- 2073 Shrawan · 10 marks
Resultant force and moment of 3D system at O
Short answer
Write each force as a vector, sum forces, and add r × F plus the 500 kN·m couple. Resultant R = (−70.71i + 66.57j − 162.43k) kN, magnitude 189.2 kN. Moment about O is M_O = (120.3i − 307.2k) kN·m, magnitude 329.9 kN·m.
Determine the resultant force and moment of the following system about the point 'O' as shown in the figure below. [Figure: box with origin O(0,0,0), x, y, z axes; dimensions 4 m, 5 m, 3 m, 3 m, 2 m, 2.5 m as printed; forces: 10 kN at B along Y, 70 kN, 200 kN, 100 kN, 80 kN, 10 kN, and another 200 kN with a 2.5 M couple as marked.]
Readings from the figure (assumed where unclear): O is the origin, x = 5 m, y = 4 m, z = 3 m for the box. The forces are:
- 10 kN along +y at B (0, 4, 0).
- 70 kN along −y at (2, 4, 0) (2 m from B along the top edge).
- 200 kN acting at (0, 0, 3) along the line towards B.
- 100 kN acting at (5, 0, 3) along the line towards B.
- 80 kN along −y at (5, 0, 3).
- 10 kN along −y at (5, 0, 3) (the second 10 kN, read as vertical downward at the same corner).
- A couple formed by two 200 kN forces 2.5 m apart on the right face (parallel to z), taken as a couple of 500 kN·m about +x.
Force components (kN) and moments about O (r × F)
| Force | Point | Components | Moment about O (kN·m) |
|---|---|---|---|
| 10 | (0,4,0) | (0, 10, 0) | (0, 0, 0) |
| 70 | (2,4,0) | (0, −70, 0) | (0, 0, −140) |
| 200 | (0,0,3) | 200(0,4,−3)/5 = (0, 160, −120) | (−480, 0, 0) |
| 100 | (5,0,3) | 100(−5,4,−3)/√50 = (−70.71, 56.57, −42.43) | (−169.71, 0, 282.84) |
| 80 | (5,0,3) | (0, −80, 0) | (240, 0, −400) |
| 10 | (5,0,3) | (0, −10, 0) | (30, 0, −50) |
| Couple | (500, 0, 0) |
Resultant force
ΣFx = −70.71, ΣFy = 10 − 70 + 160 + 56.57 − 80 − 10 = 66.57, ΣFz = −120 − 42.43 = −162.43 R = (−70.71 i + 66.57 j − 162.43 k) kN, |R| = √(70.71² + 66.57² + 162.43²) = 189.2 kN
Resultant moment about O
Mx = −480 − 169.71 + 240 + 30 + 500 = 120.29 My = 0 Mz = −140 + 282.84 − 400 − 50 = −307.16 M_O = (120.3 i − 307.2 k) kN·m, |M_O| = 329.9 kN·m
The method is the same if a reading of the figure differs: write each force as a vector, find r × F, and add.
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