Two particles of equal mass \(m\) are at the vertices of the base of an equilateral triangle. The triangle's center of mass is midway between the base and the third vertex. What's the mass at the third vertex?

Short Answer

Expert verified
The mass at the third vertex is twice the mass at the base vertices, i.e., \(M = 2m\).

Step by step solution

01

Understand the problem

Determine the position of the center of mass. We know the center of mass is midway between the base and the third vertex of an equilateral triangle. Therefore, the height of the triangle can be divided into two equal parts.
02

Set up the equation

For simplicity, let's use scalar quantities and assume that the height of the equilateral triangle is \(2a\), so the distances to the center of mass from the base and the third vertex are \(a\) each. According to the definition of center of mass, the total mass times the position of the center of mass equals the sum of each mass times its distance from the point of reference. Apply this to the triangle, we get: \[m \cdot a + m \cdot a + M \cdot a = (2m + M) \cdot a\]
03

Solve the equation

Solve the equation obtained from step 2 for the unknown mass \(M\). This equation simplifies to: \[2m = M\] This implies that the mass at the third vertex is twice the mass at the vertices of the base.

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