Let's solve the problem step by step: --- ## **Given Data** - **Mass at B (\(m_B\))** = 20 kg, **radius (\(r_B\))** = 70 mm = .07 m - **Mass at C (\(m_C\))** = 14 kg, **radius (\(r_C\))** = 70 mm = .07 m - **Mass at A (\(m_A\))** = ?, **radius (\(r_A\))** = 90 mm = .09 m - **Mass at D (\(m_D\))** = ?, **radius (\(r_D\))** = 90 mm = .09 m - **Angle between B and C** = 120° - **Angle between B and A** = 200° (measured in the same direction) - **Axial distances:** - A to B = 120 mm = .12 m - B to C = 240 mm = .24 m - Let A be at \(x = \), B at .12 m, C at .36 m, D at \(x_D\). --- ## **Step 1: Dynamic Balancing Conditions** Complete dynamic balance requires: 1. **Primary balance**: The vector sum of the centrifugal forces (m·r) must be zero. 2. **Secondary balance**: The vector sum of the moments (m·r·x) about a reference plane must be zero. Let’s fix A as the reference plane. --- ## **Step 2: Resolving Masses into Components** Let’s assign positions: - **Plane A:** Reference (angle °) - **Plane B:** 200° from A - **Plane C:** 200° + 120° = 320° from A - **Plane D:** θ from A (to be determined) Let: - \( F = m \cdot r \) - \( x \) = distance from reference (A) ### **Tabulate the Data** | Plane | Mass (kg) | r (m) | F = m·r (kg·m) | x (m) | Angle (deg) | |-------|-----------|-------|----------------|-------|-------------| | A | \(m_A\) | .09 | \(.09 m_A\)| | | | B | 20 | .07 | 1.4 | .12 | 200 | | C | 14 | .07 | .98 | .36 | 320 | | D | \(m_D\) | .09 | \(.09 m_D\)| \(x_D\)| θ | --- ## **Step 3: Primary Balance (Force Balance)** Sum of m·r vectors = . Express in components: \[ \sum F_x = \] \[ \sum F_y = \] ### **Component Equations** - For A (°): \( F_{A_x} = .09m_A \), \( F_{A_y} = \) - For B (200°): \( F_{B_x} = 1.4\cos(200^\circ) \), \( F_{B_y} = 1.4\sin(200^\circ) \) - For C (320°): \( F_{C_x} = .98\cos(320^\circ) \), \( F_{C_y} = .98\sin(320^\circ) \) - For D (θ): \( F_{D_x} = .09m_D\cos\theta \), \( F_{D_y} = .09m_D\sin\theta \) Sum \( x \) and \( y \) components and set to zero: \[ .09m_A + 1.4\cos(200^\circ) + .98\cos(320^\circ) + .09m_D\cos\theta = \] \[ 1.4\sin(200^\circ) + .98\sin(320^\circ) + .09m_D\sin\theta = \] --- ## **Step 4: Secondary Balance (Moment Balance)** Sum of moments about reference plane (A): \[ \sum (m \cdot r \cdot x)_x = \] \[ \sum (m \cdot r \cdot x)_y = \] - For A: \( x_A = \) - For B: \( x_B = .12 \) - For C: \( x_C = .36 \) - For D: \( x_D \) So, \[ 1.4 \cdot .12\cos(200^\circ) + .98 \cdot .36\cos(320^\circ) + .09m_D x_D \cos\theta = \] \[ 1.4 \cdot .12\sin(200^\circ) + .98 \cdot .36\sin(320^\circ) + .09m_D x_D \sin\theta = \] --- ## **Step 5: Substitute Values** Calculate cos/sin values: - \( \cos(200^\circ) = -.9397 \) - \( \sin(200^\circ) = -.342 \) - \( \cos(320^\circ) = .766 \) - \( \sin(320^\circ) = -.6428 \) ### **Primary Balance Equations** **(1) x-component:** \[ .09m_A + 1.4(-.9397) + .98(.766) + .09m_D\cos\theta = \] \[ .09m_A - 1.3156 + .7517 + .09m_D\cos\theta = \] \[ .09m_A + .09m_D\cos\theta = 1.3156 - .7517 \] \[ .09m_A + .09m_D\cos\theta = .5639 \tag{1} \] **(2) y-component:** \[ 1.4(-.342) + .98(-.6428) + .09m_D\sin\theta = \] \[ -.4788 - .630 + .09m_D\sin\theta = \] \[ .09m_D\sin\theta = 1.1088 \] \[ m_D\sin\theta = \frac{1.1088}{.09} = 12.32 \tag{2} \] --- ### **Secondary Balance Equations** **(3) x-component:** \[ 1.4 \cdot .12 (-.9397) + .98 \cdot .36 (.766) + .09m_D x_D\cos\theta = \] \[ 1.4 \cdot .12 = .168; \quad .98 \cdot .36 = .3528 \] \[ .168 \cdot (-.9397) = -.1578 \] \[ .3528 \cdot .766 = .2702 \] \[ -.1578 + .2702 + .09m_D x_D\cos\theta = \] \[ .09m_D x_D\cos\theta = -.1124 \tag{3} \] **(4) y-component:** \[ 1.4 \cdot .12(-.342) + .98 \cdot .36(-.6428) + .09m_D x_D \sin\theta = \] \[ .168 \cdot (-.342) = -.0575 \] \[ .3528 \cdot (-.6428) = -.2269 \] \[ -.0575 - .2269 + .09m_D x_D\sin\theta = \] \[ .09m_D x_D\sin\theta = .2844 \tag{4} \] --- ## **Step 6: Solve for Unknowns** From (2): \( m_D\sin\theta = 12.32 \) Let \( m_D\cos\theta = X \). Then use (1): \[ .09m_A + .09X = .5639 \] \[ .09m_A = .5639 - .09X \] \[ m_A = \frac{.5639 - .09X}{.09} \] From (3) and (4): Let \( Y = m_D x_D \). (3): \( .09Y\cos\theta = -.1124 \implies Y\cos\theta = -1.249 \) (4): \( .09Y\sin\theta = .2844 \implies Y\sin\theta = 3.16 \) Now, \( Y = \sqrt{(Y\cos\theta)^2 + (Y\sin\theta)^2} = \sqrt{(-1.249)^2 + (3.16)^2} = \sqrt{1.561 + 9.986} = \sqrt{11.547} = 3.398 \) So \( m_D x_D = 3.398 \) Now, \( \tan\theta = \frac{Y\sin\theta}{Y\cos\theta} = \frac{3.16}{-1.249} = -2.53 \implies \theta = 112.1^\circ \) (since sin is positive and cos is negative, this is in second quadrant) Now from earlier, \( m_D\sin\theta = 12.32 \). So \( m_D = \frac{12.32}{\sin(112.1^\circ)} \) \[ \sin(112.1^\circ) = .927 \] \[ m_D = \frac{12.32}{.927} = 13.29\, \text{kg} \] Now, \( X = m_D\cos\theta = 13.29 \cdot \cos(112.1^\circ) = 13.29 \cdot (-.374) = -4.97 \) Plug into \( m_A \): \[ m_A = \frac{.5639 - .09(-4.97)}{.09} = \frac{.5639 + .447}{.09} = \frac{1.0109}{.09} = 11.23\, \text{kg} \] Now, \( m_D x_D = 3.398 \implies x_D = \frac{3.398}{13.29} = .256\, \text{m} = 256\, \text{mm} \) --- ## **Step 7: Final Answers** ### **1. Magnitude of Masses at A and D** - \( m_A = 11.23 \) kg - \( m_D = 13.29 \) kg ### **2. Distance between Planes A and D** - \( x_D = 256 \) mm ### **3. Angular Position of Mass D** - \( \theta = 112.1^\circ \) (measured from A in the same sense as between B and C) --- ## **Summary Table** | Parameter | Value | |-------------------|--------------| | \( m_A \) | 11.23 kg | | \( m_D \) | 13.29 kg | | \( x_{AD} \) | 256 mm | | \( \theta_D \) | 112.1° | --- ## **Boxed Final Answers** \[ \boxed{ \begin{align*} 1. &\ \text{The magnitude of the masses at A and D:} \quad m_A = 11.23\,\text{kg}, \quad m_D = 13.29\,\text{kg} \\ 2. &\ \text{The distance between planes A and D:} \quad x_{AD} = 256\,\text{mm} \\ 3. &\ \text{The angular position of the mass at D:} \quad \theta_D = 112.1^\circ \\ \end{align*} } \]