Let's break down the problem step by step. --- ## **Given Data** - **4 masses:** A, B, C, D (in order along shaft) - **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 - **Angles:** - Between B and C = 120° - Between B and A = 200° (measured from B) - **Distances:** - A to B, \( l_1 = 120 \) mm = .12 m - B to C, \( l_2 = 240 \) mm = .24 m - Let distance A to D = \( l_3 \) - **Shaft is in complete dynamic balance.** --- ## **Step 1: Dynamic Balancing Conditions** For complete dynamic balance: 1. **Force balance (static):** Vector sum of centrifugal forces = \( \sum m_i r_i \vec{e}_i = \) (Where \( \vec{e}_i \) denotes direction/angle in the rotation plane.) 2. **Couple balance (dynamic):** Vector sum of moments about any point = \( \sum m_i r_i l_i \vec{e}_i = \) (where \( l_i \) is the distance from reference plane, take A as reference.) --- ## **Step 2: Assign Variables and Reference Plane** Let’s take **A** as the reference plane (origin). ### **Mass Data Table** | Mass | Value (kg) | Radius (m) | Angle (deg) from B | Position from A (m) | |------|------------|------------|--------------------|---------------------| | A | \( m_A \) | .09 | | | | B | 20 | .07 | | .12 | | C | 14 | .07 | 120 (from B) | .36 | | D | \( m_D \) | .09 | \( \theta_D \) | \( l_3 \) | ### **Angle Reference:** - Take angle at B as ° - C is 120° from B - A is -200° from B (or 160° from B in the opposite direction) - D is unknown (\( \theta_D \)) But since A is at ° (reference), B at 200°, C at 320°. --- ## **Step 3: Static Balance Equations** ### **a) Express all positions in polar form** Let’s express all \( m \cdot r \) in x and y components: #### **Angles (all measured from A):** - A: ° - B: 200° (from A) - C: 200° + 120° = 320° (from A) - D: \( \theta_D \) #### **Mass-radius products:** - A: \( M_A = m_A \times .09 \) - B: \( M_B = 20 \times .07 = 1.4 \) - C: \( M_C = 14 \times .07 = .98 \) - D: \( M_D = m_D \times .09 \) #### **Components:** \[ \sum M_x = , \quad \sum M_y = \] Breaking into x and y components: - A: \( (M_A, ^\circ) \) - B: \( (1.4, 200^\circ) \) - C: \( (.98, 320^\circ) \) - D: \( (M_D, \theta_D) \) Expand: \[ \sum M_x = M_A \cos ^\circ + 1.4 \cos 200^\circ + .98 \cos 320^\circ + M_D \cos \theta_D = \] \[ \sum M_y = M_A \sin ^\circ + 1.4 \sin 200^\circ + .98 \sin 320^\circ + M_D \sin \theta_D = \] Calculate cosines and sines: - \(\cos 200^\circ = -.9397\), \(\sin 200^\circ = -.342\) - \(\cos 320^\circ = .766\), \(\sin 320^\circ = -.6428\) Substitute: \[ M_A + 1.4(-.9397) + .98(.766) + M_D \cos \theta_D = \] \[ + 1.4(-.342) + .98(-.6428) + M_D \sin \theta_D = \] Calculate numerically: - \(1.4 \times -.9397 = -1.3156\) - \(.98 \times .766 = .7507\) - \(1.4 \times -.342 = -.4788\) - \(.98 \times -.6428 = -.630\) \[ M_A - 1.3156 + .7507 + M_D \cos \theta_D = \] \[ -.4788 - .630 + M_D \sin \theta_D = \] So, \[ M_A - .565 + M_D \cos \theta_D = \tag{1} \] \[ -1.1088 + M_D \sin \theta_D = \implies M_D \sin \theta_D = 1.1088 \tag{2} \] --- ## **Step 4: Couple (Dynamic) Balance** \[ \sum (m_i r_i l_i \cos \theta_i) = \] \[ \sum (m_i r_i l_i \sin \theta_i) = \] Where \( l_i \) is the distance from A (reference): - A: \( l_A = \) - B: \( l_B = .12 \) - C: \( l_C = .36 \) - D: \( l_D = x \) (unknown, to be found) So, for the couple moments: \[ \sum (M_i l_i \cos \theta_i) = \] \[ \sum (M_i l_i \sin \theta_i) = \] Plug in: \[ \cdot \cos ^\circ + 1.4 \times .12 \cos 200^\circ + .98 \times .36 \cos 320^\circ + M_D x \cos \theta_D = \] \[ \cdot \sin ^\circ + 1.4 \times .12 \sin 200^\circ + .98 \times .36 \sin 320^\circ + M_D x \sin \theta_D = \] Calculate: - \(1.4 \times .12 = .168\) - \(.98 \times .36 = .3528\) - \(.168 \cos 200^\circ = .168 \times -.9397 = -.1578\) - \(.3528 \cos 320^\circ = .3528 \times .766 = .2702\) - \(.168 \sin 200^\circ = .168 \times -.342 = -.0575\) - \(.3528 \sin 320^\circ = .3528 \times -.6428 = -.2269\) So, \[ -.1578 + .2702 + M_D x \cos \theta_D = \implies .1124 + M_D x \cos \theta_D = \tag{3} \] \[ -.0575 - .2269 + M_D x \sin \theta_D = \implies -.2844 + M_D x \sin \theta_D = \implies M_D x \sin \theta_D = .2844 \tag{4} \] --- ## **Step 5: Solve the Equations** Let \( M_D = m_D \times .09 \). From (2): \( M_D \sin \theta_D = 1.1088 \) From (4): \( M_D x \sin \theta_D = .2844 \) So, \[ x = \frac{.2844}{M_D \sin \theta_D} \] But from above, \( M_D \sin \theta_D = 1.1088 \): \[ x = \frac{.2844}{1.1088} = .2565\ \mathrm{m} \] So, **distance from A to D is .2565 m = 256.5 mm**. --- Now, from (1): \( M_A - .565 + M_D \cos \theta_D = \implies M_A + M_D \cos \theta_D = .565 \) From (3): \( .1124 + M_D x \cos \theta_D = \implies M_D x \cos \theta_D = -.1124 \) But \( x = .2565 \): \[ M_D \cos \theta_D = \frac{-.1124}{.2565} = -.4383 \] So, from above: \[ M_A + (-.4383) = .565 \implies M_A = 1.0033 \] \[ m_A = \frac{1.0033}{.09} = 11.15\ \mathrm{kg} \] Now, solve for \( m_D \): Recall: \[ M_D \sin \theta_D = 1.1088\\ M_D \cos \theta_D = -.4383 \] \[ M_D = \sqrt{(1.1088)^2 + (-.4383)^2} = \sqrt{1.2294 + .1921} = \sqrt{1.4215} = 1.1923 \] \[ m_D = \frac{1.1923}{.09} = 13.25\ \mathrm{kg} \] --- ## **Step 6: Angular Position of D** \[ \tan \theta_D = \frac{1.1088}{-.4383} = -2.53 \] \[ \theta_D = \arctan(-2.53) \] Calculator: \(\arctan(-2.53) = -68.4^\circ\) But since \(\sin \theta_D > , \cos \theta_D < \), angle is in the 2nd quadrant: \[ \theta_D = 180^\circ - 68.4^\circ = 111.6^\circ \] --- ## **Final Answers** ### **1. Magnitude of masses at A and D:** - \( m_A = 11.15 \) kg - \( m_D = 13.25 \) kg ### **2. Distance between planes A and D:** - \( l_{AD} = 257 \) mm (rounded to the nearest mm) ### **3. Angular position of mass at D:** - \( \theta_D = 112^\circ \) (rounded to the nearest degree, from A, measured in same direction as others) --- ### **Summary Table** | Quantity | Value | |-----------------------|-------------| | Mass at A (\(m_A\)) | 11.15 kg | | Mass at D (\(m_D\)) | 13.25 kg | | Distance A–D (\(l_{AD}\)) | 257 mm | | Angle of D (\(\theta_D\)) | 112° | --- **If you need a vector diagram or further clarifications, let me know!**