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Situation 6. Determine the resultant force R and its line of action for the following force-couple systems: (a) F = 300 N,F, = 0, and C = –150 N – m (b) F = 200 N,F, = -200 N, and C = 800 N – m (c) F = -600 N, F, = –400N, and C = 0 (d) F = -600 N, F, = 800 N, and C = -24000 N – m %3D Note: Typically during our discussions, CW positive is used for 2D. However for the problem figure notice that CCW is considered positive. This is probably because the problem used right thumb rule on the z-axis.

Situation 6. Determine the resultant force R and its line of action for the following force-couple systems: (a) F = 300 N,F, = 0, and C = –150 N – m (b) F = 200 N,F, = -200 N, and C = 800 N – m (c) F = -600 N, F, = –400N, and C = 0 (d) F = -600 N, F, = 800 N, and C = -24000 N – m %3D Note: Typically during our discussions, CW positive is used for 2D. However for the problem figure notice that CCW is considered positive. This is probably because the problem used right thumb rule on the z-axis.

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Situation 6. Determine the resultant force R and its line of action for the following force-couple systems: y (a) F = 300 N, F, = 0, and C = –150 N – m (b) Fx = 200 N, F = -200 N, and C = 800 N – m (c) F = -600 N, F, = -400N, and C = 0 (d) F = -600 N, F, = 800 N, and C =-24000 N – m Fy F, Note: Typically during our discussions, CW positive is used for 2D. However for the problem figure notice that CCW is considered positive. This is probably because the problem used right thumb rule on the z-axis. Situtation 7. Determine the resultant force of the stack of bricks on the beam and its location from the left side. 200N/m 3m 3 m Situation 8. The value of R, = EF, Ry = EF, and EMo for five force systems lying in the xy-plane are listed in the following table. Point O is the origin of the coordinate system, and positive moments are clockwise. Determine the resultant for each force system and show it on a sketch of the coordinate system. Part Ry 200 N 200 N 400 N - 600 N EMO 400 N-m - 400 N-m 600 N-m - 900 N-m - 200 N-m 1 300 N 400 N 4 5
Transcribed Image Text:Situation 6. Determine the resultant force R and its line of action for the following force-couple systems: y (a) F = 300 N, F, = 0, and C = –150 N – m (b) Fx = 200 N, F = -200 N, and C = 800 N – m (c) F = -600 N, F, = -400N, and C = 0 (d) F = -600 N, F, = 800 N, and C =-24000 N – m Fy F, Note: Typically during our discussions, CW positive is used for 2D. However for the problem figure notice that CCW is considered positive. This is probably because the problem used right thumb rule on the z-axis. Situtation 7. Determine the resultant force of the stack of bricks on the beam and its location from the left side. 200N/m 3m 3 m Situation 8. The value of R, = EF, Ry = EF, and EMo for five force systems lying in the xy-plane are listed in the following table. Point O is the origin of the coordinate system, and positive moments are clockwise. Determine the resultant for each force system and show it on a sketch of the coordinate system. Part Ry 200 N 200 N 400 N - 600 N EMO 400 N-m - 400 N-m 600 N-m - 900 N-m - 200 N-m 1 300 N 400 N 4 5
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