Three weights are attached to the corners of an equilateral triangle (side length 10 cm) made from stiff but very light wire. Measured from the bottom left corner, what is the center of mass of this object if M1 = 0.1 kg, M2 = 0.2 kg, and M3 = 0.3 kg?

To find the center of mass of an object, you need to consider both the mass and position of each individual component. In this case, you have three weights attached to the corners of an equilateral triangle.

First, let's find the positions of each weight with respect to the bottom left corner. Since it is an equilateral triangle, each side length is 10 cm.

The position of M1 (0.1 kg) is at the bottom left corner, so its position is (0, 0).

The position of M2 (0.2 kg) can be calculated using trigonometry. Since it is located at one vertex of the equilateral triangle, its position is (x, y).
By dividing the triangle into two 30-60-90 triangles, we can find that x = 10*cos(30) cm (since the adjacent side is half of the hypotenuse) and y = 10*sin(30) cm (since the opposite side is half of the hypotenuse multiplied by √3).

Similarly, the position of M3 (0.3 kg) can be calculated as (x, y) using the same trigonometric principles.

Now, let's find the weighted average position of the three masses. We need to consider both the mass and position of each weight. The center of mass (x_cm, y_cm) can be calculated using the formula:

x_cm = (m1*x1 + m2*x2 + m3*x3) / (m1 + m2 + m3)
y_cm = (m1*y1 + m2*y2 + m3*y3) / (m1 + m2 + m3)

Plugging in the values:

x_cm = (0.1*0 + 0.2*x2 + 0.3*x3) / (0.1 + 0.2 + 0.3)
y_cm = (0.1*0 + 0.2*y2 + 0.3*y3) / (0.1 + 0.2 + 0.3)

Now, substitute the values of x2, y2, x3, and y3 derived from the trigonometry earlier, and calculate the center of mass.