Question
Ind the center of mass of the given system of point masses lying on the \( x \)-axis. \[ \begin{array}{r}m_{1}=6, m_{2}=3, m_{3}=4 \\ x_{1}=-5, x_{2}=0, x_{3}=4\end{array} \] \( x=\square \)
Ask by Coleman Rogers. in the United States
Feb 28,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The center of mass is at \( x = -\frac{14}{13} \).
Solution
To find the center of mass x for the system, use the formula:
x = (m₁·x₁ + m₂·x₂ + m₃·x₃) / (m₁ + m₂ + m₃)
Substitute in the given values:
x = [6·(–5) + 3·(0) + 4·(4)] / (6 + 3 + 4)
= (–30 + 0 + 16) / 13
= (–14) / 13
Thus, the center of mass is:
x = –14/13
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution The Deep Dive
The center of mass \( x \) for a system of point masses is calculated using the formula: \[ x = \frac{\sum m_i x_i}{\sum m_i} \] For the given masses, we can plug in the values: 1. Calculate the total mass: \[ \sum m_i = m_1 + m_2 + m_3 = 6 + 3 + 4 = 13 \] 2. Calculate the weighted sum of positions: \[ \sum m_i x_i = m_1 x_1 + m_2 x_2 + m_3 x_3 = (6)(-5) + (3)(0) + (4)(4) = -30 + 0 + 16 = -14 \] Now, substituting these sums into the center of mass formula: \[ x = \frac{-14}{13} \] Thus, the center of mass of the system is: \[ x \approx -1.08 \]
