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8. Determine the sum of the moments of the three loads about A. SAN 9. Determine the moment of force F3 about point A on the beam. |F-400 Ib -250 lb 250 F- s00 Ib 150 30 200 10. Find the x and y components of the resultant force and the resultant couple moment at point O. "s00 N 300 N 200 N - m 11. Determine the x and y components of the equivalent resultant force and specify the distance where this force acts. 500 N 400 N 12. Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. 450N 02 m 200N m 15m 2m 15m 200N

8. Determine the sum of the moments of the three loads about A. SAN 9. Determine the moment of force F3 about point A on the beam. |F-400 Ib -250 lb 250 F- s00 Ib 150 30 200 10. Find the x and y components of the resultant force and the resultant couple moment at point O. "s00 N 300 N 200 N - m 11. Determine the x and y components of the equivalent resultant force and specify the distance where this force acts. 500 N 400 N 12. Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. 450N 02 m 200N m 15m 2m 15m 200N

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**ENGR 2301** **13.** Three horizontal forces are applied as shown to a vertical cast iron arm. Determine the resultant of the forces and the distance from the ground to its line of action when \( P = 200 \, N \). *Diagram Overview:* - A vertical cast iron arm is shown with three forces applied at different points along its height. - Force \( P \) is applied at the top (Point A). - Other forces, \( B = 600 \, N \) and \( C = 400 \, N \), are applied at 150 mm intervals descending the arm. **14.** Determine the location (measured from A) of the equivalent resultant of the distributed load. *Diagram Overview:* - A beam is depicted horizontally from point A to point B with distributed loads. - The beam is 12 ft in total length, divided equally into two 6 ft sections. - A distributed load of \( 50 \, lb/ft \) is on the first section, increasing to \( 100 \, lb/ft \) on the second section. **Answer:** 1-7: DDBDD, BB 8: 1 kN·m 9: M3 = 3000 ft·lb \(\sim\) 10: - \( (F_R)_x = 400 \, N (+\rightarrow) \) - \( (F_R)_y = -600 \, N (+\uparrow) \) - \( (M_R)_O = -2000 \, N \cdot m \) 11: - \( F_x = -300 \, N (+\rightarrow) \) - \( F_y = -800 \, N (+\uparrow) \) - \( d = 3 \, m \) 12: - \( F_R = 294 \, N, \theta = 40.1^\circ \) - \( (M_R)_O = 39.6 \, N \cdot m \text{ (clockwise)} \) 13: - \( R = 800 \, N \leftarrow, y = 0.1875 \, m = 187.5 \, mm \) 14: - 9.33 ft
Transcribed Image Text:**ENGR 2301** **13.** Three horizontal forces are applied as shown to a vertical cast iron arm. Determine the resultant of the forces and the distance from the ground to its line of action when \( P = 200 \, N \). *Diagram Overview:* - A vertical cast iron arm is shown with three forces applied at different points along its height. - Force \( P \) is applied at the top (Point A). - Other forces, \( B = 600 \, N \) and \( C = 400 \, N \), are applied at 150 mm intervals descending the arm. **14.** Determine the location (measured from A) of the equivalent resultant of the distributed load. *Diagram Overview:* - A beam is depicted horizontally from point A to point B with distributed loads. - The beam is 12 ft in total length, divided equally into two 6 ft sections. - A distributed load of \( 50 \, lb/ft \) is on the first section, increasing to \( 100 \, lb/ft \) on the second section. **Answer:** 1-7: DDBDD, BB 8: 1 kN·m 9: M3 = 3000 ft·lb \(\sim\) 10: - \( (F_R)_x = 400 \, N (+\rightarrow) \) - \( (F_R)_y = -600 \, N (+\uparrow) \) - \( (M_R)_O = -2000 \, N \cdot m \) 11: - \( F_x = -300 \, N (+\rightarrow) \) - \( F_y = -800 \, N (+\uparrow) \) - \( d = 3 \, m \) 12: - \( F_R = 294 \, N, \theta = 40.1^\circ \) - \( (M_R)_O = 39.6 \, N \cdot m \text{ (clockwise)} \) 13: - \( R = 800 \, N \leftarrow, y = 0.1875 \, m = 187.5 \, mm \) 14: - 9.33 ft
**ENGR 2301** **Free-Response Questions** 8. Determine the sum of the moments of the three loads about A. - **Diagram:** Shows a bent beam with three loads: a 6 kN force at the bottom, a 4 kN force at the middle, and a 5 kN force at the top, with given distances between them: 0.5 m, 1.5 m, and 0.5 m, respectively. 9. Determine the moment of force F₃ about point A on the beam. - **Diagram:** Depicts a beam with forces: F₃=250 lb applied at an angle of 30°, F₁=400 lb, and F₂=500 lb. Distances along the beam include 200, 250, and 150 units. 10. Find the x and y components of the resultant force and the resultant couple moment at point O. - **Diagram:** A horizontal beam with forces 300 N and 500 N acting perpendicular to it, and a couple moment of 200 N·m. Each force is applied 2 m from the reference point O. 11. Determine the x and y components of the equivalent resultant force, and specify the distance where this force acts. - **Diagram:** A beam with a force of 400 N acting down at point O and a 500 N force angled between the center and end. Measurements are given as 2 m apart. 12. Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. - **Diagram:** Displays a beam acted upon by a 450 N force at 30° from the y-axis and a horizontal 200 N force, along with a couple moment of 200 N·m. The distances on the beam are 0.2 m, 1.5 m, and 2 m from point O.
Transcribed Image Text:**ENGR 2301** **Free-Response Questions** 8. Determine the sum of the moments of the three loads about A. - **Diagram:** Shows a bent beam with three loads: a 6 kN force at the bottom, a 4 kN force at the middle, and a 5 kN force at the top, with given distances between them: 0.5 m, 1.5 m, and 0.5 m, respectively. 9. Determine the moment of force F₃ about point A on the beam. - **Diagram:** Depicts a beam with forces: F₃=250 lb applied at an angle of 30°, F₁=400 lb, and F₂=500 lb. Distances along the beam include 200, 250, and 150 units. 10. Find the x and y components of the resultant force and the resultant couple moment at point O. - **Diagram:** A horizontal beam with forces 300 N and 500 N acting perpendicular to it, and a couple moment of 200 N·m. Each force is applied 2 m from the reference point O. 11. Determine the x and y components of the equivalent resultant force, and specify the distance where this force acts. - **Diagram:** A beam with a force of 400 N acting down at point O and a 500 N force angled between the center and end. Measurements are given as 2 m apart. 12. Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. - **Diagram:** Displays a beam acted upon by a 450 N force at 30° from the y-axis and a horizontal 200 N force, along with a couple moment of 200 N·m. The distances on the beam are 0.2 m, 1.5 m, and 2 m from point O.
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