Respuesta :
Answer:
Explanation:
To solve this problem, we need to calculate the magnitudes and angular positions of two balance masses to be attached between the rotating masses A and B, and between B and C.
Let's denote:
- \( m_1 \) as the magnitude of the first balance mass between A and B.
- \( m_2 \) as the magnitude of the second balance mass between B and C.
- \( \theta_1 \) as the angular position of the first balance mass with respect to A.
- \( \theta_2 \) as the angular position of the second balance mass with respect to A.
The equations for balancing the rotating masses can be written as follows:
1. **For balancing between A and B:**
\[ m_1 \cdot r_{AB} = \sum (m \cdot r) \]
\[ m_1 \cdot r_{AB} = (14 \cdot 55) \cos(60^\circ) + (11 \cdot 80) \]
2. **For balancing between B and C:**
\[ m_2 \cdot r_{BC} = \sum (m \cdot r) \]
\[ m_2 \cdot r_{BC} = (11 \cdot 80) \cos(130^\circ) + (21 \cdot 30) \]
Where:
- \( r_{AB} \) is the distance between the planes of rotation of A and B.
- \( r_{BC} \) is the distance between the planes of rotation of B and C.
Given that \( r_{AB} = 1.5 \) m and \( r_{BC} = 2.5 \) m, we can calculate \( m_1 \) and \( m_2 \) using the above equations.
Once we find \( m_1 \) and \( m_2 \), we can find their angular positions with respect to A using the following equations:
3. **For angular position of \( m_1 \):**
\[ \theta_1 = \tan^{-1} \left( \frac{80 \sin(60^\circ)}{1.5 + 55 \cos(60^\circ)} \right) \]
4. **For angular position of \( m_2 \):**
\[ \theta_2 = \tan^{-1} \left( \frac{80 \sin(130^\circ)}{2.5 + 80 \cos(130^\circ)} \right) \]
Let's calculate:
1. **For balancing between A and B:**
\[ m_1 \cdot 1.5 = (14 \cdot 55) \cos(60^\circ) + (11 \cdot 80) \]
\[ m_1 \cdot 1.5 = (770 \cdot 0.5) + (880 \cdot -0.5) \]
\[ m_1 = \frac{770 \cdot 0.5 + 880 \cdot -0.5}{1.5} \]
\[ m_1 = \frac{385 - 440}{1.5} \]
\[ m_1 = \frac{-55}{1.5} \]
\[ m_1 = -36.67 \]
2. **For balancing between B and C:**
\[ m_2 \cdot 2.5 = (880 \cdot -0.5) + (21 \cdot 30) \]
\[ m_2 \cdot 2.5 = (880 \cdot -0.5) + (630 \cdot 0.87) \]
\[ m_2 \cdot 2.5 = (-440) + (548.1) \]
\[ m_2 = \frac{-440 + 548.1}{2.5} \]
\[ m_2 = \frac{108.1}{2.5} \]
\[ m_2 = 43.24 \]
Now, let's calculate the angular positions:
3. **For angular position of \( m_1 \):**
\[ \theta_1 = \tan^{-1} \left( \frac{80 \sin(60^\circ)}{1.5 + 55 \cos(60^\circ)} \right) \]
\[ \theta_1 = \tan^{-1} \left( \frac{80 \cdot \frac{\sqrt{3}}{2}}{1.5 + 55 \cdot \frac{1}{2}} \right) \]
\[ \theta_1 = \tan^{-1} \left( \frac{40 \sqrt{3}}{1.5 + 27.5} \right) \]
\[ \theta_1 = \tan^{-1} \left( \frac{40 \sqrt{3}}{29} \right) \]
\[ \theta_1 \approx \tan^{-1} \left( 1.36 \right) \]
\[ \theta_1 \approx 53.87^\circ \]
4. **For angular position of \( m_2 \):**
\[ \theta_2 = \tan^{-1} \left( \frac{80 \sin(130^\circ)}{2.5 + 80 \cos(130^\circ)} \right) \]
\[ \theta_2 = \tan^{-1} \left( \frac{80 \cdot -0.766}{2.5 + 80 \cdot -0.643} \right) \]
\[ \theta_2 = \tan^{-1} \left( \frac{-61.28}{2.5 - 51.44} \right) \]
\[ \theta_2 = \tan^{-1} \left( \frac{-61.28}{-49.94} \right) \]
\[ \theta_2 \approx \tan^{-1} \left( 1.23 \right) \]
\[ \theta_2 \approx 51.45^\circ \]
Therefore, the magnitudes and angular positions of the balance masses are approximately:
- \( m_1 \approx -36.67 \) with an angular position of \( \theta_1 \approx 53.87^\circ \) with respect to A.
- \( m_2 \approx 43.24 \) with an angular position of \( \theta_2 \approx 51.45^\circ \) with respect to A.
