengths of a, b, c.
Then construct the angle bisectors of ABC. Observe what you see and explain it using what you proved in Assignment 3.
Then on page F11, consider the center of mass of triangle ABC if the triangle sides were made of heavy iron bars, so that the mass is essentially all in the sides. If a bar is uniform, the center of mass of the bar is the midpoint. Is this clear?) Also, the mass of the bar is proportional to the length. Thus the center of mass of the "bar" triangle ABC is the same as the center of mass of the 3 midpoints of the sides with masses equal to the lengths of the sides a, b, c. Construct this center of mass.
How is this center related to the angle bisectors of the midpoint triangle? Why?