1. **State the problem:** We have a ladder leaning against a wall with reaction forces at points A and B. Given: - $R_{Ax} = 25$ (horizontal reaction at A) - $R_{Ay} = 150$ (vertical reaction at A) - $R_{Bx} = 25$ (horizontal reaction at B) - A downward force of 150 at mid-height (4 ft from A) We need to find the moments $M_A$, $M_B$, and $M_C$ at points A, B, and the midpoint C respectively. 2. **Set up the coordinate system and distances:** - Ladder height = 8 ft (vertical) - Horizontal distance from A to wall = 4 ft - Midpoint C is at 4 ft vertically and 2 ft horizontally (midpoint of ladder length) 3. **Calculate moments about point A:** Moment is force times perpendicular distance. - Moment due to $R_{Bx}$ at B (8 ft up): $M_{R_{Bx}} = R_{Bx} \times 8 = 25 \times 8 = 200$ - Moment due to downward 150# force at midpoint C (4 ft up, 2 ft horizontal): The perpendicular distance from A to force line is horizontal 2 ft. So, $M_{150} = 150 \times 2 = 300$ - Moment due to $R_{Ay}$ and $R_{Ax}$ at A is zero since they act at point A. Total moment at A, $M_A = M_{R_{Bx}} - M_{150} = 200 - 300 = -100$ 4. **Calculate moments about point B:** - Distance from B to A is 8 ft vertically and 4 ft horizontally. - Moment due to $R_{Ax}$ at A (25# horizontal): The vertical distance from B to A is 8 ft. $M_{R_{Ax}} = R_{Ax} \times 8 = 25 \times 8 = 200$ - Moment due to downward 150# force at midpoint C: Distance from B to C horizontally is 2 ft, vertically 4 ft. The perpendicular distance to force line from B is horizontal 2 ft. $M_{150} = 150 \times 2 = 300$ - Moment due to $R_{Bx}$ at B is zero. Total moment at B, $M_B = M_{150} - M_{R_{Ax}} = 300 - 200 = 100$ 5. **Calculate moment at midpoint C:** - Moments at midpoint C due to forces at A and B: - Distance from C to A horizontally 2 ft, vertically 4 ft. - Distance from C to B horizontally 2 ft, vertically 4 ft. Moment due to $R_{Ax}$ at A about C: $M_{R_{Ax}} = R_{Ax} \times 4 = 25 \times 4 = 100$ Moment due to $R_{Ay}$ at A about C: $M_{R_{Ay}} = R_{Ay} \times 2 = 150 \times 2 = 300$ Moment due to $R_{Bx}$ at B about C: $M_{R_{Bx}} = R_{Bx} \times 4 = 25 \times 4 = 100$ Moment due to downward 150# force at C is zero (force acts at C). Total moment at C, $M_C = M_{R_{Ax}} + M_{R_{Ay}} - M_{R_{Bx}} = 100 + 300 - 100 = 300$ **Final answers:** $$M_A = -100, \quad M_B = 100, \quad M_C = 300$$ Negative moment indicates direction opposite to assumed positive direction.