Let's solve the problem step by step using statics principles. We'll denote \( a = .5 \) m and \( M = 60 \) kg. We will use \( g = 9.81 \) m/s² for gravity. ## **Step 1: Free Body Diagram and Forces** ### **Key Points:** - \( A \): Pin support (reactions \( A_x \), \( A_y \)) - \( E/F \): Roller support (vertical reaction \( E_y \)) - \( B \): Cable attached, force \( T \) in cable (acts at angle) - Cable passes over pulleys at \( D \) and \( C \), hangs mass \( M \) at \( C \) - Beam length \( = 3a \) - Vertical drop \( DA = 1.25a \) ### **Forces:** - **At C:** Weight \( W = Mg \) acts downward - **At B:** Cable tension \( T \) acts along \( \overrightarrow{DB} \) ## **Step 2: Cable Tension** At equilibrium, the tension \( T \) in the cable is equal everywhere due to frictionless pulleys. At point C, the cable supports mass \( M \): \[ T = Mg = 60 \times 9.81 = \boxed{588.6~\text{N}} \] ## **Step 3: Geometry for Cable Direction at B** - \( DB \): From D to B: horizontal \( a \), vertical \( -1.25a \) - Direction cosines of \( DB \): \[ \vec{DB} = (a, -1.25a) \] \[ |\vec{DB}| = a \sqrt{1^2 + 1.25^2} = a \sqrt{2.5625} = 1.6a \] - Unit vector along \( DB \): \[ \mathbf{u_{DB}} = \left( \frac{1}{1.6}, -\frac{1.25}{1.6} \right) = (.625, -.781) \] - Components of \( T \) at point B: \[ T_x = T \times .625 = 588.6 \times .625 = 367.9~\text{N} \] \[ T_y = T \times (-.781) = 588.6 \times (-.781) = -458.9~\text{N} \] ## **Step 4: Equilibrium Equations** Let’s denote: - \( A_x \): Reaction at A in x-direction - \( A_y \): Reaction at A in y-direction - \( E_y \): Reaction at E in y-direction #### **Sum of Forces in \( x \):** \[ \sum F_x = : \quad A_x - T_x = \implies A_x = T_x = \boxed{367.9~\text{N}} \] #### **Sum of Forces in \( y \):** \[ \sum F_y = : \quad A_y + E_y + T_y - W = \] \[ A_y + E_y - 458.9 - 588.6 = \] \[ A_y + E_y = 458.9 + 588.6 = 1047.5~\text{N} \] #### **Sum of Moments about A:** Take moments about point A (clockwise is positive): - \( E_y \) acts at \( x = 2a \) - \( T \) at B acts at \( x = a \) (vertical component only) - \( W \) acts at \( x = 3a \) \[ \sum M_A = : \quad - T_y \cdot a + E_y \cdot 2a - W \cdot 3a = \] \[ -(-458.9)a + 2a E_y - 588.6 \cdot 3a = \] \[ 458.9a + 2a E_y - 1765.8a = \] \[ 2a E_y = 1765.8a - 458.9a = 1306.9a \] \[ E_y = \frac{1306.9a}{2a} = 653.45~\text{N} \] Now, plug \( E_y \) into \( A_y + E_y = 1047.5 \): \[ A_y = 1047.5 - 653.45 = 394.05~\text{N} \] --- ## **Step 5: Final Answer for Support Reactions** \[ \boxed{ \begin{align*} A_x &= 367.9~\text{N} \\ A_y &= 394.1~\text{N} \\ E_y &= 653.5~\text{N} \end{align*} } \] --- ## **Step 6: Internal Forces at Midpoint of AB and BE** Let’s take midpoints: - Midpoint of AB: at \( x = .5a \) from A - Midpoint of BE: at \( x = 1.5a \) from A ### **Section at Midpoint of AB (\( x = .5a \)):** #### **Normal Force (\( N_{AB} \)):** No horizontal loads between A and B other than the reaction at A, so: \[ N_{AB} = -A_x = -367.9~\text{N} \] (Negative: compression if left face is considered) #### **Shear Force (\( V_{AB} \)):** Shear is due to vertical reactions up to this point: \[ V_{AB} = A_y = 394.1~\text{N} \] #### **Moment (\( M_{AB} \)):** \[ M_{AB} = A_y \cdot .5a \] \[ M_{AB} = 394.1 \cdot .25 = 98.5~\text{N}\cdot\text{m} \] --- ### **Section at Midpoint of BE (\( x = 1.5a \)):** #### **Normal Force (\( N_{BE} \)):** Only horizontal load is still \( A_x \): \[ N_{BE} = -A_x = -367.9~\text{N} \] #### **Shear Force (\( V_{BE} \)):** Sum vertical forces to the left of the cut: \[ V_{BE} = A_y + T_y \] \[ V_{BE} = 394.1 + (-458.9) = -64.8~\text{N} \] #### **Moment (\( M_{BE} \)):** Take moments about the cut at \( x = 1.5a \): \[ M_{BE} = A_y \cdot 1.5a + T_y \cdot (1.5a - a) \] \[ = A_y \cdot 1.5a + T_y \cdot .5a \] \[ = 394.1 \cdot .75 + (-458.9) \cdot .25 = 295.6 - 114.7 = 180.9~\text{N}\cdot\text{m} \] --- ## **Step 7: Final Answers (Numerical, for \( a = .5 \) m)** **Support Reactions:** - \( \boxed{A_x = 367.9~\text{N}} \) - \( \boxed{A_y = 394.1~\text{N}} \) - \( \boxed{E_y = 653.5~\text{N}} \) **Internal Forces at Midpoints:** - At midpoint of AB: - \( \boxed{N_{AB} = -367.9~\text{N}} \) - \( \boxed{V_{AB} = 394.1~\text{N}} \) - \( \boxed{M_{AB} = 98.5~\text{N}\cdot\text{m}} \) - At midpoint of BE: - \( \boxed{N_{BE} = -367.9~\text{N}} \) - \( \boxed{V_{BE} = -64.8~\text{N}} \) - \( \boxed{M_{BE} = 180.9~\text{N}\cdot\text{m}} \) --- ### **Summary Table** | Quantity | Value (N or N·m) | |---------------|------------------| | \( A_x \) | 367.9 | | \( A_y \) | 394.1 | | \( E_y \) | 653.5 | | \( N_{AB} \) | -367.9 | | \( V_{AB} \) | 394.1 | | \( M_{AB} \) | 98.5 | | \( N_{BE} \) | -367.9 | | \( V_{BE} \) | -64.8 | | \( M_{BE} \) | 180.9 | --- **Let me know if you need step-by-step diagrams or further clarifications!**