Let’s solve the problem step by step: --- ### **Given Data** - Mass of load \( m = 550 \) kg - Gravitational acceleration \( g = 9.81 \) m/s² - \( a = 51.0 \) mm = 0.051 m - \( b = 204 \) mm = 0.204 m - \( c = 306 \) mm = 0.306 m - \( d = 102 \) mm = 0.102 m - \( e = 357 \) mm = 0.357 m - Center of mass at \( H \) --- ### **Step 1: Calculate the Weight** \[ W = m \cdot g = 550 \times 9.81 = 5395.5 \text{ N} \] --- ### **Step 2: Free-Body Diagram** - The weight acts downward at \( H \). - Pin reaction at \( E \) has components \( E_x \) (right) and \( E_y \) (up). --- ### **Step 3: Geometry and Coordinates** Let’s assign a coordinate system (origin at \( D \)): - \( D = (0,0) \) - \( E = (0, e) \) - \( H = (a+2b, c+d) \) So, - \( H_x = a+2b = 0.051 + 2 \times 0.204 = 0.459 \) m - \( H_y = c+d = 0.306 + 0.102 = 0.408 \) m --- ### **Step 4: Summing Moments about Point \( E \)** Take moments about \( E \): \[ \sum M_E = 0 \] The only force causing a moment about \( E \) is the weight at \( H \): - The moment arm from \( E \) to \( H \) is: - \( \Delta x = H_x - E_x = 0.459 - 0 = 0.459 \) m - \( \Delta y = H_y - E_y = 0.408 - 0.357 = 0.051 \) m The moment caused by the weight about \( E \) is: \[ M_E = (0.459)W_y - (0.051)W_x \] But the weight acts downward, so its force is only in the \( y \)-direction: \[ M_E = 0.459 \times (-5395.5) \] \[ M_E = -2477.5 \text{ Nm} \] For equilibrium, this should be balanced by the moment generated by the reaction at \( D \), which is a pin (so it can have \( D_x \) and \( D_y \)), but since we are only solving for \( E \), let's write the equilibrium equations: --- ### **Step 5: Equilibrium Equations** \[ \sum F_x = 0: \quad E_x + D_x = 0 \] \[ \sum F_y = 0: \quad E_y + D_y - W = 0 \] \[ \sum M_E = 0 \] But since \( D \) is at the origin, let's use moments about \( D \): - \( r_{ED} = (0, e) \) - \( r_{HD} = (a+2b, c+d) \) The force \( E \) acts at \( E \), \( W \) at \( H \). Take moments about \( D \): \[ \sum M_D = 0 \] Moment from \( E \): - \( \vec{r}_{DE} = (0, e) \) - \( \vec{E} = (E_x, E_y) \) Moment = \( r_x F_y - r_y F_x \): \[ M_{DE} = 0 \times E_y - e \times E_x = -e E_x \] Moment from \( W \): - \( \vec{r}_{DH} = (a+2b, c+d) \) - \( \vec{W} = (0, -W) \) Moment = \( (a+2b) \times -W \) \[ M_{DH} = (a+2b) \times (-W) \] Sum moments: \[ -e E_x + (a+2b)(-W) = 0 \] \[ -e E_x = (a+2b) W \] \[ E_x = -\frac{(a+2b) W}{e} \] Substitute numbers: \[ a+2b = 0.051 + 2 \times 0.204 = 0.459 \text{ m} \] \[ e = 0.357 \text{ m} \] \[ W = 5395.5 \text{ N} \] \[ E_x = -\frac{0.459 \times 5395.5}{0.357} \] \[ E_x = -\frac{2477.5}{0.357} \] \[ E_x = -6941 \text{ N} \] --- ### **Step 6: Vertical Equilibrium** \[ E_y + D_y - W = 0 \] Also, \( D_y \) can be found by summing moments about another point, but since only \( E \) is asked, let's use: \[ E_y = W = 5395.5 \text{ N} \] --- ### **Step 7: Final Answer** \[ E_x, E_y = -6940, \ 5390\ \text{N} \] (Rounded to three significant figures) --- ## **Final Answer** \[ \boxed{-6940,\ 5390\ \text{N}} \] --- Let me know if you want the full details for \( D \) or other checks!