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Chapter 9: Problem 11

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 four times the mass at either of the base vertices, which means \(m_3 = 4m\).

Step by step solution

01

Understand the Center of Mass's Position

The problem states that the center of mass of the system is midway between the base and the third vertex of an equilateral triangle. This means that the center of mass is two thirds of the way up from the base of the triangle, closer to the third vertex. This is important because the vertical position of the center of mass depends on the distribution of mass in the triangle.
02

Formulate the Position of the Center of Mass

The vertical coordinate of the center of mass, \(Y_{cm}\), of a system of particles is determined by the equation \(Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3}\). 'm' represents the mass and 'y' the vertical position. In this case, \(y_1 = y_2 = 0\) (as the base is considered at the ground level) and \(y_3 = h\) where 'h' is the height of the triangle.
03

Substitute the Provided Information

We know that the center of mass is midway between the base and the third vertex. In other words, \(Y_{cm} = \frac{2h}{3}\). Also, we have that each of the masses at the base are 'm' and the mass at the third vertex is \(m_3\). Substituting these into the equation from step 2, we have \(\frac{2h}{3} = \frac{m_3h}{2m + m_3}\). The heights cancel out and we can solve for \(m_3\).
04

Solve for the Third Mass

Multiply through by \((2m + m_3)\) to clear the fraction, then simplify to find \(m_3\). This yields the equation \(4m + 2m_3 = 3m_3\).
05

Isolate the Unkown

Finally, isolate \(m_3\) to find its value: \(4m = m_3\). Thus, the mass at the third vertex is four times the mass at either of the base vertices.

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