Let's **break down and solve** the static equilibrium of the six-bar linkage step by step. --- ## **Step 1: Extract and Organize the Data** **Given:** - \( P = 320\, \text{N} \) (at H, left, horizontal) - \( R \) (at C, down, vertical, unknown) - \( T = 450\, \text{Nmm} = .45\, \text{Nm} \) (at A, CCW) - \( M = 260\, \text{Nmm} = .26\, \text{Nm} \) (at G, CCW) - Each small square \( = 50\, \text{mm} = .05\, \text{m} \) --- ## **Step 2: Find Coordinates of Key Points** Let’s assign coordinates (using lower-left corner as origin): - **A:** (10, 2) → (.5, .1) m - **B:** (8, 3) → (.4, .15) m - **C:** (8, 7) → (.4, .35) m - **D:** (6, 5) → (.3, .25) m - **E:** (6, 3) → (.3, .15) m - **F:** (4, 6) → (.2, .3) m - **G:** (4, 10) → (.2, .5) m - **H:** (, 2) → (, .1) m --- ## **Step 3: Free-Body Diagrams and Static Equilibrium** ### **A. Identify Forces at Each Joint** - **A:** Pin (reactions \(A_x, A_y\)), input torque \(T\) - **G:** Pin (reactions \(G_x, G_y\)), resistance moment \(M\) - **H:** Pin (reactions \(H_x, H_y\)), input force \(P\) - **C:** Pin (reactions \(C_x, C_y\)), output force \(R\) (vertical, down) ### **B. Write Equilibrium Equations** #### **For the Whole Linkage:** \[ \sum F_x = ,\quad \sum F_y = ,\quad \sum M = \] Let’s apply moments about a convenient point, say point A, to solve for R. #### **1. Horizontal and Vertical Forces** \[ \sum F_x = : \quad A_x + G_x + H_x + C_x - P = \] \[ \sum F_y = : \quad A_y + G_y + H_y + C_y - R = \] #### **2. Moments about Point A** Take moments about A (counterclockwise positive): \[ \sum M_A = \] The moment due to each force is \( F \cdot d_\perp \), where \( d_\perp \) is the perpendicular distance from A. - **P at H:** \( P \) acts at H (, .1) \[ \vec{r}_{AH} = ( - .5, .1 - .1) = (-.5, ) \] \( P \) acts left, so \( \vec{F}_P = (-320, ) \) Moment = \( r_x F_y - r_y F_x = (-.5)() - ()(-320) = \) But need to check direction: since P acts left, line of action is along x, so moment arm is the y-distance from A to H (which is ), so **no moment about A**. - **R at C:** \( R \) acts downward at C (.4, .35) \[ \vec{r}_{AC} = (.4-.5, .35-.1) = (-.1, .25) \] \( R \) acts downward (\( -R \) in y) Moment about A: \( -.1 \cdot (-R) - .25 \cdot = .1R \) - **T at A:** T is applied as a torque at A (CCW is positive): \( +.45 \) Nm - **M at G:** M is applied as a moment at G (CCW is positive): \( +.26 \) Nm --- ### **C. Write the Moment Equation** \[ \sum M_A = = (Moment~by~R) + T + M + (Moment~by~P) \] But as seen, P at H is collinear with A-H (same y-coordinate), so no moment arm. Thus, \[ = (.1)R + .45 + .26 \] \[ .1R = -.71 \] \[ R = -7.1~\text{N} \] But a negative R would mean the assumed direction is incorrect (i.e., it should be up, not down). But the diagram indicates R is downward. Let's check moment arms carefully. #### **Moment Arm for P** The force P at H (, .1) acts in the negative x direction. Its moment arm about A (.5, .1) is the vertical distance between A and H (which is ), so P creates no moment about A. #### **Moment Arm for R** The force R at C (.4, .35), vertical downward. The moment arm is the horizontal distance from A to C = .4 - .5 = -.1 m. So: \[ M_R = -.1 \cdot R \quad \text{(CW is negative)} \] Thus the equation is: \[ = -.1R + .45 + .26 \] \[ -.1R = -.71 \] \[ R = 7.1~\text{N} \] --- ## **Step 4: Final Answer** ### **Reaction force \( R \) at C:** \[ \boxed{R = 7.10~\text{N}~\text{(downward)}} \] --- ## **Step 5: Reaction Forces at Other Joints** ### **A. At H (Pin):** - Only P (320 N, left) acts at H, so the reaction at H must be 320 N, right. ### **B. At A (Pin):** - The sum of all forces in the system must be zero. Because the only vertical force is R (down), and no other vertical forces are shown, the reaction at A in y must be 7.10 N upward. - The sum of all horizontal forces: P at H (320 N left) must be balanced by a reaction at A (320 N right). --- ## **Summary Table** | Joint | Reaction Forces (N) | |-------|---------------------| | A | \( A_x = 320 \), \( A_y = 7.10 \) | | H | \( H_x = 320 \), \( H_y = \) | | C | \( R = 7.10 \) (downward) | | G | N/A (just moment M) | --- ## **Step 6: Conclusion** - **Force \( R \) at C:** \( \boxed{7.10~\text{N}~\text{downward}} \) - **Reaction at A:** \( 320~\text{N}~\text{right},~7.10~\text{N}~\text{upward} \) - **Reaction at H:** \( 320~\text{N}~\text{left},~~\text{N} \) - **Moments:** Already balanced by T and M. --- **If you need the full force breakdown at every joint (internal linkage forces), the solution will become much more involved with individual FBDs for each bar. Let me know if you want that detailed breakdown!**