6161103 3.4 three dimensional force systems
- 1. 3.4 Three-Dimensional Force Systems For particle equilibrium ∑F = 0 Resolving into i, j, k components ∑Fxi + ∑Fyj + ∑Fzk = 0 Three scalar equations representing algebraic sums of the x, y, z forces ∑Fxi = 0 ∑Fyj = 0 ∑Fzk = 0
- 2. 3.4 Three-Dimensional Force Systems Make use of the three scalar equations to solve for unknowns such as angles or magnitudes of forces
- 3. 3.4 Three-Dimensional Force Systems Ring at A subjected to force from hook and forces from each of the three chains Hook force = weight of the electromagnet and the load, denoted as W Three scalars equations applied to FBD to determine FB, FC and FD
- 4. 3.4 Three-Dimensional Force Systems Procedure for Analysis Free-body Diagram - Establish the z, y, z axes in any suitable orientation - Label all known and unknown force magnitudes and directions - Sense of a force with unknown magnitude can be assumed
- 5. 3.4 Three-Dimensional Force Systems Procedure for Analysis Equations of Equilibrium - Apply ∑Fx = 0, ∑Fy = 0 and ∑Fz = 0 when forces can be easily resolved into x, y, z components - When geometry appears difficult, express each force as a Cartesian vector. Substitute vectors into ∑F = 0 and set i, j, k components = 0 - Negative results indicate that the sense of the force is opposite to that shown in the FBD.
- 6. 3.4 Three-Dimensional Force Systems Example 3.5 A 90N load is suspended from the hook. The load is supported by two cables and a spring having a stiffness k = 500N/m. Determine the force in the cables and the stretch of the spring for equilibrium. Cable AD lies in the x-y plane and cable AC lies in the x-z plane.
- 7. 3.4 Three-Dimensional Force Systems Solution FBD at Point A - Point A chosen as the forces are concurrent at this point
- 8. 3.4 Three-Dimensional Force Systems Solution Equations of Equilibrium, ∑Fx = 0; FDsin30° - (4/5)FC = 0 ∑Fy = 0; -FDcos30° + FB = 0 ∑Fz = 0; (3/5)FC – 90N = 0 Solving, FC = 150N FD = 240N FB = 208N
- 9. 3.4 Three-Dimensional Force Systems Solution For the stretch of the spring, FB = ksAB 208N = 500N/m(sAB) sAB = 0.416m
- 10. 3.4 Three-Dimensional Force Systems Example 3.6 Determine the magnitude and coordinate direction angles of force F that are required for equilibrium of the particle O.
- 11. 3.4 Three-Dimensional Force Systems Solution FBD at Point O - Four forces acting on particle O
- 12. 3.4 Three-Dimensional Force Systems Solution Equations of Equilibrium Expressing each forces in Cartesian vectors, F1 = {400j} N F2 = {-800k} N F3 = F3(rB / rB) = {-200i – 300j + 600k } N F = Fxi + Fyj + Fzk
- 13. 3.4 Three-Dimensional Force Systems Solution For equilibrium, ∑F = 0; F1 + F2 + F3 + F = 0 400j - 800k - 200i – 300j + 600k + Fxi + Fyj + Fzk = 0 ∑Fx = 0; - 200 + Fx = 0 Fx = 200N ∑Fy = 0; 400 – 300 + Fy = 0 Fy = -100N ∑Fz = 0; - 800 + 600 + Fz = 0 Fz = 200N
- 14. 3.4 Three-Dimensional Force Systems Solution r r r r F = {200i − 100 j − 200k }N r F = (200 )2 + (− 100 )2 + (200 )2 = 300 N r r F 200 r 100 r 200 r uF = r = i− j− k F 300 300 300 −1 200 α = cos = 48.2o 300 − 100 β = cos −1 = 109o 300 γ = cos −1 200 o = 48.2 300
- 15. 3.4 Three-Dimensional Force Systems Example 3.7 Determine the force developed in each cable used to support the 40kN (≈ 4 tonne) crate.
- 16. 3.4 Three-Dimensional Force Systems Solution FBD at Point A - To expose all three unknown forces in the cables
- 17. 3.4 Three-Dimensional Force Systems Solution Equations of Equilibrium Expressing each forces in Cartesian vectors, FB = FB(rB / rB) = -0.318FBi – 0.424FBj + 0.848FBk FC = FC (rC / rC) = -0.318FCi – 0.424FCj + 0.848FCk FD = FDi W = -40k
- 18. 3.4 Three-Dimensional Force Systems Solution For equilibrium, ∑F = 0; FB + FC + FD + W = 0 -0.318FBi – 0.424FBj + 0.848FBk - 0.318FCi – 0.424FCj + 0.848FCk + FDi - 40k =0 ∑Fx = 0; -0.318FB - 0.318FC + FD = 0 ∑Fy = 0; – 0.424FB – 0.424FC = 0 ∑Fz = 0; 0.848FB + 0.848FC - 40 = 0
- 19. 3.4 Three-Dimensional Force Systems Solution Solving, FB = FC = 23.6kN FD = 15.0kN
- 20. 3.4 Three-Dimensional Force Systems Example 3.6 The 100kg crate is supported by three cords, one of which is connected to a spring. Determine the tension in cords AC and AD and stretch of the spring.
- 21. 3.4 Three-Dimensional Force Systems Solution FBD at Point A - Weight of the crate = 100 (9.81) = 981 N - To expose all three unknown forces in the cables
- 22. 3.4 Three-Dimensional Force Systems Solution Equations of Equilibrium Expressing each forces in Cartesian vectors, FB = FBi FC = FCcos120°i + FCcos135°j – FCcos60°k FD = -0.333FDi + 0.667FDj + 0.667FDk W = -981k
- 23. 3.4 Three-Dimensional Force Systems Solution For equilibrium, ∑F = 0; FB + FC + FD + W = 0 FBi + FCcos120°i + FCcos135°j – FCcos60°k -0.333FDi + 0.667FDj + 0.667FDk - 981k =0 ∑Fx = 0; FB + FCcos120° - 0.333FD = 0 ∑Fy = 0; FCcos135° + 0.667FD = 0 ∑Fz = 0; FCcos60° + 0.667FD - 981 = 0
- 24. 3.4 Three-Dimensional Force Systems Solution Solving, FB = 693.7N FC = 813N FD = 693.7N For the stretch of the spring, FB = ks 693.7N = 1500s s = 0.462m