1. Replace the force F having a magnitude of F = 20 lb and
acting at point C by an equivalent force and couple moment
at point O. The force has coordinate direction angles of α =
60°, β = 120°, γ = 45°. Express the result as a Cartesian
vector.
Ans. F = 10i – 10i + 14.14k lb, M = 297.99i + 15.15j –
200k lb-in.
2. A force F = 80 N acts vertically downward on the z
bracket. Determine the moment of this force about the bolt
(z –axis), which is directed at 15o from the vertical.
Ans. M = - 23.18i – 23.18j – 6.21k N-m.
3. Replace the force F having a magnitude of F = 50
lb and acting at point A by an equivalent force and
couple moment at point C.
Ans. F = 100/7i + 150/7j – 300/7k lb, Ans. M = 1928.57i + 428.57j – 428.57k lb–ft.
C
4. Replace the loading system acting on the beam by an
equivalent resultant force and couple moment at point O.
Ans. F = 294.3 N @ 139,86o C.W. w.r.t x-axis. M = -39.57
N-m
5. Determine the centroid (x, y) of the shaded area.
Ans. X = 4.57 m, Y = 0.8 m.
6. Locate the centroid (x, y) of the shaded area.
Ans. X = 6 m, Y = 2.8 m.
7. Determine the centroid (x, y) of the shaded area.
Ans. X = 2 in, Y = 2.849 in.
8. Locate the centroid ( ,
) of the thin plate
Ans. X = -0.2624 m, Y = 0.2624 m.
9. Locate the centroid (x, y) for the strut’s cross-sectional
area.
Ans. X = 0 mm, Y = 53.414mm.
10. Determine the magnitude of the moment of the force Fc
about an axis formed by points D and A of the door.
MDA = 94.462 N-m
D