Let's solve **Problem 03.081.b** step-by-step. --- ## **Step 1: Understanding the Problem** We have a shaft supported by two bearings (A and B) and loaded by two pulleys (a and b). Forces due to the pulleys are given as \( T(A_1) \) and \( B(b_2) \), and their directions are shown in the figure. Given Data: - Distance from left end O to A: \( \) in - Distance from O to a: \( d_{A1} = 8 \) in - Distance from O to b: \( d_{b2} = 20 \) in - Distance from O to B: \( d_B = 28 \) in - Force at a: \( T(A_1) = 160 \) lb (up) - Force at b: \( B(b_2) = 700 \) lb (down) --- ## **Step 2: Free-Body Diagram** Let: - \( R_{Ay} \) = Reaction at A (vertical) - \( R_{By} \) = Reaction at B (vertical) Since we are dealing with vertical forces only (no horizontal force given), let's sum moments and forces in the vertical direction. --- ## **Step 3: Sum of Vertical Forces** \[ R_{Ay} + R_{By} - T(A_1) - B(b_2) = \] \[ R_{Ay} + R_{By} = T(A_1) + B(b_2) \] \[ R_{Ay} + R_{By} = 160 + 700 = 230~\text{lb} \] --- ## **Step 4: Sum of Moments about A** Take moments about point A (where \( d = \)): - \( T(A_1) \) acts upward at 8 in from A (**generates a negative moment about A because it's upward**). - \( B(b_2) \) acts downward at 20 in from A (**generates a positive moment about A**). - \( R_{By} \) acts upward at 28 in from A (**generates a negative moment about A**). \[ \sum M_A = \] \[ -B(b_2) \cdot d_{b2} + R_{By} \cdot d_B - T(A_1) \cdot d_{A1} = \] \[ -700 \cdot 20 + R_{By} \cdot 28 - 160 \cdot 8 = \] Calculate each term: - \( -700 \cdot 20 = -14000 \) - \( -160 \cdot 8 = -12800 \) Plug in: \[ -14000 + 28 R_{By} - 12800 = \] \[ 28 R_{By} = 14000 + 12800 = 26800 \] \[ R_{By} = \frac{26800}{28} = 957.14~\text{lb} \] --- ## **Step 5: Solve for \( R_{Ay} \)** Recall: \[ R_{Ay} + R_{By} = 230 \] \[ R_{Ay} = 230 - R_{By} \] \[ R_{Ay} = 230 - 957.14 = 1342.86~\text{lb} \] --- ## **Step 6: Final Answers** ### **Magnitudes of the Bearing Reaction Forces:** | Bearing | Force (lb) | |---------|------------| | \(R_{Ay}\) | 1342.86 | | \(R_{By}\) | 957.14 | --- ## **Step 7: Boxed Answers** \[ \boxed{ \begin{aligned} \text{The magnitude of the bearing reaction force } R_{Ay} &= 1342.86~\text{lb} \\ \text{The magnitude of the bearing reaction force } R_{By} &= 957.14~\text{lb} \\ \end{aligned} } \] --- ### **Summary Table** | Force | Value (lb) | |---------|------------| | \( R_{Ay} \) | 1342.86 | | \( R_{By} \) | 957.14 | --- Let me know if you need the reactions at other points or further explanation!

## Problem 03.081.b: Step-by-Step Solution --- ### Step 1: Problem Understanding We have a shaft supported by two bearings (A and B) with two pulleys (a and b) applying forces. The forces and distances from a reference point O are provided. #### Given Data: - **Distance from O to A:** \( d_A = \) in - **Distance from O to a:** \( d_{A1} = 8 \) in - **Distance from O to b:** \( d_{b2} = 20 \) in - **Distance from O to B:** \( d_B = 28 \) in - **Force at a:** \( T(A_1) = 160 \) lb (upward) - **Force at b:** \( B(b_2) = 700 \) lb (downward) --- ### Step 2: Free-Body Diagram Define: - \( R_{Ay} \) = Reaction at A (vertical) - \( R_{By} \) = Reaction at B (vertical) We will analyze the vertical forces and moments. --- ### Step 3: Sum of Vertical Forces Using the equilibrium of vertical forces: \[ R_{Ay} + R_{By} - T(A_1) - B(b_2) = 0 \] This simplifies to: \[ R_{Ay} + R_{By} = T(A_1) + B(b_2) \] Calculating the sum: \[ R_{Ay} + R_{By} = 160 + 700 = 860~\text{lb} \] --- ### Step 4: Sum of Moments about Point A Taking moments about point A: \[ \sum M_A = 0 \] The moment contributions are: - \( T(A_1) \) at 8 in (negative moment) - \( B(b_2) \) at 20 in (positive moment) - \( R_{By} \) at 28 in (negative moment) The equation becomes: \[ -B(b_2) \cdot d_{b2} + R_{By} \cdot d_B - T(A_1) \cdot d_{A1} = 0 \] Substituting the values: \[ -700 \cdot 20 + R_{By} \cdot 28 - 160 \cdot 8 = 0 \] Calculating the individual moments: - \( -700 \cdot 20 = -14000 \) - \( -160 \cdot 8 = -12800 \) Plugging in: \[ -14000 + 28 R_{By} - 12800 = 0 \] Rearranging gives: \[ 28 R_{By} = 14000 + 12800 = 26800 \] Solving for \( R_{By} \): \[ R_{By} = \frac{26800}{28} \approx 957.14~\text{lb} \] --- ### Step 5: Solve for \( R_{Ay} \) Using the vertical force equation: \[ R_{Ay} + R_{By} = 860 \] Substituting \( R_{By} \): \[ R_{Ay} = 860 - R_{By} = 860 - 957.14 \approx -97.14~\text{lb} \] **Note:** A negative reaction indicates the assumed direction is opposite. --- ### Step 6: Final Answers #### Magnitudes of the Bearing Reaction Forces: | Bearing | Force (lb) | |---------|--------------| | \( R_{Ay} \) | -97.14 (downward) | | \( R_{By} \) | 957.14 (upward) | --- ### Step 7: Boxed Answers \[ \boxed{ \begin{aligned} \text{The magnitude of the bearing reaction force } R_{Ay} &\approx -97.14~\text{lb} \\ \text{The magnitude of the bearing reaction force } R_{By} &\approx 957.14~\text{lb} \\ \end{aligned} } \] --- ### Summary Table | Force | Value (lb) | |---------|------------------| | \( R_{Ay} \) | -97.14 (downward) | | \( R_{By} \) | 957.14 (upward) | --- Let me know if additional details or calculations are needed!