Q1. The forces F1 and F2 were: F1 = (6i + 5j + 4k)N and F2 = (-6i + 4j - 2k)N act through point A with...

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Q1. The forces F1 and F2 were:
F1 = (6i + 5j + 4k)N and
F2 = (-6i + 4j - 2k)N act through point A with position vector (3i + 2j + 3k)m. A third force F3 = (16i + 5j + 2k)N act through the point B with position vector (-6i + 4j + k)m.
(a) Show that the system of 3 forces, F1, F2, and F3 is equivalent to a single force F and find F (10mks)
(b) The work done by F in moving a particle from B to A (5mks)
(c) The magnitude of the moment of F about the origin (2mks)

SnapAndSolve AI · Accepted Answer

Show that the system of 3 forces, F1, F2, and F3 is equivalent to a single force F and find F:To find the equivalent single force F, sum the forces F1, F2, and F3: F = F1 + F2 + F3 = (6i + 5j + 4k) + (-6i + 4j - 2k) + (16i + 5j + 2k) F = (6 - 6 + 16)i + (5 + 4 + 5)j + (4 - 2 + 2)k F = 16i + 14j + 4k N The work done by F in moving a particle from B to A:Work done, W = F • d, where d is the displacement vector from B to A. d = A - B = (3i + 2j + 3k) - (-6i + 4j + k) = (3 + 6)i + (2 - 4)j + (3 - 1)k d = 9i - 2j + 2k W = (16i + 14j + 4k) • (9i - 2j + 2k) W = (16 * 9) + (14 * -2) + (4 * 2) W = 144 - 28 + 8 W = 124 J The magnitude of the moment of F about the origin:The moment of a force about a point is given by M = r × F, where r is the position vector of the point of application of the force. Using the position vector of point A (since F acts through A): r = (3i + 2j + 3k) M = r × F = (3i + 2j + 3k) × (16i + 14j + 4k) Using the cross product formula: M = [(2*4 - 3*14)i - (3*4 - 3*16)j + (3*14 - 2*16)k] M = (8 - 42)i - (12 - 48)j + (42 - 32)k M = -34i + 36j + 10k The magnitude of M is |M| = √((-34)^2 + 36^2 + 10^2) |M| = √(1156 + 1296 + 100) |M| = √2552 |M| ≈ 50.52 Nm

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Q1. The forces F1 and F2 were: F1 = (6i + 5j + 4k)N and F2 = (-6i + 4j - 2k)N act through point A with...

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