Let's solve the problem step by step: --- ### **Step 1: Draw the Free-Body Diagram and Set Up Equilibrium Equations** **Explanation:** - Identify all the forces acting on the frame. - At joint A, a force T = 50 N is applied at a 15° angle. - Supports at C and D are fixed, so they provide reactions. - We are required to find the forces at B on member BD, so focus on joint B. #### **Free-body forces at A:** - \( T_x = T \cos 15^\circ \) - \( T_y = T \sin 15^\circ \) #### **Member geometry:** - Coordinates: - A (0, 0) - B (2, 2) - C (6, 4) - D (4, 0) --- ### **Step 2: Analyze Joint B** **Explanation:** - At joint B, three forces are acting: the force in member AB, force in member BC, and force in member BD (which we need). - Use **method of joints** and resolve forces at B into their horizontal and vertical components. #### **Find the direction cosines for BD:** - \( \Delta x = 4 - 2 = 2 \) - \( \Delta y = 0 - 2 = -2 \) - Length BD = \( \sqrt{(2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2} \) - Direction cosines: - \( \cos \theta_x = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \) - \( \cos \theta_y = \frac{-2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}} \) #### **Let Force in BD = F_{BD} = (F_{BD,x}, F_{BD,y})** --- ### **Step 3: Apply Equilibrium at Joint B** **Explanation:** - Set the sum of forces at joint B to zero in both x and y directions. - Assume all members are 2-force members (forces along their axis). #### **Write equilibrium equations:** \[ \sum F_x = 0: \quad F_{AB,x} + F_{BC,x} + F_{BD,x} = 0 \] \[ \sum F_y = 0: \quad F_{AB,y} + F_{BC,y} + F_{BD,y} = 0 \] But since we want the forces at B on member BD, focus on \( F_{BD} \). The force T at A is transmitted through the members. #### **Calculate F_{BD} using moment about C (to eliminate reactions at C):** Take moment about C: - \( T \) acts at A: Distance from C is \( \sqrt{(6-0)^2 + (4-0)^2} = \sqrt{36+16} = \sqrt{52} \) - But more simply, sum moments about C for the whole frame. Let's use the geometry to get the vertical and horizontal force at B on BD. #### **Write equations for horizontal and vertical components at B:** At joint B, the force in BD must balance the effect of T. From geometry, the member BD is at a 45° downward angle from B (since \( \Delta x = 2, \Delta y = -2 \)), so any force in BD has equal x and y components, but y is negative. Let the force in BD at B be \( F_{BD} \) directed from B to D. So, - \( F_{BD,x} = F_{BD} \cdot \frac{1}{\sqrt{2}} \) - \( F_{BD,y} = F_{BD} \cdot \left(-\frac{1}{\sqrt{2}}\right) \) Using joint equilibrium at B, the sum of forces in y direction (assuming only vertical member BD and force from AB): \[ T \sin 15^\circ = F_{BD} \cdot \left(-\frac{1}{\sqrt{2}}\right) \] \[ F_{BD} = -T \sin 15^\circ \cdot \sqrt{2} \] Plug in \( T = 50 \) N and \( \sin 15^\circ = 0.2588 \): \[ F_{BD} = -50 \times 0.2588 \times 1.414 \approx -18.3 \text{ N} \] **Negative sign** indicates the direction is opposite to assumed. #### **Components at B:** - **Horizontal:** \( F_{BD,x} = -18.3 \times \frac{1}{\sqrt{2}} = -12.9 \) N (to the left) - **Vertical:** \( F_{BD,y} = -18.3 \times \left(-\frac{1}{\sqrt{2}}\right) = +12.9 \) N (upward) --- ## **Final Answer:** **The forces at B on member BD are:** - **Horizontal component:** \( -12.9 \) N (to the left) - **Vertical component:** \( +12.9 \) N (upward) --- ### **Summary:** 1. **Drew the free-body diagram and set equilibrium equations.** 2. **Analyzed joint B using geometry and force directions.** 3. **Solved for forces using equilibrium, finding the force in member BD at B is 18.3 N (with components ±12.9 N).**