Let's solve the problem step by step: ## **Step 1: Understand the Problem** We are to **calculate the moment about point A** due to the 150-N force shown in the figure. ### **Given:** - Force \( F = 150 \) N - The force is applied at point B, perpendicular to the segment BC. - The dimensions are given in mm and some as ratios (3-4-5 triangle). ## **Step 2: Find the Perpendicular Distance from A to the Line of Action of Force** The moment (torque) about point A is given by: \[ M_A = F \times d \] Where: - \( F \) = Magnitude of the force (150 N) - \( d \) = Perpendicular distance from point A to the line of action of the force ### **Analyzing the Geometry** #### 1. **Coordinates:** Let’s place point A at the origin (0,0). - **Point C:** Given as 260 mm along the x-axis → (260, 0) - **Point B:** We need to find coordinates. From the triangle: - AC = 12 units - CB = 5 units - The force is perpendicular to BC. #### 2. **Find Location of Point B:** \( \triangle ABC \) is a right triangle at C with AC = 12 units, CB = 5 units, and AB = 13 units (from 5-12-13 right triangle property). Let’s scale to mm: - AC = (12/5) × CB - CB = 5 units (but what's the actual length in mm?) From C to A is 260 mm. So, 12 units = 260 mm → 1 unit = \( 260/12 \) mm = 21.67 mm/unit So, - AC = 260 mm - CB = 5 × 21.67 = 108.33 mm #### 3. **Find the Perpendicular Distance (d):** The moment arm is the perpendicular distance from A to the line of action of the 150 N force (applied vertically at B). We can use the area method: Area of triangle ABC: \[ \text{Area} = \frac{1}{2} \times AB \times h \] where \( h \) is the perpendicular from A to BC. But more directly, we can calculate the perpendicular distance from A to BC: \[ d = AC \times \sin(\theta) \] where \( \theta \) is the angle between AC and AB. Or, as BC is 5 units, and the triangle is 5-12-13: - The altitude from A to BC (moment arm) is (12 × 5) / 13 units. Convert this to mm: \[ \text{Moment arm} = \frac{12 \times 5}{13} = \frac{60}{13} \text{ units} \] In mm: \[ 1 \text{ unit} = 21.67 \text{ mm} \] So, \[ \text{Moment arm} = \frac{60}{13} \times 21.67 = 100 \text{ mm} \] ## **Step 3: Calculate the Moment** \[ M_A = F \times d = 150 \text{ N} \times 0.1 \text{ m} = 15 \text{ Nm} \] ## **Step 4: Direction of Moment** The force tends to rotate the system **counterclockwise about A**, so the moment is **positive**. ## **Final Answer** \[ \boxed{M_A = 15~\text{N}\cdot\text{m}~(\text{counterclockwise})} \] --- ### **Summary Table** | Step | Value | Calculation | |------|----------------------------|---------------------------------| | Force (\(F\)) | 150 N | Given | | Perpendicular Distance (\(d\)) | 100 mm = 0.1 m | See above | | Moment (\(M_A\)) | \(150 \times 0.1 = 15\) N·m | \(F \times d\) | | Direction | Counterclockwise | From force orientation | --- **Final Answer:** \[ \boxed{15~\text{N}\cdot\text{m}~\text{(counterclockwise)}} \]