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- 1. 3.3 Coplanar SystemsA particle is subjected to coplanar forces inthe x-y planeResolve into i and j components forequilibrium ∑Fx = 0 ∑Fy = 0Scalar equations of equilibriumrequire that the algebraic sumof the x and y components toequal o zero
- 2. 3.3 Coplanar SystemsScalar Notation- Sense of direction = an algebraic signthat corresponds to the arrowheaddirection of the component along each axis- For unknown magnitude, assumearrowhead sense of the force- Since magnitude of the force is alwayspositive, if the scalar is negative, the forceis acting in the opposite direction
- 3. 3.3 Coplanar SystemsExampleConsider the free-body diagram of theparticle subjected to two forces Assume unknown force F acts to the right for equilibrium ∑Fx = 0 ; + F + 10N = 0 F = -10N Force F acts towards the left for equilibrium
- 4. 3.3 Coplanar Systems The chain exerts three forces on the ring at A. The ring will not move, or will move with constant velocity, provided the summation of the forces along the y axis is zero With any force known, the magnitude of other two forces are found by equations of equilibrium
- 5. 3.3 Coplanar Systems Procedure for Analysis1. Free-Body Diagram - Establish the x, y axes in any suitable orientation - Label all the unknown and known forces magnitudes and directions - Sense of the unknown force can be assumed
- 6. 3.3 Coplanar Systems Procedure for Analysis2) Equations of Equilibrium - Given two unknown with a spring, apply F = ks to find spring force using deformation of spring - If the solution yields a negative result, the sense of force is the reserve of that shown in the free-body diagram
- 7. 3.3 Coplanar Systems Procedure for Analysis2) Equations of Equilibrium - Apply the equations of equilibrium ∑Fx = 0 ∑Fy = 0 - Components are positive if they are directed along the positive negative axis and negative, if directed along the negative axis
- 8. 3.3 Coplanar SystemsExample 3.2Determine the tension incables AB and AD forequilibrium of the 250kgengine.
- 9. 3.3 Coplanar SystemsSolutionFBD at Point A- Initially, two forces acting, forces of cables AB and AD- Engine Weight = (250kg)(9.81m/s2) = 2.452kN supported by cable CA- Finally, three forces acting, forces TB and TD and engine weight on cable CA
- 10. 3.3 Coplanar SystemsSolution+→ ∑Fx = 0; TBcos30° - TD = 0+↑ ∑Fy = 0; TBsin30° - 2.452kN = 0Solving, TB = 4.90kN TD = 4.25kN*Note: Neglect the weights of the cables since they are small compared to the weight of the engine
- 11. 3.3 Coplanar SystemsExample 3.3If the sack at A has a weightof 20N (≈ 2kg), determinethe weight of the sack at Band the force in each cordneeded to hold the system inthe equilibrium positionshown.
- 12. 3.3 Coplanar SystemsSolutionFBD at Point E- Three forces acting, forces of cables EG and EC and the weight of the sack on cable EA
- 13. 3.3 Coplanar SystemsSolution+→ ∑Fx = 0; TEGsin30° - TECcos45° = 0+↑ ∑Fy = 0; TEGcos30° - TECsin45° - 20N = 0Solving, TEC = 38.6kN TEG = 54.6kN*Note: use equilibrium at the ring to determine tension in CD and weight of B with TEC known
- 14. 3.3 Coplanar SystemsSolutionFBD at Point C- Three forces acting, forces by cable CD and EC (known) and weight of sack B on cable CB
- 15. 3.3 Coplanar SystemsSolution+→ ∑Fx = 0; 38.6cos30° - (4/5)TCD = 0+↑ ∑Fy = 0; (3/5)TCD – 38.6sin45°N – WB = 0Solving, TCD = 34.1kN WB = 47.8kN*Note: components of TCD are proportional to the slope of the cord by the 3-4-5 triangle
- 16. 3.3 Coplanar SystemsExample 3.4Determine the required length of the cord ACso that the 8kg lamp is suspended. Theundeformed length of thespring AB is l’AB = 0.4m,and the spring has astiffness of kAB = 300N/m.
- 17. 3.3 Coplanar SystemsSolutionFBD at Point A- Three forces acting, force by cable AC, force in spring AB and weight of the lamp- If force on cable AB is known, stretch of the spring is found by F = ks
- 18. 3.3 Coplanar SystemsSolution+→ ∑Fx = 0; TAB – TACcos30° = 0+↑ ∑Fy = 0; TABsin30° – 78.5N = 0Solving, TAC = 157.0kN TAB = 136.0kN
- 19. 3.3 Coplanar SystemsSolutionTAB = kABsAB; 136.0N = 300N/m(sAB) sAB = 0.453NFor stretched length, lAB = l’AB+ sAB lAB = 0.4m + 0.453m = 0.853mFor horizontal distance BC, 2m = lACcos30° + 0.853m lAC = 1.32m

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